MAS111: Mathematics Core II
Solutions to Problems booklet – Chapters 1–4
Chapter 1
1. We use the equation
y − y1
x − x1
=
,
y2 − y1
x2 − x1
with (x1 , y1 ) = (−3, 1) and (x2 , y2 ) = (1, 2), and get
x − (−3)
y−1
=
,
2−1
1 − (−3)
which rearranges to 4(y − 1) = x + 3 or x − 4y + 7 = 0.
We can now see whether (9, 4) lies on the line, by checking that 9 − 4.4 + 7 = 0,
which is true, so the point lies on the line.
2. The line 7x−y = 6 has gradient 7. We need the line through (−2, 3) with gradient 7.
We use the form y = mx + c with m = 7, so that the line is of the form y = 7x + c,
and (−2, 3) lies on it, so c = 17. So the line is y = 7x + 17.
The perpendicular line has gradient −1/7, and (0, 0) lies on it, so the line is y = − 71 x.
3. We need to solve the simultaneous equations
y = 5x − 7,
y = −3x + 1,
but this is easy, as we simply need 5x−7 = −3x+1, which gives x = 1. Substituting
back, we get y = 5.1 − 7 = −2, so the point is (1, −2).
2
y
1
y = 5x − 7
x
−1
0
1
2
−1
y = −3x + 1
−2
4. Notice that A and B lie on the circle, and the line joining them passes through the
b
b
, and the gradient of BP is a−1
. Their
centre (0, 0). The gradient of AP is a+1
product is
b
b2
b
.
= 2
,
a+1 a−1
a −1
but as (a, b) lies on the circle, we have a2 + b2 = 1, i.e., b2 = 1 − a2 . So the product
of gradients is, in fact, −1, and so the two lines are perpendicular.
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5. if (x, y) is equidistant, then we need
p
p
(x + 1)2 + (y − 2)2 = (x − 2)2 + (y − 3)2 .
Squaring and expanding, this gives
x2 + 2x + 1 + y 2 − 4x + 4 = x2 − 4x + 4 + y 2 − 6y + 9,
and cancelling and rearranging gives 6x − 10y = 8, or 3x − 5y − 4 = 0.
Alternatively, the slope of the line segment joining P and Q is −5/3 and the midpoint
is ( 12 , − 12 ), and geometrically it should be clear that we need the perpendicular
bisector of this segment. The slope of the bisector should be 3/5, and it must pass
through the midpoint, so the equation should be
1
1
3
y+ =
x−
,
2
5
2
and this rearranges to the given equation.
6. We argue as in Q5. We need
p
p
2 (x + 1)2 + (y − 2)2 = 3 (x − 3)2 + (y − 1)2 .
Squaring and expanding gives
4(x2 + 2x + 1 + y 2 − 4y + 4) = 9(x2 − 6x + 9 − y 2 − 2y + 1),
or
5x2 − 62x + 5y 2 − 2y + 70 = 0.
(This is the equation of a circle. The locus of points P satisfying a|P A| = b|P B| is
always a circle, unless a = b, when we get a line.)
7. In the usual way,
d2 = (496t + 22 − 500t − 10)2 + (−310t − 23 + 312t + 12)2
= (12 − 4t)2 + (2t − 11)2
= 20t2 − 140t + 265
= 20(t − 27 )2 + 20,
which has a minimum at t =
√
d = 20.
7
2
(i.e., at t = 3.5 hours), when d2 = 20, so that
8. (a) Since the line has equation which can be rearranged to y = − ab x − cb , the line
has gradient −a/b.
(b) So the gradient of the perpendicular line is b/a. Thus the perpendicular line is
y = ab x + k for some k. But P lies on the line, so
b
y0 = x0 + k,
a
i.e., k = y0 − bx0 /a. Then the perpendicular line through P has equation
y = ab x + y0 − ab x0 ; rearranging, this is bx − bx0 + ay0 − ay = 0.
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(c) We can intersect the original line and this perpendicular to find the coordinates
of Q: we need
ax + by + c = 0,
bx − bx0 + ay0 − ay = 0.
Multiplying the first equation by a and the second by b, and adding, we get
a2 x + aby + ac + b2 x − b2 x0 + aby0 − aby = 0,
or
(a2 + b2 )x = b2 x0 − aby0 − ac,
so that the x-coordinate of Q is given by
x=
b2 x0 − aby0 − ac
.
a2 + b2
Substituting back into (say) the original equation of the line, we get
a
b2 x0 − aby0 − ac
+ by + c = 0,
a2 + b2
and rearranging multiplying up, we get
a(b2 x0 − aby0 − ac) + (a2 + b2 )by + (a2 + b2 )c = 0;
note that the terms a2 c cancel, and that the remaining terms are all divisible
by b. Dividing by b (we really ought to treat the case b = 0 separately!), we
get:
abx0 − a2 y0 + (a2 + b2 )y + bc = 0,
and so the y-coordinate of Q is
y=
a2 y0 − abx0 − bc
.
a2 + b2
(Note, in passing, the similarity between the expressions for the x- and ycoordinate – I hope this suggests that an alternative way to get the solution for
y having got the solution for x is just to argue that the problem is symmetrical
when exchanging x and y, and a and b. This sort of symmetrical argument can
save a lot of time sometimes!) So
2
b x0 − aby0 − ac a2 y0 − abx0 − bc
.
,
Q=
a2 + b2
a2 + b2
(d) Now we can work out the distance between Q and
2
(a + b2 )x0 (a2 + b2 )y0
P = (x0 , y0) =
,
a2 + b2
a2 + b2
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in the usual way as
s
2 2
2
(a + b2 )y0 − (a2 y0 − abx0 − bc)
(a2 + b2 )x0 − (b2 x0 − aby0 − ac)
|P Q| =
+
a2 + b2
a2 + b2
s
2 2
2
b y0 + abx0 + bc
a2 x0 + aby0 + ac
=
+
a2 + b2
a2 + b2
s
2 2
a(ax0 + by0 + c)
b(ax0 + by0 + c)
=
+
a2 + b2
a2 + b2
s
2
(ax0 + by0 + c)
(a2 + b2 )
=
a2 + b2
|ax0 + by0 + c|
√
=
,
a2 + b2
as required.
9. Firstly,
|AB| =
p
(x2 − x1 )2 + (y2 − y1 )2 .
The line joining A and B has equation
y−y1
y2 −y1
=
x−x1
,
x2 −x1
which rearranges to
(y2 − y1 )x − (x2 − x1 )y − x1 (y2 − y1 ) + y1 (x2 − x1 ) = 0,
which is in the form we can apply the distance of a point from a line formula. So
the distance of C from AB is therefore
|(y2 − y1 )x3 − (x2 − x1 )y3 − x1 (y2 − y1 ) + y1 (x2 − x1 )|
p
(x2 − x1 )2 + (y2 − y1 )2
|y2x3 − y1 x3 − x2 y3 + x1 y3 − x1 y2 + x1 y1 + y1 x2 − y1 x1 |
p
=
(x2 − x1 )2 + (y2 − y1 )2
|y2x3 − y1 x3 − x2 y3 + x1 y3 − x1 y2 + y1 x2 |
p
=
.
(x2 − x1 )2 + (y2 − y1 )2
d=
Using the usual formula of
1
2
× base × height, we get that the area is
1
1
|AB|d = |(x2 y3 − x3 y2 ) − (x1 y3 − x3 y1 ) + (x1 y2 − x2 y1 )|
2
2
as required.
(We remark that this can be simply expressed as adeterminant;
if you know about
1 x1 y1 these already, you should also verify that this is 21 1 x2 y2 . But we will see this
1 x3 y3 in Chapter 5.)
10. (a) We have
r=
p
ρ2 + z 2 ,
φ = tan−1 (z/ρ),
with θ being the same for both.
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(b) Now z = r cos φ (as in the notes), and ρ = r sin φ, and again θ is the same for
both.
11. The natural way to specify the solution is as the set of points
   
  

2
1
0
2+s
3 + 1 s + 1 = 3 + s + t .
5
0
1
5+t
So we have x = 2+s, and so s = x−2; similarly, z = 5+t, so t = z −5. Substituting
into the middle coordinate, we get
y = 3 + s + t = 3 + (x − 2) + (z − 5),
so the equation for this plane is x − y + z = 4.
12. The normal to the plane ax + by + cz + d = 0 is in the direction (a, b, c), so the
parametric equation of the normal through the given point consists of all points
(x0 + ta, y0 + tb, z0 + tc).
This lies on the plane when a(x0 + ta) + b(y0 + tb) + c(z0 + tc) + d = 0, i.e., when
ax0 + by0 + cz0 + d = −t(a2 + b2 + c2 ), which gives the desired value of t.
So the line segment from (x0 , y0 , z0 ) to the point
√ of intersection is the vector t(a, b, c),
with the given value of t. Its length is |t| a2 + b2 + c2 , and this means that the
desired distance is
√
|ax0 + by0 + cz0 + d|
√
|t| a2 + b2 + c2 =
.
a2 + b2 + c2
Note the similarity with Q8.
13. The normal is simply given by the line (1, 1, 1) + (1, 1, 1)t, so all points of the form
(t + 1, t + 1, t + 1). This is also described by the equation x = y = z.
The previous question tells us that the distance is given by
√
3
|1 × 1 + 1 × 1 + 1 × 1|
√
= √ = 3.
12 + 12 + 12
3
There are several ways to see that (0, 0, 0) is the nearest point in the plane to
(1, 1, 1). Since (0, 0, 0) is obviously on the plane, and the distance between (0, 0, 0)
and (1, 1, 1) is the same as the distance from (1, 1, 1) to the plane, then (0, 0, 0) must
be the closest point. Alternatively, you can find the point of intersection between
the normal line and the plane, for example.
Chapter 2
1. It’s usually worth thinking of good ways to find solutions, rather than blindly eliminating variables mindlessly. Here, we know we’d like to end up with x =?, and we
can get this by taking the second equation from the first, to get x = −1.
We also know we’d like to get y =?, and we see that if we take the second from the
third, we get 2y = 1, so that y = 1/2. Now some back-substitution gives z = 5/2.
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We can deduce that the three vectors are linearly independent.
Historical Note.
first course I ever marked the exams for in Sheffield had exactly
 The
2 1 1
the matrix 1 1 1. Every single student mindlessly eliminated variables, taking
1 3 1
about twice as long as was necessary, and, in several cases, making errors that might
not have crept in with a more reasoned approach.
2. (a) We assume that there is some linear relationship
 
 
 
 
3
2
0
0
−2 x + 1 y + −7 z = 0 ,
1
3
−7
0
and see whether there is any solution other than x = y = z = 0. For this, we
put the system into augmented matrix form:


3 2 0 0
 −2 1 −7 0  ,
1 3 −7 0
and reduce it. To get a 1 in the top left (convenient for pivoting), we can either
add the second row to the first, or swap the first and third rows. Let’s do the
first, to get


1 3 −7 0
 −2 1 −7 0  ,
1 3 −7 0
but now we see that the first and
the top left entry gives

1
 0
0
third equations are identical. So pivoting on

3 −7 0
7 −21 0  ,
0 0 0
and there are nontrivial solutions; if z = t is a parameter, the second row gives
7y − 21z = 0, so that y = 3t. The top row gives x + 3y − 7z = 0, so that
x = −2t. This not only shows that there is a nontrivial relationship, but also
gives what it is:
 
     
3
2
0
0







−2 −2 + 3 1 + −7 = 0 .
1
3
−7
0
So the vectors are linearly dependent.
Notice that although we put the final column of zeros in for illustration, none
of the row operations were ever going to change it. We might as well have left
it out, and we’ll omit it in the next part.
(b) Again, we assume that there is some linear relationship
 
 
 
 
3
2
1
0
−2 x + 1 y +  4  z = 0 ,
1
3
−3
0
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and see whether there is any solution other than x = y = z = 0. For this, we
put the system into augmented matrix form, but since the constants are all 0,
we
 to include them (see the remark at the end of the last part):
 won’t bother
3 2 1
−2 1 4 . Now we’ll add the second row to the first again to get a 1 in
1 3 −3




1 3 5
1 3 5
the top left: −2 1 4  and then pivot on it: 0 7 14 . This is in
1 3 −3
0 0 −8
row echelon form, and we can read off from the bottom row that −8z = 0, so
z = 0; the middle row give 7y + 14z = 0, so y = 0, and then the top row gives
x = 0. These vectors are therefore linearly independent.
3. There are lots of ways to do this, and the answer is not unique! It avoids fractions if
we try to get a 1 in the pivot position – we could halve the first row (equation) but
this introduces halves; better is to exchange the first and second rows (equations).
Given that the aim of reducing the matrix is often to solve equations, even better
would be to get a top row of 1 0 0, and we can achieve this by first taking row 2
from row 1 to get


1 0 0
 1 1 1 .
1 3 1
Then we can pivot on the top left element, subtracting multiples of the first row
from the others to get


1 0 0
 0 1 1 .
0 3 1
Then we notice that the first nonzero entry in the second row is already a 1, so we
pivot on this, subtracting three times the middle row from the bottom one to get


1 0 0
0 1 1  ,
0 0 −2
and this is in row echelon form.
As remarked already, there are lots of ways to do this, and you will get different
answers depending on how you do this. All answers will have three nonzero rows at
the end of the process.
4. For the first set of equations, the augmented

1 2 1
 2 3 1
1 1 −1
matrix is

2
4 .
3
We can reduce this to row echelon form with row operations. First eliminate x, i.e.,
subtract multiples of the first equation (row) from the others to get:


1 2
1 2
 0 −1 −1 0  .
0 −1 −2 1
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Then we can pivot on the first nonzero entry
row from the third, to get

1 2
1
 0 −1 −1
0 0 −1
in the second row, taking the second

2
0 .
1
This is now in row echelon form. Converting back into equations, the third row is
−z = 1, so that z = −1. Then the second row is −y − z = 0, so y = 1. Finally, the
first equation gives x + 2y + z = 2, and this gives x = 1.
The second set of equations is similar. The augmented matrix is


1 1
1 2
 1 −1 2 4  .
0 2 −1 2
We pivot on the first entry of the top row (i.e., we subtract multiples of the top row
to make the other entries in that column become 0):


1 1
1 2
 0 −2 1 2  .
0 2 −1 2
Now pivot on the first nonzero entry

1
 0
0
of the middle row, to get:

1 1 2
−2 1 2  .
0 0 4
This is in row echelon form. Now the final row says 0 = 4, which is obviously
impossible, and so the system has no solutions.
1 1 2
, which is already in echelon
5. (a) The first set corresponds to the matrix
0 1 3
form. We get y = 3 from the bottom line, and x + y = 2 from the top;
substituting back in, we see x = −1.


1 1 1 1
(b) The augmented matrix here is  0 1 1 1 . Pivoting on the first nonzero
1 0 0 1

1 1
1 1
1 1 . Adding the middle row to the
entry of the top row gives  0 1
0 −1 −1 0


1 1 1 1
last one gives  0 1 1 1 , and the bottom equation is 0 = 1, which is
0 0 0 1
impossible. So there are no solutions.
(c) 
This
1
 0
1
is the same
 as the last part, except for the constants. The matrix is
1 1 3
1 1 2 , and the same method as the previous part results in the
0 0 1

1 1 1 3
matrix  0 1 1 2 . So there are just two equations, x + y + z = 3 and
0 0 0 0
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y + z = 2, so we can read off that x = 1. The only other equation is y + z = 2;
this represents a line, and we can parametrise the line by setting y = t, and
then z = 2 − t. So the general solution is (x, y, z) = (1, t, 2 − t).
(d) Now we have three equations in four

1 1
 1 0
2 1
variables. The matrix is

0 0 2
1 1 1 .
1 1 2
Pivoting on the first entry in row 1 gives


1 1 0 0 2
 0 −1 1 1 −1  ,
0 −1 1 1 −2
and taking the middle row from

1
 0
0
which has no solution.
the bottom gives

1 0 0 2
−1 1 1 −1  ,
0 0 0 −1
(e) This is similar to the previous part; the

1 1 0
 1 0 1
2 1 1
matrix is

0 1
1 1 ,
1 2
and the same reduction steps result in


1 1 0 0 1
 0 −1 1 1 0  .
0 0 0 0 0
This time, we have two equations, x + y = 1 and −y + z + w = 0. With
four variables and two equations, we expect that these will form a plane in
4-dimensional space, parametrised by two parameters. We could put y = λ,
z = µ, and read off that x = 1 − λ and w = λ − µ. So the general solution is
 
 
  
  
0
−1
1
1−λ
x
0
1
 y   λ  0
 
 
  
 =
 z   µ  = 0 + λ  0  + µ  1  .
−1
1
0
λ−µ
w
6. We solve the system of equations using row reduction. The augmented matrix is


1 1 1 1
 1 −1 2 0  .
2 4 1 4
Pivoting on the top left entry gives


1
1 1
1
 0 −2 1 −1  .
0 2 −1 2
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Adding the middle row to the bottom gives


1 1 1 1
 0 −2 1 −1  .
0 0 0 1
Now the last equation corresponds to 0 = 1, which is clearly false. So there is no
point of intersection, and no common solution. However, each pair of equations
intersect in a line, since they are not parallel (no left-hand side is a multiple of any
other). Since there is no common solution, the three lines must all be parallel.
The normal vectors to the planes are
 
 
1
1
1 , −1 ,
1
2
 
2
4 .
1
We want to see that these three vectors are linearly dependent, so we want to find
some nontrivial solution to
 
 
 
 
1
1
2
0
1 α + −1 β + 4 γ = 0 .
1
2
1
0
This corresponds to the augmented matrix


1 1 2 0
 1 −1 4 0  .
1 2 1 0
(Notice that the final column of zeros isn’t really necessary – all the row operations
will leave it unchanged.) Pivoting on the top left element gives:


1 1
2 0
 0 −2 2 0  .
0 1 −1 0
Then we can scale the middle row, and add a multiple to the bottom row to get


1 1 2 0
 0 1 −1 0  .
0 0 0 0
This corresponds to the equations
α + β + 2γ = 0
β−γ =0
and this has solution
 
 
α
−3
β  = c  1  ,
γ
1
and so we get a nontrivial relationship
       
1
1
2
0







−3 1 + −1 + 4 = 0 .
1
2
1
0
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7. (a) We need to find values of x, y and z, such that xa1 + ya2 + za3 = b. So we
need to solve
2x + 3y + z = 7
3x + 7y − 6z = −2
5x + 8y + z = λ.
Put this into augmented matrix form:


2 3 1
7
 3 7 −6 −2  .
5 8 1
λ
We can row-reduce this in various ways. I like to avoid fractions, so I’ll start
by taking the first from the second, and swapping the top two rows, to get


1 4 −7 −9
 2 3 1
7 .
5 8 1
λ
Now I have a 1 in the top left,
two rows/equations:

1
 0
0
I can easily take multiples of it from the other

4 −7
−9
−5 15
25  .
−12 36 λ + 45
Now divide the middle row/equation by −5:


−9
1 4 −7
 0 1 −3
−5  .
0 −12 36 λ + 45
Add 12 times the middle row to the bottom, to get


1 4 −7
−9
 0 1 −3
−5  .
0 0 0 λ − 15
The bottom equation needs λ = 15 in order to work.
(b) In a similar way, we should find that any value of λ works (the three vectors
are linearly independent).
(c) No value of λ works.
8. Pivoting on the top left entry gives

1 2
1
3 0
 0 −9 −9 −18 9

 0 −5 −5 −10 5
0 −3 −3 −6 3
Next, we scale the second row, to get

1 2
 0 1

 0 −5
0 −3
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

.


1
3
0
1
2 −1 
,
−5 −10 5 
−3 −6 3
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and pivot on the first nonzero entry of the second row, to get


1 0 −1 −1 2
 0 1 1
2 −1 

,
 0 0 0
0
0 
0 0 0
0
0
which is in reduced row echelon form. Hence we just have two equations,
x−z−u=2
y + z + 2u = −1.
The variables which do not correspond to pivot elements are z and u, so we make
these into parameters, by putting z = λ, u = µ, and then we can read off
x=2+λ+µ
y = −1 − λ − 2µ,
so that the general solution is
 
 
  
  
1
1
2
2+λ+µ
x
−2
−1
y  −1 − λ − 2µ −1
 =   + λ  + µ .
 =
0
1
 0
z  
λ
1
0
0
µ
u
So the 
set ofsolutions is actually a plane, passing through the point with position
2
−1

vector 
 0 .
0
9. We have
Now

  
p − 2q + 2r
−p
P  q  =  2p − q + 2r  .
2p − 2q + 3r
r
(p − 2q + 2r)2 + (2p − q + 2r)2 = (p2 − 4pq + 4pr + 4q 2 − 8qr + 4r 2)
+ (4p2 − 4pq + 8pr + q 2 − 4qr + 4r 2 )
= 5p2 − 8pq + 12pr + 5q 2 − 12qr + 8r 2 .
Also,
(2p − 2q + 3r)2 = 4p2 − 8pq + 12pr + 4q 2 − 12qr + 9r 2
= (5p2 − 8pq + 12pr + 5q 2 − 12qr + 8r 2 ) + (−p2 − q 2 + r 2 ).
Since p2 + q 2 = r 2 , we see that the last bracket vanishes, and the two sides are the
same.
The same method works for the other two examples.
 But we can also see this
p
by exploiting the symmetries here. The second, P −q  is the same as the first,
r
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except that p and q are each multiplied by −1 – this changes only the terms in
pr and qr above,
and these were the same anyway on both sides of the equation.
 
−p
Similarly, P −q  is the same as the first, except that q is multiplied by −1; this
r
changes the signs of the terms in pq and qr, but the result will be unchanged.
Chapter 3
1. (a) The image is stretched in the x-direction by a factor of 2, and halved in the
y-direction.
(b) The image is rotated around the origin, and also stretched by a factor of 2 in
the y-direction.
(c) This is a shear in the y-direction, so y-axis is fixed, while the x-axis is rotated
to be at an angle of π/4.
√
√
(d) Recalling that cos π6 = 3/2, sin π6 = 1/2, so that cos(− π6 ) = 3/2 and
sin(− π6 ) = 1/2, we note that this matrix is an anticlockwise rotation by − π6 ,
i.e., a clockwise rotation by π6 .
2. These are all easy by hand, of course. These all give examples to show that matrix
multiplication is not like multiplication of real numbers! In R, there is no square
root of −1; there is no nonzero number whose square is 0; multiplication is always
commutative (i.e., cd = dc); and the product of two nonzero numbers is always
nonzero.
3. (a) We can compute 3A, which will also be a 3 × 2 matrix, but then C and 3A are
of different sizes, so we can’t write down C − 3A.
−4 −10
9
6
.
, and C − 3B =
(b) 3B =
−17 10
18 −12


−9 10
(c) AB =  3 −6.
12 −8
(d) BA cannot be computed as the number of columns of B differs from the number
of rowsin A.
17 −16
.
(e) BC =
26 −16
−9 26
(f) CB =
.
−9 10
4. We have
cos α cos β − sin α sin β
cos α sin β + sin α cos β
cos β sin β
cos α sin α
=
− sin α cos β − cos α sin β − sin α sin β + cos α cos β
− sin β cos β
− sin α cos α
cos(α + β) sin(α + β)
.
=
− sin(α + β) cos(α + β)
This means that rotating by β and then by α is the same as rotating by α + β.
n cos α sin α
cos nα sin nα
Finally, this implies that
=
.
− sin α cos α
− sin nα cos nα
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ab 0
1 a+b
. which is quite interesting – it means that 2 × 2
and
5. We get
0 1
0
1
matrix multiplication “contains” normal addition and multiplication.


a b c
6. (a) 0 0 0;
0 0 0


d e f
(b) 0 0 0 ;
0 0 0


0 0 0
(c) a b c ;
0 0 0


a 0 0
(d) d 0 0;
g 0 0


0 a 0
(e) 0 d 0;
0 g 0


b 0 0
(f)  e 0 0.
h 0 0
7. Firstly,
a11 a12
b11 b12
a11 b11 + a12 b21 a11 b12 + a12 b22
AB =
=
,
a21 a22
b21 b22
a21 b11 + a22 b21 a21 b12 + a22 b22
and
b11 b12
a11 a12
b11 a11 + b12 a21 b11 a12 + b12 a22
BA =
=
,
b21 b22
a21 a22
b21 a11 + b22 a21 b21 a12 + b22 a22
so
trace(AB) = a11 b11 + a12 b21 + a21 b12 + a22 b22
trace(BA) = b11 a11 + b12 a21 + b21 a12 + b22 a22 ,
and these are clearly the same.
More generally, the formula for matrix multiplication and the trace shows that
!
m
m
n
X
X
X
trace(AB) =
(AB)ii =
aij bji
trace(BA) =
i=1
i=1
j=1
n
X
n
X
m
X
(BA)jj =
j=1
j=1
i=1
bji aij
!
,
and these are clearly the same.
1 2
3 5
13 19
8. AB =
=
, which is clearly not symmetric.
2 3
5 7
21 31
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

3
0
9. AC = −4 2 . And
2 −4
3 −4 2
1 −1 0
5
1
T T
,
=
C A =
0 2 −4
−2 1 2
−4 −2
so C T AT = (AC)T as claimed.
10. We have
trace(A−1 BA) = trace(A−1 (BA)) = trace((BA)A−1 ) = trace(B).
1 1
1 3
, then we need to solve AX = B. We can
and B =
11. If we write A =
1 1
1 2
multiply both sides by A−1 at the front (remember that matrix multiplication is not
commutative, so we have to be careful to do exactly the same thing
to bothsides):
−2 3
A−1 (AX) = A−1 B, so X = A−1 B. So we simply work out A−1 =
, so
1 −1
1 1
1 1
−2 3
−1
.
=
X=A B=
0 0
1 1
1 −1
−2 −1
−1 1
. Then we
and D =
For the second equation, we let C =
3
4
3 −4
−1
need to solve XC = D, so that (XC)C −1 = DC
at the back this
, multiplying
−4
−1
, so
time, and get X = DC −1 . We work out C −1 =
−3 −1
11
3
−4 −1
−2 −1
−1
.
=
X = DC =
−24 −7
−3 −1
3
4
12. (a) We rewrite the equations as
5 −4
x
2
=
,
1 −2
y
4
so
−1 2
5 −4
x
.
=
4
1 −2
y
Working out ad − bc for the matrix gives −6; then the inverse is
So
− 16
−2 4
.
−1 5
−1 1 −2 4
1 12
x
2
−2
=−
=−
.
=
y
4
−3
6 −1 5
6 18
(b) Solving in a similar way to the first part would require us to invert the matrix
2 −4
, but this is not invertible. Geometrically, the two equations give
1 −2
parallel lines.
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13. There are several methods to do this. We want to get a 1 in the top left, so we
could halve the top row, or exchange the rows,
or take the second from the first,
1 2
. Now take 2 times the top row from
etc. Let’s swap the rows, to map A to
2
3
1 2
. Next, add 2 times the bottom row to the top row to
the second to get
0 −1
1 0
, and then multiply the bottom row by −1 to get the identity matrix.
get
0 −1
Equivalently,
we start with A, and premultiply by the elementary matrix E1 =
0 1
to exchange the two rows. Next, premultiply the result E1 A by E2 =
1 0 1 0
to take 2 times the top row from the second, and then premultiply the
−2 1
1 2
to add 2 times the bottom row to the top row.
result E2 E1 A by E3 =
0 1
1 0
to multiply the bottom
Finally, we multiply the result E3 E2 E1 A by E4 =
0 −1
row by −1.
So E4 E3 E2 E1 A = I. Then
A = E1−1 E2−1 E3−1 E4−1
0 1
1 0
1 −2
1 0
=
.
1 0
2 1
0 1
0 −1
We can check this:
2 1
1 0
0 1
=
1 0
2 1
1 0
1 2
1 0
1 −2
,
=
0 −1
0 −1
0 1
and
as expected.
2 1
1 0
1 2
0 −1
2 3
,
=
1 2
14. We apply row reduction methods to the augmented matrix (A|I) and reduce until
we get the augmented matrix (I|A−1 ). From (A|I), we will pivot on the top left
entry:




1
3 −1 1 0 0
1 3 −1 1 0 0
 −2 −5 1 0 1 0  −→  0 1 −1 2 1 0  .
4 11 −2 0 0 1
0 −1 2 −4 0 1
Now we pivot on the first nonzero entry in the middle row, subtracting some multiple
of this from both the top and bottom rows:




1 0 0 −1 −5 −2
1 0 2 −5 −3 0
1 0  −→  0 1 0 0
2
1 
−→  0 1 −1 2
1
0 0 1 −2 1 1
0 0 1 −2 1
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where the last matrix comes
 from pivoting
 on the third entry in bottom row. We
−1 −5 −2
2
1 .
read off that the inverse is  0
−2 1
1
15. We reduce A by taking 2 times the top row from the second, to get
1 −1 2 5
,
0 0 1 2
and then take 2 times the bottom row from the top, to get
1 −1 0 1
,
0 0 1 2
which is in reducedrow echelon form.
this, we can
To do
1 −2
1 0
. So
, and then by
tiply A by
0 1
−2 1
1 −2
1 0
1 −1 0
A=
0 1
−2 1
0 0 1
equivalently first premul-
1
,
2
and so we can take
1 −2
1 0
5 −2
P =
=
.
0 1
−2 1
−2 1
We have
1 −1 0 1
;
PA =
0 0 1 2
we can add column 1 to column 2, to get
1 0 0 1
PA =
,
0 0 1 2
and take column 1 from column 4, to get
1 0 0 0
,
0 0 1 2
and then take 2 times column 3 from column 4 to get
1 0 0 0
.
0 0 1 0
Finally, we swap columns 2 and 3 to get
1 0 0 0
.
0 1 0 0
In terms of elementary matrices, we start

1

0
E1′ = 
0
0
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17
with P A, and then postmultiply by

1 0 0
1 0 0

0 1 0
0 0 1
Spring 2016-17
to add column 1 to column 2; then we postmultiply by


1 0 0 −1
0 1 0 0 

E2′ = 
0 0 1 0 
0 0 0 1
to take column 1 from column 4. Next,

1

0
E3′ = 
0
0
we postmultiply by

0 0 0
1 0 0

0 1 −2
0 0 1
to take 2 times column 3 from column 4m and

1 0 0

0 0 1
E4′ = 
0 1 0
0 0 0
to swap columns 2 and 3. So
(P A)E1′ E2′ E3′ E4′
by

0
0

0
1
1 0 0 0
;
=
0 1 0 0
we take
Q = E1′ E2′ E3′ E4′

1 1 0
0 1 0
=
0 0 1
0 0 0

1 0 1
0 0 1
=
0 1 0
0 0 0

1
0
0
0

0 0
0
1

−1
0
.
−2
1
0
1
0
0

1
0 −1
0
0 0

1 0  0
0
0 1
0
1
0
0

1
0 0
0
0 0

1 −2 0
0
0 1
0
0
1
0
0
1
0
0

0
0

0
1
As an alternative, one can do the same that you do for the Gauss-Jordan procedure
to invert a matrix. We reduce (A|I2 ) using row operations, until we get to(P A|P
),
PA


where P A is in row echelon form. Then we can do column operations on
I4


P AQ
, and so you can read off Q. Try to think about why this

until you get
Q
works!
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Chapter 4
1. We have
a2 + bc ab + bd
A =
ac + cd bc + d2
2
a + ad ab + bd
a b
=
trace(A).A = (a + d)
ac + cd ad + d2
c d
ad − bc
0
1 0
.
=
det(A).I = (ad − bc)
0
ad − bc
0 1
2
It should be easy to see that
2
2
a + ad ab + bd
a + bc + (ad − bc)
ab + bd
2
=
A + det(A).I =
ac + cd d2 + ad
ac + cd
bc + d2 + (ad − bc)
= trace(A).A,
and the result follows.
This is a special case of the Cayley-Hamilton Theorem, which we’ll see later.
2. We get 16, 0, 1 and a2 + b2 + c2 + d2 respectively.
3. (i) Expanding along the top row:
4 1 0 0 2 −1 = 4 2 −1 − 1 0 −1 + 0 0 2 = 4 × 5 − 1 × 2 + 0 × (−4) = 18.
2 3
2 1 3 1 2 3 1 (ii) Expanding down the first column:
4 1 0 0 2 −1 = 4 2 −1 − 0 1 0 + 2 1 0 = 4 × 5 − 0 × 1 + 2 × (−1) = 18.
2 −1
3 1
3 1 2 3 1 4. (a) We can expand along the middle row, say:
1 2 3
4 5 6 = −4 2 3+5 1 3−6 1 2 = −4×(−6)+5×(−12)−6×(−6) = 0.
8 9 7 9 7 8
7 8 9
Here is an alternative method, which means that you don’t need to do any
evaluations. We will see also that the determinant is unchanged when you add
or subtract a multiple of one row to another. This means that we can subtract
the second row from the third and see that
1 2 3 1 2 3
4 5 6 = 4 5 6 .
7 8 9 3 3 3
MAS111
Now we can take the first row
1
4
3
from the second:
2 3 1 2 3
5 6 = 3 3 3 ,
3 3 3 3 3
19
Spring 2016-17
and finally take the third from the second:
1 2 3 1 2 3
3 3 3 = 3 3 3 = 0.
3 3 3 0 0 0
Note that what we are doing is changing the rows from R1 , R2 , R3 to R1 , R2 , R3 −
R2 , then to R1 , R2 − R1 , R3 − R2 , and to R1 , R2 − R1 , (R3 − R2 ) − (R2 − R1 ),
and we see that R3 − 2R2 + R1 = 0. So the linear relationship between the
rows is implying that the determinant vanishes.
(b) Here, we can expand down the first column:
1 2 3
4 5
0 4 5 = 1 0 6 = 24.
0 0 6
5. Expanding along the bottom row,
a 3
0 b
1 2
0 0
we get
0 5
a 3 0
0 2
0 b 0 .
=
d
c 3
1 2 c 0 d
For this 3 × 3 determinant, expanding down the last column gives
a 3 0
a 3
= abcd.
d 0 b 0 = dc 0
b
1 2 c 6. We can expand down the final
f
k
e m
p
column to get
a
g h i k
l 0 0
− j n 0 0
m
p
q 0 0
b
l
n
q
c
0
0
0
d
0
;
0
0
after expanding each of these down the final column, we get
k l 0
k l 0
−ei m n 0 + dj m n 0 ,
p q 0
p q 0
and the 3 × 3 determinants vanish as the final columns are all 0.
7. We take column 1 from the other two:
1 1 1 1
0
0
x y z = x y −x
z − x .
2 2 2 2 2
x y z x y − x2 z 2 − x2 We can take a factor of y − x from the second column and z − x from the third, to
get
1
1
0
0 0
0 x y −x
1
1 .
z − x = (y − x)(z − x) x
2 2
2
2
2
2
x y + x z + x
x y − x z − x Expanding along the top row, we get (y − x)(z − x)[(z + x) − (y + x)] and the result
follows.
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8. Adding 100 times the first column plus 10 times the middle to the last one will leave
the determinant unchanged; we get
1 2 3 1 2 123
4 5 6 = 4 5 456 ;
7 5 3 7 5 753
expanding down the final column, we see that all these entries are divisible by 3,
and the other entries are integers, so the determinant is divisible by 3.
9. Thinking about how determinants work, we should see that conjugating a determinant is equivalent to taking the determinant of the conjugates:
z w .
D = z w
z w
, which is the same as the original determinant, but with
But this is just D = z w
the two rows interchanged. We know that interchanging two rows in a determinant
multiplies its value by −1, so we see that D = −D. This means that D is a complex
number whose conjugate is its negation, so its real part must vanish.
10. We’ll expand down the first column (since there is a 0, that will mean one fewer
2 × 2 determinant we need to evaluate):
2 1 3
1 3
2 1
0 2 1 = 2 1 6 − 0 + 3 2 1 = 2 × 11 + 3 × (−5) = 7.
3 1 6
(a) We are exchanging two rows, and this means that the value of the determinant
is multiplied by −1, so the answer is −7.
(b) This is the transpose of the original matrix, and taking transposes leaves the
determinant unchanged, so this determinant is 7.
(c) This is the determinant in (a), but with the middle column scaled by 3. So
this determinant is −21.
11. (a) The whole determinant is multiplied by (−1)3 = −1.
(b) The determinant is doubled.
(c) The determinant is multiplied by 8.
(d) The determinant is unchanged.
12. The fourth row is the average of the first and third. So there is a linear relationship
between the rows.
13. The determinant
1 2 3
2 3 5 = 1,
3 5 7
and this is nonzero, so the result follows.
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14.

a2 + b2 + c2 + d2
0
0
0


0
a2 + b2 + c2 + d2
0
0
.
AAT = 


0
0
a2 + b2 + c2 + d2
0
2
2
2
2
0
0
0
a +b +c +d

Hence det(AAT ) = (a2 + b2 + c2 + d2 )4 . Since det(AAT ) = det A. det AT = (det A)2 ,
we see that det A = ±(a2 + b2 + c2 + d2 )2 . (In fact, the sign is +, as one can see by
comparing the coefficients of a4 , say.)




−16 7
1
5 0 0
15. We compute adj A =  9 −8 1 , and A.adj A = 0 5 0, and so det A =
1
3 −1
0 0 5
5.
16. We define


2 1 −1
A = 1 2 1  ,
3 0 −1


0 1 −1
A1 = 3 2 1  ,
3 0 −1


2 0 −1
A2 = 1 3 1  ,
3 3 −1


2 1 0
A3 = 1 2 3 .
3 0 3
Then det A = 6, det A1 = 12, det A2 = −6 and det A3 = 18. So x =
A3
A2
= −1 and z = det
= 3.
y = det
det A
det A
det A1
det A
= 2,
17. (a) det A = a2 + b2 ; det B = c2 + d2 .
ac − bd ad + bc
, so α = ac − bd and β = ad + bc.
(b) AB =
−ad − bc ac − bd
(c) Suppose that m = a2 +b2 and n = c2 +d2. Then form matrices A and B as in the
question, so that det A = m and det B = n. We know det AB = det A. det B,
and det AB = α2 + β 2 , so
(a2 + b2 )(c2 + d2 ) = mn = (ac − bd)2 + (ad + bc)2 .
(d) (a + ib)(c + id) = (ac − bd) + i(ad + bc).
18. (a) det A = z1 z 1 + w1 w 1 = (a2 + b2 ) + (c2 + d2 ).
(b) We have
AB =
z1 w2 + w1 z 2
z1 z2 − w1 w 2
−w 1 z2 − z 1 w2 −w1 w2 + z 1 z 2
=
z w
−w z
where z = z1 z2 − w1 w2 and w = z1 w2 + w1 z 2 .
(c) This gives the relation
det A. det B = (z1 z 1 + w1 w1 )(z2 z 2 + w2 w2 ) = det AB = zz + ww.
But
det A. det B = (a2 + b2 + c2 + d2 )(e2 + f 2 + g 2 + h2 );
we calculate
z = (a + ib)(e + if ) − (c + id)(g − ih) = (ae − bf − cg − dh) + i(af + be + ch − dg)
w = (a + ib)(g + ih) + (e − if )(c + id) = (ag − bh + ce + df ) + i(ah + bg − cf + de),
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so
det AB = zz + ww
= (ae − bf − cg − dh)2 + (af + be + ch − dg)2
+(ag − bh + ce + df )2 + (ah + bg − cf + de)2 ,
an identity which took Euler 13 years to prove.
This identity shows that if two integers can be written as the sum of four
squares, then so can their product. Given that all primes can be expressed as
the sum of four squares, it follows that all positive integers can be expressed
as the sum of four squares.
19. (a) As ij and ji are different, this system is not commutative, so we’ll need to be
careful about the order of multiplication. As ij = k, we get ik = i(ij) = i2 j =
−j, so ik = −j. Similarly, kj = (ij)j = ij 2 = −i.
Starting with ji = −k, we get −jk = j(ji) = −i, so jk = i; similarly −ki =
(ji)i = ji2 = −j, and so ki = j.
Finally, ijk = (ij)k = k.k = −1.
(b) Using these rules,
(a + bi + cj + dk)(e + f i + gj + hk)
= ae + af i + agj + ahk + bei + bf i2 + bgij + bhik
+cej + cf ji + cgj 2 + chjk + dek + df ki + dgkj + dhk 2
= ae + af i + agj + ahk + bei − bf + bgk − bhj
+cej − cf k − cg + chi + dek + df j − dgi − dh
= (ae − bf − cg − dh) + (af + be + ch − dg)i
+(ag − bh + ce + df )j + (ah + bg − cf + de).
(c) One should compare this with the brackets in the identity in the previous
question; they are exactly the same!
(d) This may suggest that a + bi + cj+ dk ought
z w
w = c + di in the matrix
.
−w z
i
In particular, i ought to correspond to I =
0
0 1
J=
, and k ought to correspond to K
−1 0
to correspond to z = a + bi,
0
, j ought to correspond to
i 0 i
=
,
i 0
(e) It is easy to check that I 2 = J 2 = K 2 are all minus the identity, and that
IJ = K and JI = −K. (Looking back at the arguments in (a), all the other
relationships follow from these two.)
20. (a) We can check this by expanding along any row or column. (Or one can do row
operations to add the second and third row to the top, and then seeing that
there is a factor of x + y + z.)
(b) We have


  ′ ′ ′  ′
xx + yz ′ + zy ′ xy ′ + yx′ + zz ′ xz ′ + yy ′ + zx′
x y z
x y z
z x y  z ′ x′ y ′  = zx′ + xz ′ + yy ′ zy ′ + xx′ + yz ′ yz ′ + zy ′ + xx′  ,
yx′ + zz ′ + xy ′ yy ′ + zx′ + xz ′ yz ′ + zy ′ + xx′
y ′ z ′ x′
y z x
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and this has the right form with X = xx′ + yz ′ + zy ′ , Y = xy ′ + yx′ + zz ′ and
Z = xz ′ + yy ′ + zx′ .
(c) Using det AB = det A. det B as usual, we get
(x3 + y 3 + z 3 − 3xyz)(x′3 + y ′3 + z ′3 − 3x′ y ′ z ′ ) = (X 3 + Y 3 + Z 3 − 3XY Z).
21. Write
1
1 0 0 . . . 0 0
−1 1 1 0 . . . 0 0
Dn = 0 −1 1 1 . . . 0 0 .
..
.. .. .. . .
. .
.
. .. .. .
.
.
0
0 0 0 . . . −1 1
1 1
= 2 = F2 . Expanding Dn
Then D1 = |1| = 1 = F1 , D2 = −1 1
column, we get
1 1 0 . . . 0 0
1 0 0 . . . 0
−1 1 1 . . . 0 0
0 1 1 . . . 0
Dn = 1. .. .. .. . .
.. .. − (−1) .. .. .. . .
.
. . .
. . .
. . .
. ..
0 0 0 . . . −1 1
0 0 0 . . . −1
down the first
0
0
.. .
. 1
The first determinant looks exactly like Dn , but has size n − 1, so is Dn−1 . The
second determinant can be expanded along the top row, and is just Dn−2 . So we
conclude that Dn = Dn−1 + Dn−2 . So Dn satisfies all the defining relationship as
the Fibonacci numbers, so Dn = Fn .
22. We can simply expand along any row or column (exercise).
Alternatively (and this argument
0
a
−a 0
−b −c
generalises to any odd order), we have
0 −a −b
b c = (−1)3 a 0 −c ,
b c
0
0
taking a factor of −1 from each row. But now this means det A = (−1)3 det(AT ),
but we have (−1)3 = −1 and det AT = det A, so det A = − det A, and so det A = 0.
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23. The permutations and corresponding matrices are


1 0 0
id = 0 1 0 ,
0 0 1


0 1 0
(1 2) = 1 0 0 ,
0 0 1


0 0 1
(1 3) = 0 1 0 ,
1 0 0


1 0 0
(2 3) = 0 0 1 ,
0 1 0


0 1 0
(1 2 3) = 0 0 1 ,
1 0 0


0 0 1
(1 3 2) = 1 0 0 .
0 1 0
In each case, it is easy to see that the determinant of the corresponding matrix is
equal to the sign of the permutation. (Indeed, one can use this as the definition of
the sign of a permutation for any permutation in Sn .)




0 1 0
0 0 1
Finally, A1 = 1 0 0, A2 = 0 1 0, and it is easy to verify the properties
0 0 1
1 0 0
claimed for A1 and A2 . In fact, the multplication table for the group elements of S3
is exactly mirrored by the multiplication table for the corresponding matrices – these
groups are isomorphic.
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MAS111: Mathematics Core II
Solutions to Problems booklet – Chapters 5–8
Chapter 5
1. (a) We work out the characteristic polynomial at
2 − λ
0 = (2 − λ)(3 − λ),
det(A − λI) = 0
3 − λ
0 0
so the eigenvalues are 2 and 3. For λ = 2, we consider A−2I, which is
,
0 1
1
and look for a nonzero vector x with (A − 2I)x = 0; clearly x =
works
0
−1 0
(as does any multiple). Similarly, for λ = 3, we look at A − 3I =
,
0 0
0
.
and the eigenvector is clearly
1
Notice that this always happens for diagonal matrices – the eigenvalues are the
diagonal entries, and the eigenvectors are the standard basis elements.
(b) Similarly, the eigenvalues are the diagonal entries, namely 2 and 1,since the
0 1
,
matrix is upper triangular. Again, for λ = 2, we consider A − 2I =
0 −1
1
is the eigenvector. For λ = 1, we consider A − I =
and again we see that
0
1
1 1
.
, and it is easy to see that the eigenvector is
−1
0 0
(c) Now we can work out
2 − λ
3
= (2−λ)(1−λ)−6 = λ2 −3λ−4 = (λ−4)(λ+1),
det(A−λI) = 2
1 − λ
−2 3
, and
so the eigenvalues are 4 and −1. With λ = 4, A − 4I =
2 −3
3 3
3
, and an eigenvector is
; with λ = −1, A + I =
an eigenvector is
2 2
2
1
.
−1
(d) This time there is clearly
eigenvalue of λ = 2. But there is only one
a repeated
0 1
so (A − 2I)x = 0 implies that x is (a multiple
eigenvector; A − 2I =
0 0
1
.
of)
0
1 1
, we consider
(e) Finally, if A =
1 0
1 − λ 1 = λ2 − λ − 1,
det(A − λI) = 1
−λ
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whose roots are λ =
√
1± 5
.
2
With λ =
A − λI =
the eigenvector is
−1
√
1∓ 5
2
√
1± 5
,
2
√
1∓ 5
2
1
then
1√
−1∓ 5
2
!
;
(or any multiple).


2 0 0
2. (a) If A = 0 −1 0, then it is diagonal, so the eigenvalues are the diagonal
0 0 2
 
0

entries, namely 2 (twice) and −1. The eigenvector with eigenvalue −1 is 1.
 
0
1
0



There are two natural eigenvectors with eigenvalue 2, namely 0 and 0,
0
1
 
a

but it is important to note that any vector of the form 0 is also an eigenvecb
tor with eigenvalue 2 – there is a whole 2-dimensional plane (the plane y = 0)
which gets scaled by 2 under this transformation, and you can take any two
linearly independent vectors to be eigenvectors.


1 0 0
(b) 2 3 0  is lower triangular, so the eigenvalues are the diagonal entries, 1,
1 0 −7


0 0 0
3 and −7. With λ = 1, we have A − I = 2 2 0 , and an eigenvector is
1 0 −8


 
−2 0 0
8
−8. With λ = 3, we have A − 3I =  2 0 0 , and an eigenvector
1 

1 0 −10

0
8 0 0
is 1. With λ = −7, we have A + 7I = 2 10 0, and an eigenvector is
0
1 0 0
 
0
0.
1
3. As usual, we work out det(A − λI):
−λ
1
0
−4
4 − λ
0
0
− 1. 0 = −λ det(A − λI) = −4 4 − λ
−2 2 − λ
1
2 − λ
−2
1
2−λ
= −λ(4 − λ)(2 − λ) − (−4.(2 − λ))
= −λ3 + 6λ2 − 12λ + 8
= −(λ − 2)3 .
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So the only eigenvalue is λ = 2. We have


−2 1 0
A − 2I = −4 2 0 .
−2 1 0
 
x

So the eigenvectors are all y  satisfying −2x + y = 0, a plane. We can choose
z
 
 
1
0



eigenvectors 2 and 0, but any two vectors satisfying y = 2x will do, as long
0
1
as they are no multiples of each other.
4. The characteristic polynomial turns out to be
λ3 − 5λ2 + 17λ − 13 = (λ − 1)(λ2 − 4λ + 13),
so one eigenvalue is 1 and the other two are roots of the quadratic, which the usual
formula gives
as 4±16−52
= 2 ± 3i. For the eigenvector
2

  with λ = 1, we consider
3 −5 7
1



A−I =
1 −5 9 , and find the eigenvector 2.
−4 0 4
1


2 ∓ 3i
−5
7
−6 ∓ 3i
9 , and find
For λ = 2 ± 3i, we consider A − (2 ± 3i)I =  1
−4
0
3 ∓ 3i


3 ± 3i
the eigenvectors 5 ± 3i.
4
5. (a) The characteristic polynomial is λ2 −2λ−3
eigenvalues
=(λ+1)(λ−3). So the
1
2 2
. When
, with eigenvector
are −1 and 3. When λ = −1, A + I =
−1
2 2
1
−2 2
, with eigenvector
λ = 3, then A − 3I =
.
2 −2
1
(b) If λ is an eigenvalue of A, then λ2 is an eigenvalue of A2 , so A2 has eigenvalues
1 and 9. Similarly, the eigenvalues of A3 − A are λ3 − λ, which are 0 and 24.
Finally, the eigenvalues of A2 − 6A + 2I are λ2 − 6λ + 2, which are 9 and −7.
(c) Consider
so
a
1
1
=α
+β
,
b
−1
1
α+β =a
−α + β = b,
so β = (a + b)/2 and α = (a − b)/2.
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(d) So
(a − b) 1
(a + b) 1
a
n
=A
+
A
b
−1
1
2
2
(a − b) n 1
(a + b) n 1
=
+
A
A
−1
1
2
2
(a + b) n 1
(a − b)
1
n
+
.
(−1)
3
=
−1
1
2
2
n
6. We have
A(Bu) = (AB)u = (BA)u = B(Au) = B(λu) = λ.Bu,
so Bu is also an eigenvector with eigenvalue λ.
7. We work out the characteristic polynomial in the usual way to find the eigenvalues.
We substitute λ = 10 into the equation, and find that it is indeed a root (1000 −
1800 + 960 − 160 = 0), so there is a factor (λ − 10). We find that
λ3 − 18λ2 + 96λ − 160 = (λ − 10)(λ − 4)2 ,
and so there is one other eigenvalue, λ = 4, repeated twice.
8. This was done in Chapter 1, question 9, except that there we didn’t observe that
this area could be written as a determinant.
9. (a) We have
1
Ae1 =
3
1
Ae2 =
3
1
1
= 1.e1 + 3.e2
=
3
0
2
0
2
=
= 2.e1 + 4.e2 .
4
1
4
1
Ae1 =
3
1
Ae2 =
3
1
1
=
= 2.f1 − 1.f2
0
3
2
0
2
=
= 3.f1 − 1.f2 .
4
1
4
2
4
(b) We have
2
4
(c) We have
1
3
Af1 =
=
= 3.e1 + 7.e2
1
7
1 2
1
−1
Af2 =
=
= −1.e1 − 1.e2 .
3 4
−1
−1
1 2
3 4
(d) We have
1
Af1 =
3
1
Af2 =
3
MAS111
1
3
=
= 5.f1 − 2.f2
1
7
2
1
−1
=
= −1.f1 + 0.f2 .
4
−1
−1
2
4
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Spring 2016-17
(e) Using the last part, it seems that
and f2 .
−1
1
−2
5 −1
−2 0
represents A using the vectors f1
−1 −1
= 12 M. Then a calculation shows that
−1 1
(f) det M = −2, so M =
5 −1
−1
, the same answer as in the previous part.
M AM =
−2 0
10. We first need to find the eigenvalues and eigenvectors. The characteristic polynomial
is λ2 −4λ−5 = 0, whose
= 5 and λ = −1. The corresponding eigenvectors
rootsare λ
1
1
. We form M by taking its columns to be the
and
are, respectively,
−1
1
1 1
eigenvectors, so M =
. Then M −1 AM is diagonal, and the entries are the
1 −1
eigenvalues.
11. (a) The characteristic polynomial is λ2 − λ −
2=
(λ − 2)(λ + 1), which gives the
2
; the eigenvector for λ = −1 is
first claim. The eigenvector for λ = 2 is
1
1
.
−1
1
. Then
(b) Let’s take x =
1
1 2
1
3
Ax =
=
1 0
1
1
1 2
3
5
A2 x =
=
1 0
1
3
1 2
5
11
A3 x =
=
1 0
3
5
1 2
11
21
A4 x =
=
1 0
5
11
1 2
21
43
A5 x =
=
1 0
11
21
etc.
20
10
0
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(c) If x = αv1 + βv2 , then
Ax = A(αv1 + βv2 )
= α.Av1 + β.Av2
= 2αv1 − βv2 ,
as required. Applying A again gives
A2 x = 22 αv1 + βv2 ,
and more generally,
An x = 2n αv1 + (−1)n βv2 .
As n gets large, the coefficient of 2n begins to dominate the (−1)n , so the
second term becomes negligible compared with the first. Then
An x ≈ 2n αv1 ,
which is in the direction of v1 .
(d) We have
20
10
0
10
20
30
40
It appears that the points in the graph are getting more and more in the
direction of the line in the direction of the main eigenvector.
More generally, if A is a square n × n-matrix, with eigenvalues λ1 , . . . , λn , and
corresponding eigenvectors x1 , . . . , xn , which we assume to be linearly independent,
and if |λ1 | > |λk | for all k ≥ 2, then this procedure gives a computational method for
working out the eigenvector corresponding to λ1 . Google uses this sort of method
for computing the main eigenvector of its matrix.
12. (a) (M −1 AM)2 = (M −1 AM)(M −1 AM) = M −1 A(MM −1 )AM = M −1 A2 M.
(b) The characteristic polynomial is λ2 − 4λ + 3 = (λ − 1)(λ − 3), so the eigenvalues
1
are 1 and 3. The eigenvalue λ1 = 1 corresponds to the eigenvector x1 =
,
1
1
and the eigenvalue λ2 = 3 to the eigenvector x2 =
.
3
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(c) We put M =
3 −1
1 1
1
−1
. Then
. Then M = 2
−1 1
1 3
1 3 −1
0 1
1 1
−1
M AM =
−3 4
1 3
2 −1 1
1 3 −1
1 3
=
−1
1
1 9
2
1 2 0
1 0
=
.
=
0 3
2 0 6
Let’s explain how we could have seen
this without dong lots of matrix mul1
. Then Me1 = x1 , the first eigenvector.
tiplication. Suppose that e1 =
0
Now AMe1 = Ax1 = λ1 x1 , as x1 is an eigenvector with eigenvalue λ1 . Since
Me1 = x1 , clearly e1 = M −1 x1 . So M −1 AMe1 = M −1 λ1 x1 = λ1 e1 . Similarly,
M −1 AMe2 = λ2 e2 . Alternatively,
1
λ1
−1
M AM
=
;
0
0
0
0
−1
M AM
=
.
1
λ2
λ1 0
−1
.
From this we see that M AM =
0 λ2
1 0
(d) If we write D =
, then M −1 AM = D, so A = MDM −1 . Now
0 3
A10 = MD 10 M −1
1 3 −1
1 0
1 1
.
=
0 310 2 −1 1
1 3
1 3 −1
1
0
1 1
.
=
0 59049 2 −1 1
1 3
1 3 −1
1 59049
.
=
1 177147 2 −1 1
−29523 29524
.
=
−88572 88573
13. Since A is triangular, its eigenvalues are the diagonal entries, so the characteristic
polynomial is (λ − 3)(λ − 2)(λ − 1) = 0. We need to check that (A − 3I)(A − 2I)(A −
I) = 0, and we simply multiply it out.
14. We have
A2
e =I +A+
+··· = I +A =
2!
1 0
2
B
as A = 0. Similarly, e =
.
1 1
A
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Spring 2016-17
0 1
2 1
, and (A + B)2 = I. Indeed,
. But A + B =
We have e e =
1 0
1 1
0 1
2k
2k+1
. Then
(A + B) = I and (A + B)
=
1 0
(A + B)2
1 + 2!1 + · · ·
1+···
A+B
+··· =
,
e
= I + (A + B) +
1 + ...
1 + 2!1 + · · ·
2!
A B
where the entries in the main diagonal are clearly the same, unlike those of eA eB .
In fact, if AB = BA, then it is true that eA+B = eA eB , as you might like to try to
show – you just need to use the result that ex ey = ex+y for the usual exponential
function.
e 0
D
.
15. We have e =
0 e2
2 1
1 −1
−1
Write M =
, with determinant 1, and inverse M =
. We have
1 1
−1 2
eA = eM
This is
−1 DM
= M −1 eD M.
e 0
0 e2
2e − e2 e − e2
.
and this works out as
2e2 − 2e 2e2 − e
1 −1
−1 2
2 1
,
1 1
16. If A has diagonal entries a1 , . . . , an , then eA is again diagonal, with diagonal entries
ea1 , . . . , ean . Then
det(eA ) = ea1 . . . ean = ea1 +···+an = etrace(A)
as required.
In fact, the result is true for any square matrix.
17. We can either eliminate one of the variables or we can use matrix techniques.
x(t)
, and get the matrix equation
For these, we write x(t) =
y(t)
ẋ(t) = Ax(t),
1 2
. The solution is eAt x(0). Since the eigenvalues of A are roots of
where A =
4 3
2
the
polynomial λ −4λ −5 = 0, which are 5 and −1, with eigenvectors
characteristic
−1
1
, we have the general solution
and
1
2
−t −1
5t 1
;
+ Be
x(t) = Ae
1
2
1
, we see that A = 1, B = 0, and so x(t) = e5t , y(t) = 2e5t is the
as x(0) =
2
desired solution.
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x+y
x
.
=
18. (a) A
x
y
(b) Then we have:
2
1
=
A
1
1
3
2
2 1
=
=A
A
2
1
1
5
3
1
=
=A
A3
3
2
1
8
5
1
=
=A
A4
5
3
1
1
Fn+2
=
. To prove this by induction, we can see it holds
(c) It looks like A
1
Fn+1
Fk+1
k−1 1
=
.
for n = 0 (and n = 1, . . . , 4). Suppose that for k ≥ 1: A
1
Fk
Then
Fk+2
Fk+1 + Fk
Fk+1
k 1
,
=
=
=A
A
Fk+1
Fk+1
Fk
1
n
as required.
(d) The characteristic
polynomial√ of A is λ2 − λ − 1, and the roots of x2 − x − 1 = 0
√
are λ1 = 1+2 5 and λ2 = 1−2 5 . For !λ1 , we can work out the eigenvector by
√
1− 5
1√
1√
2
, and the eigenvector is x1 = −1+ 5 .
computing A − λ1 I =
−1− 5
1
2
2
1√
Similarly, the eigenvector for λ2 is x2 = −1− 5 .
2
(e) We solve
αx1 + βx2 = α
1√
−1+ 5
2
+β
1√
−1− 5
2
,
1
if
so we can solve αx1 + βx2 =
1
α
√ !
−1 + 5
+β
2
α+β =1
√ !
−1 − 5
= 1.
2
√
Multiply the first equation by −1+2 5 , to get
√ !
√ !
√ !
−1 + 5
−1 + 5
−1 + 5
α
+β
=
2
2
2
√ !
√ !
−1 + 5
−1 − 5
α
+β
= 1.
2
2
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Spring 2016-17
Now subtract the second from the first:
√
−3 +
β. 5 =
2
√
5
.
√
Similarly, if we multiply the first equation by −1−2 5 , we get
√ !
√ !
√ !
−1 − 5
−1 − 5
−1 − 5
+β
=
α
2
2
2
√ !
√ !
−1 + 5
−1 − 5
α
+β
= 1.
2
2
Subtracting the first from the second gives
√
√
3+ 5
α. 5 =
.
2
So
1
α= √
5
√ !
3+ 5
,
2
1
β=√
5
−3 +
2
√ !
5
.
(f) An x1 = λn1 x1 , and An x2 = λn2 x2 .
Fn+2
n 1
.
in two ways. We already know that it equals
(g) Now we compute A
Fn+1
1
Now we also compute
n 1
A
= An (αx1 + βx2 )
1
= λn1 αx1 + λn2 βx2 .
We now compare the bottom components:
√
√
5
5
n −1 −
n −1 +
+ λ2 β
.
Fn+1 = λ1 α
2
2
Now note that
√
√
√
−1 + 5
1 3 + 5 −1 + 5
α
=√ .
2
2
2
5
√
1
1 1+ 5
= √ λ1
=√ .
2
5
5
√
√
√
1 −3 + 5 −1 − 5
−1 − 5
=√ .
β
2
2
2
5
√
1
1 −1 + 5
= − √ λ2 .
=√ .
2
5
5
So
1
Fn+1 = √ (λn+1
− λn+1
),
1
2
5
and so
1
Fn = √ (λn1 − λn2 ).
5
This formula is Binet’s formula for the Fibonacci numbers.
MAS111
35
Spring 2016-17
1
, and find that it is
(h) Again, we use the same matrix A, and consider A
3
Ln+2
. Now we want to solve
Ln+1
1√
1√
1
,
αx1 + βx2 = α −1+ 5 + β −1− 5 =
3
2
2
n
and a similar argument to the above shows that
Ln = λn1 + λn2 .
Note that |λ2 | < 1 < λ1 , so
√
√ λn + λn2
Ln
λn + λn
→ 5,
= n 1 n 2 √ = 5 n1
n
Fn
λ1 − λ2
(λ1 − λ2 )/ 5
as required.
Chapter 6
1. (a) Level curves for c = −3, . . . , 3. These are translations of a parabola x = y 2 in
a horizontal direction (the level curve at c passes through the x-axis at c):
y
2
1
x
−3
−2
−1
0
1
2
3
4
−1
−2
−3
(b) You should recognise the equation x2 + 2y 2 = c as an ellipse, at least if c > 0
(if c = 0, we just get the origin, and if c < 0, there will be no points). Here
are the level curves for c = 0 (just the origin), and ellipses going out from the
centre with c = 1, . . . , 10:
y
2
1
x
−3
−2
−1
0
1
2
3
−1
−2
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36
Spring 2016-17
(c) Consider xy = 0 – this is just the pair of axes. For c > 0, xy = c can be
rearranged to y = c/x, which is just a hyperbola in the first (top right) and
third (bottom left) quadrants, going further from the origin as c increases. For
c < 0, we get the same, except that this hyperbola is in the second (top left) and
fourth (bottom right) quadrants. Here are the level curves for c = −6, . . . , 6:
y
2
1
x
−3
−2
−1
0
1
2
3
−1
−2
2. We consider the level curves around (1, 1), and we see how widely spaced the curves
are in the x direction and the y direction:
2
y
1
x
−2
−1
0
1
2
3
−1
−2
At the point (1, 1), f appears to be increasing more rapidly in the x direction.
We can check as follows:
at (1, 1).
bigger than ∂f
∂y
∂f
∂x
= 1, while
3.
∂f
∂x
= 3x2 y 2 − 5yexy and
4.
∂f
∂x
= 2xy + ex cos y + cos x ln y and
∂f
∂y
= 23 y, so at (1, 1),
∂f
∂y
= 32 . So
∂f
∂x
is indeed
= 2x3 y − 5xexy .
5. z = f (x, y), where f (x, y) = 41 x2 +
∂f
∂x
∂f
∂y
= 12 x. So at (x, y) = (1, 1),
∂f
∂x
∂f
∂y
= x2 − ex sin y +
sin x
.
y
1 2
y .
12
= 12 . So the gradient is 12 .
6. Let the radius be r, the height h and the volume v. Then v = πr 2 h. So
∂v
and ∂h
= πr 2 . So
∂v
∂v
δv ≈
δr +
δh = 2πrhδr + πr 2 δh.
∂r
∂h
Thus
4
−2
δv
δr δh
3
δv
+
=
= 2 ≈2 +
=2
.
v
πr h
r
h
1000
1000
1000
∂v
∂r
= 2πrh
So the volume goes up by about 0.4%.
MAS111
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Spring 2016-17
7. This example is the prototype for a more general result, which is the 2-variable
analogue of the Chain Rule.
Note that z = r 2 , and is independent of θ. So
∂z
= 2r,
∂r
∂z
= 0.
∂θ
On the other hand, we also have:
∂z
∂x
∂z
∂y
∂x
∂r
∂y
∂r
= 2x,
= 2y,
= cos θ,
= sin θ,
∂x
∂θ
∂y
∂θ
= −r sin θ;
= r cos θ.
Now
∂z ∂x ∂z ∂y
+
= 2x cos θ + 2y sin θ
∂x ∂r ∂y ∂r
= 2(r cos θ) cos θ + 2(r sin θ) sin θ
= 2r(cos2 θ + sin2 θ)
= 2r
∂z
=
∂r
and
∂z ∂x ∂z ∂y
+
= 2x(−r sin θ) + 2y(r cos θ)
∂x ∂θ ∂y ∂θ
= 2(r cos θ)(−r sin θ) + 2(r sin θ)(r cos θ)
= 2r 2 (− cos θ sin θ + sin θ cos θ)
=0
∂z
=
∂θ
as required.
You may also like to express r and θ in terms of x and y, and work out the partial
∂r
derivatives ∂x
etc., and show that we also have equalities
∂z ∂r ∂z ∂θ
∂z
=
+
,
∂x
∂r ∂x ∂θ ∂x
∂z
∂z ∂r ∂z ∂θ
=
+
.
∂y
∂r ∂y ∂θ ∂y
You could also replace z = r 2 by other functions of x and y, and verifying that the
same result holds.
8. We had
r=
with inverse
p
ρ2 + z 2 ,
φ = tan−1 (z/ρ),
ρ = r sin φ,
z = r cos φ,
so
MAS111
∂r
∂(r, φ) ∂ρ
= ∂φ
∂(ρ, z) ∂ρ
∂r ∂z ∂φ ∂z ρ
√
2 2
= ρ−z+z
z 2 +ρ2
z
ρ2 +z 2 ρ
2
2
z +ρ
√
38
ρ2 + z 2
1
=p
=p
.
ρ2 + z 2 (ρ2 + z 2 )
ρ2 + z 2
Spring 2016-17
On the other hand,
∂(ρ, z) ∂ρ
∂r
= ∂z
∂(r, φ)
∂r
9. We have
∂ρ ∂φ ∂z ∂φ sin φ r cos φ = r.
= cos φ −r sin φ
∂x ∂x ∂x ∂r
∂θ
∂φ ∂(x, y, z) ∂y
∂y = ∂r ∂y
∂θ
∂φ ∂(r, θ, φ) ∂z ∂z ∂z ∂r ∂θ ∂φ cos θ sin φ −r sin θ sin φ r cos θ cos φ
= sin θ sin φ r cos θ sin φ r sin θ cos φ cos φ
0
−r sin φ cos θ sin φ −r sin θ sin φ
−r sin θ sin φ r cos θ cos φ
− r sin φ = cos φ sin θ sin φ r cos θ sin φ r cos θ sin φ r sin θ cos φ = cos φ × (−r 2 cos φ sin φ) − r sin φ × (r sin2 φ)
= −r 2 sin φ.
10. At the point, we have z = 2. We also have
tangent plane at (1, 1, 2) is given by
z−2=
∂z
∂x
= 2x,
∂z
∂y
= 2y. The equation of the
∂z
∂z
(1, 1)(x − 1) +
(1, 1)(y − 1),
∂x
∂y
i.e., z − 2 = 2(x − 1) + 2(y − 1), which simplifies to z = 2x + 2y − 2.
2
∂z
= 2x+y
= 2,
11. At (x, y) = (−1, 3), we have z = ln 1 = 0. At this point, we also have ∂x
∂z
1
and ∂y = 2x+y = 1. So the equation of the plane is z − 0 = 2(x + 1) + 1(y − 3), or
z = 2x + y − 1.
∂f1
∂y
∂f1
∂x
∂f1
∂y
= 0 when (x, y) = (0, 0).
∂f2
∂x
=
12. (a)
∂f1
∂x
= 2x and
(b)
∂f2
∂x
(c)
∂f3
∂x
= y − 1 and
(d)
4
= e−xy − (x + y)ye−xy = (1 − xy − y 2 )e−xy and ∂f
= e−xy − (x + y)xe−xy =
∂y
4
4
(1 − x2 − xy)e−xy . So ∂f
= ∂f
= 0 when 1 − xy − y 2 = 1 − xy − x2 = 0.
∂x
∂y
This means that x2 = y 2, so either x = y or x = −y. When x = y, we need
1 − x2 − x2 = 1 − 2x2 = 0, meaning that x = y = ± √12 , so we get stationary
points (x, y) = ( √12 , √12 ) or (x, y) = (− √12 , − √12 ). When x = −y, 1−xy−y 2 = 1,
so this never vanishes.
13.
∂2f
∂x2
MAS111
= 2x and
∂f4
∂x
= 2y,
∂2f
∂x ∂y
∂f3
∂y
= 2y. So
∂f2
∂y
=
= x − 1. So
= −2y. So
= 2x + 2y and
∂2f
∂y 2
∂f3
∂x
=
∂f3
∂y
∂f2
∂y
= 0 when (x, y) = (1, 1).
= 0 when (x, y) = (0, 0).
= 2x.
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Spring 2016-17
14. We have
∂φ
∂x
∂φ
∂y
∂2φ
∂x2
∂2φ
∂x∂y
∂2φ
∂y 2
∂f
∂x
15. We have
Then
∂2f
∂y 2
∂2f
∂x2
2
= 3x2 y + y 2 exy ,
2
= x3 + 2xyexy ,
2
= 6xy + y 4exy ,
2
= 3x2 + (2y + 2xy 3 )exy ,
2
= (2x + 4x2 y 2 )exy .
∂f
∂y
x+y
= yex+y +xyex+y = (x+1)yex+y and
= yex+y + (x + 1)yex+y = (x + 2)ye
= xex+y +xyex+y = x(y+1)ex+y .
,
∂2f
∂x ∂y
= (x + 1)(y + 1)ex+y and
= x(y + 2)ex+y . So the Taylor series about a point (x, y) = (a, b) looks like
f (a + h, b + k) = abea+b + (a + 1)bea+b h + a(b + 1)ea+b k
h2
k2
+ (a + 2)bea+b + (a + 1)(b + 1)ea+b hk + a(b + 2)ea+b .
2!
2!
16. (a) We have
∂f
(a, b)h +
∂x
∂f
f (a, b + k) = f (a, b) +
(a, b)k +
∂y
f (a + h, b) = f (a, b) +
∂2f
h2 ∂ 3 f
h3
(a,
b)
+
(a,
b)
+···
∂2x
2!
∂3x
3!
∂2f
k2 ∂ 3 f
k3
(a,
b)
+
(a,
b)
+··· .
∂2y
2!
∂3y
3!
(b) This is as in the notes (but with one extra term):
f (a+h, b+k) = f (a, b+k)+
∂f
∂2f
h2 ∂ 3 f
h3
(a, b+k)h+ 2 (a, b+k) + 3 (a, b+k) +· · · .
∂x
∂ x
2! ∂ x
3!
(c) For fx , we have
∂fx
∂ 2 fx
k2
fx (a, b + k) = fx (a, b) +
(a, b)k + 2 (a, b) + · · ·
∂y
∂ y
2!
2
k
= fx (a, b) + fxy (a, b)k + fxyy (a, b) + · · ·
2!
For fxx , we have
∂fxx
(a, b)k + · · ·
∂y
= fxx (a, b) + fxxy (a, b)k + · · · .
fxx (a, b + k) = fxx (a, b) +
Finally,
fxxx (a, b + k) = fxxx (a, b).
MAS111
40
Spring 2016-17
(d) Combining these, we get
h2
h3
+ fxxx (a, b + k) + · · ·
2!
3!
k3
k2
= f (a, b) + fy (a, b)k + fyy (a, b) + fyyy (a, b) + · · ·
2!
3!
k2
+ fx (a, b) + fxy (a, b)k + fxyy (a, b) + · · · h +
2!
2
h
+
+ (fxx (a, b) + fxxy (a, b)k + · · · )
2!
h3
+fxxx (a, b) + · · ·
3!
= f (a, b) + (fx (a, b)h + fy (a, b)k) +
1
+ fxx (a, b)h2 + 2fxy (a, b)hk + fyy (a, b)k 2 +
2
1
+ fxxx (a, b)h3 + 3fxxy (a, b)h2 k + 3fxyy (a, b)hk 2 + fyyy (a, b)k 3 + · · ·
6
f (a + h, b + k) = f (a, b + k) + fx (a, b + k)h + fxx (a, b + k)
You may begin to see the general pattern, as well as how to prove it.
17. (a)
∂ 2 f1
∂x2
2
2
∂ f1
= 0 and ∂∂yf21 = 2. We also have f1 (0, 0) = 0 at the stationary
= 2, ∂x
∂y
point. So the Taylor series about the point (x, y) = (0, 0) looks like
f1 (h, k) = 0 + 0h + 0k + 2
(b)
∂ 2 f2
∂x2
2
2
∂ f2
= 0, ∂x
= 1 and ∂∂yf22 = 0. We also have f2 (1, 1) = 1 at the stationary
∂y
point. So the Taylor series about the point (x, y) = (1, 1) looks like
f2 (h, k) = 1 + 0h + 0k + 0
(c)
∂ 2 f3
∂x2
2
h2
k2
+ 1hk + 0 + · · · = 1 + hk + · · · .
2!
2!
2
∂ f3
= 2, ∂x
= 0 and ∂∂yf23 = −2. We also have f3 (0, 0) = 0 at the stationary
∂y
point. So the Taylor series about the point (x, y) = (0, 0) looks like
f3 (h, k) = 0 + 0h + 0k + 2
(d)
k2
h2
+ 0hk + 2 + · · · = h2 + k 2 .
2!
2!
∂ 2 f4
∂x2
h2
k2
+ 0hk − 2 + · · · = h2 − k 2 .
2!
2!
∂ 2 f4
∂x ∂y
2
= (x2 y + xy 2 − 2x − 2y)e−xy and ∂∂yf24 =
q
√
1
(x2 y +x3 −2x)e−xy . We also have f4 ( √12 , √12 ) = 2e− 2 = 2e at this stationary
= (xy 2 + y 3 − 2y)e−xy ,
2
2
∂ f4
= − √32e and
point. And at this point, ∂∂xf24 = − √12e , ∂x
∂y
Taylor series about the point (x, y) = ( √12 , √12 ) looks like
∂ 2 f4
∂y 2
= − √12e . So the
r
1 h2
3
2
1 k2
+ 0h + 0k − √
− √ hk − √
+···
e
2e 2!
2e
2e 2!
1
= √ (4 − h2 − 6hk − k 2 ) + · · · .
2 2e
1
1
f4 ( √ + h, √ + k) =
2
2
MAS111
41
Spring 2016-17
q
√
1
Similarly, f4 (− √12 , − √12 ) = − 2e− 2 = − 2e at this stationary point. And at
2
2
2
∂ f4
= √32e and ∂∂yf24 =
this point, ∂∂xf24 = √12e , ∂x
∂y
the point (x, y) = (− √12 , − √12 ) looks like
√1 .
2e
So the Taylor series about
r
1
1
1 k2
2
1 h2
3
f4 (− √ + h, − √ + k) = −
+ 0h + 0k + √
+ √ hk + √
+···
e
2
2
2e 2!
2e
2e 2!
1
= √ (−4 + h2 + 6hk + k 2 ) + · · · .
2 2e
18.
∂f
∂x
∂f
∂y
= −e−y sin x and
= −e−y cos x and
Hence
∂2f
∂x2
+
19.
∂2f
∂y 2
∂2f
∂x2
∂2f
∂y 2
= −e−y cos x.
= e−y cos x.
= 0.
∂
∂2g ∂2g
+ 2 =
2
∂x
∂y
∂x
∂f
∂y
∂
+
∂y
∂f
∂2f
∂2f
−
=
−
= 0,
∂x
∂x∂y ∂y∂x
by equality of the mixed partials. So g is harmonic.
If f (x, y) = e−y cos x, then we want g such that
∂g
= −e−y cos x,
∂x
∂g
= e−y sin x.
∂y
So g(x, y) = −e−y sin x works.
Chapter 7
1. The equation is (x + 1)2 + (y − 2)2 = r 2 . The distance of (−1, 2) from 4x − 3y = 15
is given by Q8; it is
|4.(−1) − 3.2 − 15|
25
√
= 5,
=
5
16 + 9
so the equation of the circle is
(x + 1)2 + (y − 2)2 = 25.
2. We can write the equation of the circle as
(x − 4)2 + (y − 3)2 = 1,
so it is a circle centred at (4, 3) with radius 1.
MAS111
42
Spring 2016-17
y
x
There are lots of ways of doing this question. Here is one: Since (4, 3) is a distance
of 5 from the origin, and the circle has radius 1, the two circles we need must have
radii 4 = 5 − 1 and 6 = 5 + 1, so must be
x2 + y 2 = 16
and
x2 + y 2 = 36.
3. (a) Note that the focus is on the x-axis, and that the focus and directrix are 2a = 6
apart. So a = 3. We write Y 2 = 4aX; the vertex is at (3, 0), i.e., y = 0 when
x = 3. So we need the translation X = x − 3, Y = y, so that the equation is
y 2 = 4.3(x − 3), i.e., y 2 = 12x − 36.
(b) Here, the focus is on the y-axis, and the parabola goes downwards. The vertex
and focus are a = 2 apart. So we have X 2 = −4aY ; the vertex is at (x, y) =
(0, 4), so we need X = x, Y = y − 4, and the equation is x2 = −8(y − 4) =
32 − 8y.
4. We note that
(1 − t2 )2
4t2
1 − 2t2 + t4 + 4t2
x2 y 2
+
=
+
=
= 1,
a2
b2
(1 + t2 )2 (1 + t2 )2
1 + 2t2 + t4
so this is a parametric representation.
5. Note that if a = b = 1, this gives all points on the circle. It should also be clear
that if t ∈ Q, then (x, y) is a rational point on the circle (i.e., its coordinates are in
Q), and this gives a way to find all Pythagorean triangles. For example, if t = 21 , we
get the rational point ( 35 , 54 ), which shows that ( 53 )2 + ( 45 )2 = 1, or that 32 + 42 = 52 .
With t = 23 , we get the triangle (5, 12, 13), and with t = 41 , we get (15, 8, 17). Try
some other rational numbers for yourself!
2
6. Starting from x4 − y 2 = 1, we differentiate implicitly with respect to x to get
dy
dy
2x
x
− 2y dx
= 0, and rearranging gives dx
= 4y
. At the given point, then, the gradient
4
(of the tangent) is 5/6, and so the gradient of the normal is −6/5. Since the given
point lies on both, we soon see that the tangent has equation 5x − 6y = 8 and the
normal has equation 24x + 20y = 75.
7. (a) We complete the square to get 2(x + 1)2 − 2 + (y − 1)2 − 1 + 2 = 0, or
2(x + 1)2 + (y − 1)2 = 1.
MAS111
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Spring 2016-17
This is an ellipse, centred at (−1, 1), √with major axis of length 1 (in the ydirection) and minor axis of length 1/ 2 (in the x-direction).
y
2
1
x
−2
−1
1
0
(b) Again, we complete the square to get 2(x − 1)2 − 2 − (y − 1)2 + 1 = 3, or
2(x − 1)2 − (y − 1)2 = 4.
This is a hyperbola, centred at (1, 1).
4
y
2
x
−2
2
0
4
−2
8. Recall that
ex + e−x
,
2
ex − e−x
sinh x =
,
2
cosh x =
so that
2 sinh x cosh x =
2(ex − e−x )(ex + e−x )
e2x − e−2x
=
= sinh 2x,
4
2
and that
cosh2 x + sinh2 x =
2e2x + 2e−2x
(ex + e−x )2 + (ex − e−x )2
=
= cosh 2x.
4
2
Then, recalling also that cosh2 x − sinh2 x = 1,
2 cosh2 x = (cosh2 x − sinh2 x) + (cosh2 x + sinh2 x) = 1 + cosh 2x,
and similarly
2 sinh2 x = (cosh2 x + sinh2 x) − (cosh2 x − sinh2 x) = cosh 2x − 1.
Thus
MAS111
Z
2
cosh x dx =
1
2
Z
1 + cosh 2x dx = 21 (x + 12 sinh 2x).
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Spring 2016-17
9. The first answer is the integral
Za
cosh x dx = [sinh x]a0 = sinh a.
0
The volume is given by
Za
2
π cosh x dx = π
0
a sinh 2a
,
+
2
4
using the result of Q8.
The arc length is given by
Za q
1+
dy 2
( dx
)
dx =
Za p
2
1 + sinh x dx =
0
0
Za
cosh x dx = sinh a,
0
and the surface area by
Za
0
Za
q
sinh 2a
dy 2
2
2πy 1 + ( dx ) dx = 2π cosh x dx = π a +
.
2
0
10. If y = cosh x = (ex + e−x )/2, then clearly 2y = z + 1/z. Multiplyingp
through by
2
z gives the quadratic z − 2yz + 1 = 0, whose p
solutions are z = y ± y 2 − 1, as
required. But we put z = ex , and so x = ln(y ± y 2 − 1).
We can use the formula for difference of two squares, (a + b)(a − b) = a2 − b2 , to see
that
p
p
(y + y 2 − 1)(y − y 2 − 1) = y 2 − (y 2 − 1) = 1.
So
ln(y −
p
y 2 − 1) = ln
1
p
y + y2 − 1
!
= − ln(y +
p
y 2 − 1),
and the final result follows (as of course it must, as cosh is even, so the two values
of cosh−1 y must sum to 0).
p
p
For sinh, the same analysis gives z = y ± y 2 + 1. But as y 2 + 1 > y, if we choose
the minus sign, the value of z is negative, and so we cannot take its logarithm
(at least,
p the answer is not a real number). So the only possible value for x is
ln(y + y 2 + 1). This reflects the graph – for each real y-value, there is only one
x-value giving that y-value.
dx
= cosh u, and so dx = cosh u.du. Also, 1 + x2 = 1 + sinh2 u =
11. If x = sinh u, then du
cosh2 u, so our integral becomes:
Z
Z
Z
1
1
√
dx =
. cosh u du = 1 du = u + C = sinh−1 x + C.
cosh u
1 + x2
The same method works for the second integral, when you substitute x = a sinh u.
MAS111
45
Spring 2016-17
12. We write x2 + x + 1 = (x + 12 )2 + 43 , so that the integral is
the previous equation, we substitute x +
the integral is equal to
Z1
1
2
=
√
3
2
R1
0
√
1
(x+ 12 )2 + 34
dx; as in
sinh u, and the method gives that
h
√ i3/2
1
dx = sinh−1 (2x/ 3)
1/2
x2 + x + 1
0
√
√
= sinh−1 ( 3) − sinh−1 (1/ 3)
√
√
√
= ln( 3 + 2) − ln(1/ 3 + 2/ 3)
√
√
= ln( 3 + 2) − ln( 3)
√
= ln(1 + 2/ 3).
√
Here we use the result from Q10 that sinh−1 x = ln(x + x2 + 1).
√
13. Since | sinh x| < | cosh x| for all x, we see that | tanh x| < 1 for all x. Thinking about
the graphs of sinh x and cosh x, we get that the range of tanh is the interval (−1, 1).
This is also the domain of tanh−1 .
Next, notice that
d
d
tanh x =
dx
dx
sinh x
cosh x
1
cosh2 x − sinh2
=
= sech2 x.
2
cosh x
cosh2 x
=
dy
So if y = tanh−1 x, we get x = tanh y, and differentiating gives 1 = sech2 y. dx
, and
dy
2
2
2
2
1
so dx = sech2 y . But cosh y − sinh y = 1, and so 1 − tanh y = sech y, and so
We can use this to see that
Zx
0
But
Zx
0
1
1
dy
=
.
=
2
dx
1 − x2
1 − tanh y
dt
−1 x
=
tanh
t
= tanh−1 x.
2
0
1−t
dt
=
1 − t2
Zx
1
2
0
1
1
+
1−t 1+t
dt
1
=
(ln(1 + t) − ln(1 − t))
2
x
1+t
1
ln
=
2
1−t 0
1
1+x
= ln
.
2
1−x
x
0
As an alternative, if y = tanh−1 x, then
x = tanh y =
ey − e−y
e2y − 1
=
.
ey + e−y
e2y + 1
Then x(e2y + 1) = e2y − 1, so e2y (1 − x) = 1 + x and e2y =
result on taking logs.
MAS111
46
1+x
,
1−x
and this gives the
Spring 2016-17
Chapter 8
1. We work out the eigenvalues in the usual way; the characteristic polynomial is
λ(λ 
− 1)(λ
are 0, 1 and 6. The corresponding eigenvectors
 −6), so
 the eigenvalues
 
−1
−2
1





are −2 ,
1 , and 2. We can check that
1
0
5
 T  
−2
−1
−2  1  = (−1) × (−2) + (−2) × 1 + 1 × 0 = 0
0
1
 T  
−1
1
−2 2 = (−1) × 1 + (−2) × 2 + 1 × 5 = 0
1
5
 T  
1
−2
 1  2 = (−2) × 1 + 1 × 2 + 0 × 5 = 0,
5
0
so every pair of eigenvectors are orthogonal.
2. We have
(A + AT )T = AT + (AT )T = AT + A,
so A + AT is symmetric, and the claim follows. For the second part, we write
A − AT
A + AT
+
;
A=
2
2
we have already seen that the first matrix is symmetric, but the second is antisymmetric, since
(A − AT )T = AT − A = −(A − AT ).
3. We have (AB)−1 = B −1 A−1 = B T AT = (AB)T , since A and B are orthogonal, and
using standard results about transposes and inverses of products.
4. In each case, we need to work out the eigenvalues and eigenvectors; we then normalise the eigenvectors to have length 1, and form M by taking these normalised
eigenvectors as the columns. M will automatically be orthogonal as long as the
eigenvalues are distinct, and the columns have length 1.
1
(a) For this A, the eigenvalues are 3 and −1, with corresponding eigenvectors
1!
1
1
√
√
√
1
2
and
. These both have length 2, so we need to take M = √12
.
√1
−1
−
2
2
(b) For this A, the characteristic polynomial isλ2
− 5λ =0, sothe eigenvalues are
2
1
5 and 0, with corresponding eigenvectors
and
. Both these have
1
−2
!
√2
√1
√
5
5
.
length 5, so the desired matrix M is M = √1
√2
−
5
5
MAS111
47
Spring 2016-17
5. We have

 
1 1
2
x
2
2
2



x + 2xy + y + 4xz − 2yz + 2z = (x y z) 1 1 −1
y .
2 −1 2
z
6. We have
x
3 1
.
3x + 2xy + 3y = (x y)
y
1 3
2
2
3 1
has eigenvalues 2 and 4, with corresponding eigenvectors proThe matrix
1 3
!
− √12 √12
1
−1
. We set M =
and
portional to
, so that the columns of M
√1
√1
1
1
2
2
are the normalised eigenvectors of A. We have
2 0
T
−1
,
M AM = M AM =
0 4
and we write D for this matrix. Then A = MDM −1 = MDM T , and we can write
our quadratic form as
vT Av = vT MDM T v = (M T v)T D(M T v).
Note that
MT v =
− √12
√1
2
√1
2
√1
2
! √ (y − x)/√2
x
.
=
y
(x + y)/ 2
Then
vT Av = (M T v)T D(M T v)
√ √
√ 2 0
(y − x)/√2
= (y − x)/ 2 (x + y)/ 2
0 4
(x + y)/ 2
2
2
y−x
x+y
=2 √
+4 √
2
2
2
2
= (x − y) + 2(x + y) .
7. We can either consider the corresponding matrix, and work out when its eigenvalues
are all positive, or we can complete the square:
x2 + 2y 2 + 2z 2 + 2λxy + 2xz = (x + λy + z)2 + (2 − λ2 )y 2 − 2λyz + z 2
= (x + λy + z)2 + (z − λy)2 + (2 − 2λ2 )y 2.
If 2 − 2λ2 > 0, i.e., −1 < λ < 1, then this is the sum of nonnegative squares, so
is positive definite. However, if λ ≤ −1 or λ ≥ 1, the coefficient 2 − 2λ2 ≤ 0,
and we can pick any non-zero value for y, choose z so that z = λy, and x so that
x + λy + z = 0, and then the quadratic form takes on a non-positive value, so that
it cannot be positive definite.
8. We worked out the stationary points in Chapter 6, Q12
MAS111
48
Spring 2016-17
(a) f (x, y) has a unique stationary point at (0, 0). Its partial derivatives are
∂f
= 2x,
∂x
∂f
= 2y.
∂y
Then the second partial derivatives are:
∂2f
= 2,
∂x2
∂2f
∂2f
= 0,
= 2.
∂x ∂y
∂y 2
2 0
So the Hessian at the point is
, which has two positive eigenvalues.
0 2
Thus the point is a minimum.
(b) f (x, y) has a unique stationary point at (1, 1). Its partial derivatives are
∂f
= x − 1.
∂y
∂f
= y − 1,
∂x
Then the second partial derivatives are:
∂2f
= 0,
∂x2
∂2f
∂2f
= 1,
= 0.
∂x ∂y
∂y 2
0 1
, which has one positive and one negative
So the Hessian at the point is
1 0
eigenvalue. Thus the point is a saddle point.
(c) f (x, y) has a unique stationary point at (0, 0). Its partial derivatives are
∂f
= 2x,
∂x
∂f
= −2y.
∂y
Then the second partial derivatives are:
∂2f
∂2f
= 0,
= −2.
∂x ∂y
∂y 2
2 0
, which has one positive and one negSo the Hessian at the point is
0 −2
ative eigenvalue. Thus the point is a saddle point.
∂2f
= 2,
∂x2
(d) f (x, y) has two stationary points, one at ( √12 , √12 ), and the other at (− √12 , − √12 ).
We have
∂2f
∂2f
∂x2
= (xy 2 + y 3 − 2y)e−xy ,
∂2f
∂x ∂y
= (x2 y + xy 2 − 2x − 2y)e−xy and
= (x2 y + x3 − 2x)e−xy . At the first stationary point, we have
∂y 2
∂2f
∂x ∂y
= − √32e and
∂2f
∂y 2
= − √12e . So the Hessian matrix is
− √12e − √32e
− √32e − √12e
!
1
= −√
2e
∂2f
∂x2
= − √12e ,
1 3
.
3 1
The Hessian has one positive and one negative eigenvalue, so this is a saddle
point again. The other stationary point is left for the reader, but is again a
saddle point.
MAS111
49
Spring 2016-17
MAS111: Mathematics Core II
Solutions to Problems booklet – Chapters 9–11
Chapter 9
1. This is a geometric series with first term a = 32 and ratio r =
series converges and
∞
2
X
a
2
3
=
=
1 = 1.
n
3
1
−
r
1
−
3
n=1
1
3
< 1. Hence the
2. The series can be rewritten as
2 3 4
2
2
2
+
+
··· .
3
3
3
2
1+ +
3
So the series is a geometric series with first term a = 1 and r =
series converges and
1+
8
16
a
1
2 4
+ +
+
+··· =
=
3 9 27 81
1−r
1−
3. The ratio test tells us that
In our case, an =
2
3n
so
2
3
2
3
< 1. Hence the
= 3.
P∞
an converges if
an+1 < 1.
ρ = lim n→∞
an n=1
2 n
1
2
3n+1 3
ρ = lim 2 = lim n+1 · = lim < 1.
n→∞ 3
n→∞ 3
n→∞
2
3n
Hence the series converges.
P
2
4. We know that ∞
n=1 1/n converges. We also know that log(n) > 1 for n ≥ 3 so
1
<1
log(n)
for n ≥ 3. Hence
∞
X
n=2
∞
∞
X
X
1
1
1
=1+
≤1+
2
2
n log(n)
n log(n)
n2
n=3
n=2
which converges. Hence
∞
X
n=2
converges by the comparison test.
MAS111
n2
50
1
log(n)
Spring 2016-17
5. The ratio test tells us that
In our case, an =
2n
n2
so
P∞
an diverges if
an+1 > 1.
ρ = lim n→∞
an n=1
2n+1 n+1
2
an+1 n2 2
2 n
(n+1)
= lim n = lim ρ = lim · = 2 lim n→∞
n→∞ (n + 1)2 an n→∞ 2n2 n→∞ (n + 1)2 2n = 2 lim
n→∞
1
1+
2
n
+
1
n2
= 2 > 1.
Hence the series diverges.
P
2
2 n
2
6. We know that ∞
n=1 1/n converges. Note that n 3 > n so
1
n2 3n
This means that
<
1
.
n2
∞
∞
X
X
1
1
<
n2 3n
n2
n=1
n=1
which converges. Hence the series converges by the comparison test.
P
1
7. Let sN = N
n=1 n2 . Working to four decimal places, we have the following table:
N
1 2
3
4
5
6
7
sN
1 1.25 1.3611 1.4236 1.4636 1.4914 1.5118
Thus s7 is 1.5 to 1 d.p.. The next term to be added is 1/64 = 0.015625, which does
not change the first decimal place. However, the next few partial sums, to 4 decimal
places, are s8 = 1.5274, s9 = 1.5398, s10 = 1.5498, s11 = 1.5580. So s11 = 1.6 to 1
d.p., and the first decimal place has changed.
P
8. We know that a series ∞
n=0 an does not converge if |an | 6→ 0 as n → ∞. In our
2 n
case an = 1+i
so
n n 2 − 2i n 1
−
i
2
2
=
= |(1 − i)n |
|an | =
·
= 1+i
1+i 1−i 2
iθ n n niθ = r e = |r n |
= re
where r = |1 − i| and θ is the argument of 1 − i. But
p
√
r = |1 − i| = 12 + (−1)2 = 2.
So
n
5
= ∞.
lim |an | = lim
n→∞
n→∞ 4
Hence |an | 6→ 0 and so the series does not converge.
MAS111
51
Spring 2016-17
9. Here is a table of partial sums, to 4 decimal places:
N
1 2
sN
1 0.5 0.8333 0.5833 0.7833 0.6167 0.7595 0.6345 0.7456 0.6456
3
4
5
6
7
8
9
10
s1 , s3 , s5 is a decreasing sequence, since 1/2 > 1/3, 1/4 > 1/5, etc. And s2 , s4 , s6 is
an increasing sequence, since 1/3 > 1/4, 1/5 > 1/6, etc. The ‘odd’ partial sums are
all greater than all the ‘even’ partial sums. Hence s2 , s4 , s6 is an increasing sequence,
bounded above (by any odd partial sum), so converges to a limit. Likewise, s1 , s3 ,
s5 is a decreasing sequence, bounded below (by any even partial sum), so converges
to a limit. These limits are the same, since sn − sn−1 = an → 0. This common limit
is then the sum of the series, which converges. The limit is in between successive
partial sums sn−1 and sn , within |an | = 1/n of each of them. If we keep going until
1/n < 0.00005, i.e. until n > 20, 000, then we will have 4 d.p. correct. (So, the
convergence is slow.)
P
P∞ | cos(n)|
1
10. (a) We know that the series ∞
has posin=1 n2 converges. The series
n=1
n2
tive terms, soPits partial sums form an increasing sequence. They are bounded
1
by the sum ∞
since | cos(n)|
≤ n12 for all n, so the sequence of partial
n2
n2
P
P∞ |n=1
cos(n)
converges. This means that ∞
converges absosums of n=1 cos(n)|
n=1 n2
n2
lutely,
and we are told that absolute convergence implies convergence. Thus
P∞ cos(n)
converges.
n=1 n2
P
1
(b) We know that the harmonic series ∞
n=1 n diverges. So the series
∞ ∞
X
(−1)n+1 X
1
=
n
n
n=1
diverges i.e., the series
For an =
(−1)n+1
n2
P∞
n=1
n=1
(−1)n+1
n
does not converge absolutely.
We have
(−1)2n++1+1 (−1)2n+2+1
1
1
+
=
−
2
2
2
(2n + 1)
(2n + 2)
(2n + 1)
(2n + 2)2
2n + 3
6n + 3
(2n + 2)2 − (2n + 1)2
=
<
=
(2n + 1)2 (2n + 2)2
(2n + 1)2 (2n + 2)2
(2n + 1)2 (2n + 2)2
3
3
3(2n + 1)
=
<
.
=
(2n + 1)2 (2n + 2)2
2(2n + 1)(n + 1)
2(n + 1)2
a2n+1 + a2n+2 =
This means that
∞
X
(−1)n+1
n=1
n
=
∞ X
n=0
1
1
−
2n + 1 2n + 2
≤
∞
X
n=0
∞
3
3X 1
=
2(n + 1)2
2 k=1 k 2
P
(−1)n+1
converges
which converges. Thus, the alternating harmonic series ∞
n=1
n
by the comparison test but is not absolutely convergent.
11. If k ≤ 1 then nk ≤ n for n ≥ 1 which implies that
1
1
≥ .
k
n
n
MAS111
52
Spring 2016-17
So
∞
∞
X
X
1
1
≥
k
n
n
n=1
n=1
which diverges. Hence, by the comparison test,
∞
X
1
nk
n=1
diverges if k ≤ 1.
If k > 1 then, from Figure 1, we can bound
the graph of x−k . i.e.,
P∞
1
n=1 nk
from above by the area under
∞
1−k ∞
Z
∞
X
1
1
1
x
1
=1+
<1+
dx = 1 +
+ lim k−1
k
k
n
x
1 − k x=1
k − 1 x→∞ x
n=1
1
=1+
1
k−1
P
1
as k − 1 > 0. Thus ∞
n=1 nk converges by the comparison test if k > 1.
P∞ 1
Hence n=1 nk converges if and only if k > 1.
y
1
xk
The graph of f (x) =
1
area = 1 ×
1
2k
1
nk
1
Figure 1: Bounding
12. We have
P∞
1
n=1 nk
2
3
4
n−1
n
1
nk
x
from above using the area under the graph of y =
1
.
xk
5n
1
> >0
n
n
for all n. Thus, the series is bounded below by the divergent harmonic series.
Namely,
∞
∞
X
X
1
5n
≥
,
n
n
n=1
n=1
P
5n
and so the series ∞
n=1 n diverges by the comparison test.
MAS111
53
Spring 2016-17
13. The denominator n! looks like nn for big n, so we guess that the series diverges.
3
Using the ratio test with general term an = nn! gives
an+1
lim
= lim
n→∞ an
n→∞
(n+1)3
(n+1)!
n3
n!
2
=
(n+1)3
(n+1)(n!)
lim
n3
n→∞
n!
(n + 1)3
(n + 1)2
=
lim
n→∞ (n + 1)n3
n→∞
n3
= lim
2
1
1
n + 2n + 1
= lim + 2 + 3 = 0 < 1.
3
n→∞ n
n→∞
n
n
n
= lim
Hence the series
∞
X
an =
n=1
converges by the ratio test.
∞
X
n3
n=1
n!
14. We are given that
∞
X
1
1
1
1
1
1
π2
1
=
+
+
+
+
+
+
·
·
·
=
n2
12 22 32 42 52 6
6
n=1
So1
π2
1
1
1
1
1
1
= 2 + 2 + 2 + 2 + 2 + 2 + ...
6
1
2
3
4
5 6 1
1
1
1
1
1
1
1
=
+
+
+
+ ... +
+
+
+
+ ...
12 32 52 72
22 42 62 82
∞
∞
∞
∞
∞
∞
X
X
X
X
X
1
1X 1
1
1
1
1
=
+
=
+
=
+
2
2
2
2
2
(2n
+
1)
(2n)
(2n
+
1)
4n
(2n
+
1)
4 n=1 n2
n=0
n=1
n=0
n=1
n=0
=
∞
X
n=0
i.e.,
1
1 π2
+
. .
(2n + 1)2 4 6
∞
∞
X
1 π2
π2
1
1
π2 X
=
+
.
⇒
=
.
2
2
6
(2n
+
1)
4
6
(2n
+
1)
8
n=0
n=0
P
1
15. We showed that ∞
n=2 nk converges if and only if k > 1 by comparing theRseries by an
∞
1
dx
improper integral. So, we compare the series to the improper integral 2 x log
x
2
in the same way. The function x(log(x)) is strictly increasing and positive for
1
x > 1, so the function f (x) = x log(x)
is strictly decreasing and positive on (1, ∞).
Moreover, limx→∞ f (x) = 0 because x(log(x))2 diverges as x → ∞. From Figure 2
we see that
∞
X
n=2
1
∞
Z
∞
X
1
1
1
= 2 log(2) +
≤ 2 log(2) +
dx.
2
2
2
n(log(n))
n(log(n))
x(log(x))
n=3
2
why is this rearrangement okay?
MAS111
54
Spring 2016-17
y
The graph of f (x) =
1
x(log x)2
1
(log 2)−2
2
area = 1 ×
1
n(log n)2
1
Figure 2: Bounding
1
.
x(log(x))2
P∞
2
1
n=1 n(log(n)2
3
4
n=2
n
x
from above using the area under the graph of f (x) =
Using the substitution u = log(x) ⇒ du =
∞
X
n−1
1
n(log n)2
2(log(2))2
1
≤
n(log(n))2
+
Z∞
dx
x
and so
1
1
dx =
+
2
x(log(x))
2(log(2))2
2
Z∞
1
du
u2
log(2)
∞
1
1
1
1
1
=
+ −
− lim
+
2
2
x→∞ 2u
2(log(2))
2u u=log(2) 2(log(2))
2 log(2)
1
1
=
+
< ∞.
2(log(2))2 log(4)
=
Thus the series
∞
X
n=2
converges by the comparison text.
1
n(log(n))2
16. The radius of convergence for the series is
where an =
2n
.
n!
So
an R = lim n→ an+1 n n
2 2 (n + 1)! n + 1
n! = ∞.
= lim R = lim 2n+1 = lim ·
n+1 n→∞ n!
n→∞ 2 n→∞ 2
(n+1)!
Thus, the radius of convergence is infinite and the series converges for all x ∈
(−∞, ∞).
17. The radius of convergence for the series is
MAS111
an R = lim n→ an+1 55
Spring 2016-17
where an = n1 . So
1 n + 1
1 n R = lim 1 = lim = lim 1 + = 1.
n→∞ n→∞ n n→∞ n
n+1
Thus, the radius of convergence is 1, and the series converges for all x ∈ (−1, 1). At
x = 1 the series
∞
∞
X
xn X 1
=
n
n
n=1
n=1
which is the divergent harmonic series. At x = −1, the series is
∞
X
xn
n=1
n
=
∞
X
(−1)n
n=1
n
=−
∞
X
xn
n=1
n
=
∞
X
(−1)n+1
n=1
n
which converges by Exercises 9 or 10.
P
xn
Hence the series ∞
n=1 n converges on the interval [−1, 1).
18. The radius of convergence for the series is
where an =
10n
.
n!
So
an R = lim n→ an+1 n n
10 10 (n + 1)! n + 1
n! = ∞.
= lim R = lim 10n+1 = lim ·
n+1 n→∞ n!
n→∞ n→∞ 10 10
(n+1)!
So, radius of convergence is infinite and the series converges for all x ∈ (−∞, ∞).
Thus the interval of convergence is (−∞, ∞).
19. We note that for x fixed, our series converges if the ratio
ρ = lim
n→∞
where an =
have
(−1)n (x−1)2n+1
.
n3n
an+1
<1
an
So, for x 6= 1 (at x = 1 the series clearly converges, we
(−1)n+1 (x−1)2n+3 (n+1)3n+1
(−1)n+1 (x − 1)2n+3
n3n
·
ρ = lim (−1)n (x−1)2n+1 = lim n+1
n
2n+1
n→∞
n→∞
(n
+
1)3
(−1)
(x
−
1)
n3n
n(x − 1)2 (x − 1)2
n 1
= lim −
lim −
=
= (x − 1)2 < 1
n→∞
n→∞ n + 1 3(n + 1) 3
3
√
√
⇔ (x − 1)2 < 3 ⇔ 1 − 3 < x < 1 + 3.
Our series diverges if
an+1
>1
an
by the ratio test. Reasoning as above, and √
reversing the direction
of the inequalities
√
tells us that our series diverges if x > 1 + 3 or x < 1 − 3.
ρ = lim
n→∞
MAS111
56
Spring 2016-17
At x = 1 +
√
3
∞
X
(−1)n (x − 1)2n+1
n=1
n3n
=
2n+1
∞
X
(−1)n 3 2
n3n
n=1
√
∞
∞
X
(−1)n 3n 3 √ X (−1)n
=
= 3
n3n
n
n=1
n=1
which is a scalar multiple of the alternating harmonic series which is convergent by
Exercises 9 or 10.
√
At x = 1 − 3
√
√
∞
∞
∞
∞
X
√ X
(−1)n (x − 1)2n+1 X (−1)n (− 3)2n+1 X (−1)n (−1)2n+1 ( 3)2n+1
(−1)n
=
=
=
−
3
n3n
n3n
n3n
n
n=1
n=1
n=1
n=1
which is again a scalar multiple of the convergent alternating harmonic series.
Thus the series
∞
X
(−1)n (x − 1)2n+1
n3n
n=1
converges on the interval [1 −
√
3, 1 +
√
3].
20. The radius of convergence of the series
∞
X
(n!)2 n
x
(2n)!
n=1
is
2
an R = lim n→∞ an+1 (n!)
where an = (2n)!
. So
(n!)2 (2n)! (n!)2 (2(n + 1))! (n!)2 (2n + 2)(2n + 1)((2n)!) R = lim ((n+1)!)2 = lim = lim ·
·
2 (n!)2
n→∞
n→∞ (2n)! ((n + 1)!)2 n→∞ (2n)!
(n
+
1)
(2(n+1))! 2 + n1 2n + 1 2(n + 1)(2n + 1) = 2 lim = 2 lim = 4.
= lim n→∞ 1 + 1 n→∞ n + 1 n→∞
(n + 1)2
n
Finally, at x = ±4 we have
4n (n!)2
(n!)2 n
x =±
.
(2n)!
(2n)!
Experimentation shows that this is greater than 1 in absolute value for small values
n (n!)2
of n. If 4 (2n)!
> 1 for all n then
∞
X
(n!)2 n
x
(2n)!
n=1
diverges at x = ±4 because the terms of a convergent series necessarily tend to zero.
n (n!)2
> 1 for all n by induction. We have b1 = 2 > 1 so the
We prove that bn = 4(2n)!
base case holds. We assume the inductive hypothesis that bk > 1. We have
bk+1 =
MAS111
4(k + 1)2 4k (k!)2
4(k + 1)2
2k + 2
4k+1 ((k + 1)!)2
=
=
bk =
bk > bk > 1
(2(k + 1))!
(2k + 2)(2k + 1)((2k)!)
2(k + 1)(2k + 1)
2k + 1
57
Spring 2016-17
by the inductive hypothesis. Therefore bn > 1 for all n and so the series
∞
X
(n!)2 n
x
(2n)!
n=1
diverges at x = ±4, and the interval of convergence for the series is (−4, 4).
21. We first note that
∞
X
x(x + 1)n
2n
n=1
=x
∞
X
(x + 1)n
n=1
2n
.
P
(x+1)n
converges. We
So the left hand side converges for x fixed provided that ∞
n=1
2n
now set y = x + 1.
P
P∞ yn
n
The radius of convergence of the power series ∞
n=1 an y =
n=1 2n is given by
an = lim R = lim n→∞ n→∞ an+1
Hence
1
2n
1
2n+1
∞
X
x(x + 1)n
= lim 2 = 2.
n→∞
2n
n=1
converges when |y| = |x + 1| < 2 i.e., when x ∈ (−3, 1). At the end points we have
the series
∞
X
(−3)(−2)n
n=1
2n
= −3
∞
X
(−1)n
and
n=1
∞
X
(1)(2)n
n=1
2n
= −3
∞
X
1n
n=1
which both diverge.
Hence the interval of convergence for the series
∞
X
x(x + 1)n
n=1
2n
is (−3, 1).
22. We have the approximation
sin(x) ≈ x −
x3
x5
+
.
6
120
This gives
Z1
−1
1
Z1
sin(x)
x2
x4
x3
x5
dx = 2 1 −
+
dx = 2 x −
+
x
6
120
18 600 0
0
0
1
1
=2 1−
≈ 1.8922.
+
18 600
sin(x)
dx ≈ 2
x
Z1
The approximation
sin(x) ≈ x −
MAS111
58
x3
x5
x7
+
−
6
120 5040
Spring 2016-17
gives
Z1
Z1
Z1
x2
x4
x6
+
−
dx
6
120 5040
0
0
1
5
3
1
x
x7
1
1
x
≈ 1.8921.
= 2 1−
+
−
+
−
= 2 x−
18 600 35280 0
18 600 35280
sin(x)
dx ≈ 2
x
−1
sin(x)
dx = 2
x
1−
Assuming
R 1 sin(x) that term-by-term integration of the series is legal, the infinite series for
dx has terms of alternating sign and strictly decreasing in absolute value.
−1
x
Therefore, as in ExerciseR 9, the sum of the series is between any two successive
1
partial sums. Therefore −1 sin(x)
dx must be 1.892 to three decimal places.
x
23. We already know that
∞
X xn
x2 x3
+
+··· =
e = 1+x+
2!
3!
n!
n=1
x
for all x. Hence
2
e−x = 1 + (−x2 ) +
∞
∞
X (−x2 )n X (−1)n x2n
(−x2 )2 (−x2 )3
+
+··· =
=
2!
3!
n!
n!
n=0
n=0
for all x. So
−x2
f (x) = xe
∞
X (−1)n x2n+1
x5 x7
−
+··· =
= x−x +
2!
3!
n!
n=0
3
2
for all x. Hence2 the Maclaurin series of f (x) = xe−x is
∞
X
(−1)n x2n+1
n!
n=0
.
24. We already know that
∞
X
1
= 1 − x + x2 − x3 + · · · =
(−1)n xn
1+x
n=0
for |x| < 1. Integrating3 gives
log(1 + x) = x −
∞
X (−1)n xn+1
x2 x3 x4
+
−
+··· =
2
3
4
n+1
n=0
for |x| < 1. At x = −1 the series
∞
X
(−1)n xn+1
n=0
2
3
n+1
=
∞
X
(−1)n (−1)n+1
n=0
n+1
∞
X
1
=−
n
n=0
why is this the Maclaurin series?
why can we integrate term by term?
MAS111
59
Spring 2016-17
which is minus the harmonic series, which diverges. So the radius of convergence
for the Maclaurin4 series of log(1 + x) is 1.
At x = 1 we get
∞
X
(−1)n xn+1
n=0
=
n+1
∞
X
(−1)n
n+1
n=0
which converges (see Exercises 9 or 10). Hence, the Maclaurin series of log(1 + x)
converges at x = 1.
Hence5 the Maclaurin series of f (x) = log(1 + x) is
∞
X
(−1)n x2n+1
n!
n=0
,
the radius of convergence is R = 1, and the interval of convergence is (0, 2].
25. We know that
∞
X xn
x2 x3
e =1+x+
+
+··· =
2!
3!
n!
n=0
x
∞
∞
ex
1 X xn X xn−1
1
x
x2 x3
⇒
=
=
= +1+ +
+
+ ....
x
x n=0 n!
n!
x
2! 3!
4!
n=0
Differentiating6 gives
∞
X
xex − ex
1
1
2x 3x2
1
nxn−1
d ex
=
=
−
+
+
+
+
·
·
·
=
−
+
.
dx x
x2
x2 2!
3!
4!
x2 n=1 (n + 1)!
Setting x = 1 in the above equality gives
0 = −1 +
∞
X
n=1
1
2
3
4
n
⇒
+ + + +··· = 1
(n + 1)!
2! 3! 4! 5!
26. We have
∞
X
1
= 1 + x + x2 + x3 + x4 + · · · =
xn
1−x
n=0
for |x| < 1. Differentiating7 , we get
∞
X
1
2
3
=
1
+
2x
+
3x
+
4x
+
·
·
·
=
nxn−1 .
(1 − x)2
n=1
for |x| < 1. Setting x =
1
2
gives
∞
X
n
=
4=
.
2
n−1
2
1 − 12
n=1
1
4
why
why
6
why
7
why
5
does the power series we wrote down agree with the Maclaurin series?
is this the Maclaurin series?
is this okay?
is this okay?
MAS111
60
Spring 2016-17
Multiplying the left and right hand sides by
∞
Hence, the series
P∞
n
n=1 2n
1
2
gives
∞
X n
1X n
2=
=
.
2 n=1 2n−1
2n
n=1
converges and its exact value is 2.
Chapter 10
1. The Fundamental Theorem of Calculus tells us that
 x

Z
d 
g(t) dt = g(x).
dx
t=0
sin(t3 )
For g(t) = e
we get
f ′ (x) =

d 
dx
Zx
t=0

3
3
esin(t ) dt = esin(x ) .
2. The Fundamental Theorem of Calculus tells us that
 x

Z
d 
g(t) dt = g(x).
dx
t=0
For g(t) =
t2
e
1+t
we get
f ′ (x) =

d 
dx
Zx
t2
2
e
ex
dt =
.
1+t
1+x
t=0
So at x = 0 we have

2
f ′ (0) =
e0
= 1.
1+0
3. The Fundamental Theorem of Calculus tells us that
 x

Z
1
d 
1
dt =
.
3
dx
1+t
1 + x3
t=0
So the chain rule implies that
 g(x)

Z
d 
1
g ′ (x)
=
dt
.
dx
1 + t3
1 + (g(x))3
t=0
3
For g(x) = x we get
f ′ (x) =

d 
dx
Zx3
t=0
MAS111

2
1
 = 3x .
dt
1 + t3
1 + x9
61
Spring 2016-17
4. The Fundamental Theorem of Calculus tells us that
 x

Z
d 
sin(t3 ) dt = sin(x3 ).
dx
t=0
So the product rule for differentiation implies that
 x

 x

Zx
Z
Z
d 
d
g ′ (x) =
x sin(t3 ) dt =
(x)
sin(t3 ) dt + x  sin(t3 ) dt
dx
dx
t=1
=
Zx
t=1
t=0
sin(t3 ) dt + x sin(x3 ).
t=1
So at x = 1 we get
′
g (1) =
Z1
t=1
sin(t3 ) dt + 1 × sin(13 ) = 0 + sin(1) = sin(1).
5. Using the hint we have
f ′ (x) =

d 
dx
Zx
t=0

f (t) dt = f (x)
by the Fundamental Theorem of Calculus. A function equals its derivative if and
only if it is a scalar multiple of the exponential function i.e., f ′ (x) = f (x) if and
only if f (x) = Cex for some constant C. At x = 0 we have
f (0) = 1 +
Z0
f (t) dt = 1.
0
i.e., f (0) = C = 1. Hence the unique function that satisfies the integral identity is
f (x) = ex .
6. The first statement is incorrect because f (x) is not defined at x = 0. We can see
Rb
directly that 0 x12 dx is infinite and that integrating over any interval that contains
zero is therefore infinite (NOT a positive number).
The second statement is incorrect because the Fundamental Theorem of Calculus
is a statement about continuous functions. The function f (x) is not a continuous
function on the interval [−1, 1] (it is NOT continuous at zero), and the conclusion
does not hold (the function is not integrable).
7. Setting x = sin(u) gives dx = cos(u)du; x = 0 gives u = 0, and x = 1 gives u = π2 .
So
Z1
0
√
1
dx =
1 − x2
by substitution.
MAS111
π
π
π
Z2
Z2
Z2
0
1
du =
cos(u) p
1 − sin2 (u)
62
0
cos(u)
1
du =
cos(u)
dx =
π
.
2
0
Spring 2016-17
8. Setting x = sinh(u) ⇒ dx = cosh(u)du, x = 0 ⇒ u = 0, and x = 1 ⇒ u =
arcsinh(1). So
Z1
0
1
√
dx =
1 + x2
arcsinh(1)
Z
=
arcsinh(1)
Z
1
du =
cosh(u) q
1 + sinh2 (u)
0
arcsinh(1)
Z
cosh(u)
1
du
cosh(u)
0
du = arcsinh(1).
0
9. Setting x = tan(u) ⇒ dx = sec2 (u)du, x = 0 ⇒ u = 0, and x = 1 ⇒ u = π4 . So
Z1
√
0
1
dx =
1 + x2
π
π
π
Z4
Z4
Z4
0
1
sec2 (u) p
du =
1 + tan2 (u)
π
4
sec2 (u)
0
= [log | tan(u) + sec(u)|]0 = log(1 +
√
1
du =
sec(u)
sec(u) du
0
2) − log(0 + 1) = log(1 +
√
2).
We can quickly see that our answer is consistent with Question 8 because
log(1 +
√
√
elog(1+
√
√
2)
− e− log(1+ 2)
2) = arcsinh(1) ⇔ sinh(log(1 + 2)) = 1 ⇔
=1
2
√
1 + 2 − 1+1√2
√
√
√
1
√ = 2 ⇔ (1 + 2)2 − 1 = 2(1 + 2)
=1 ⇔ 1+ 2−
⇔
2
1+ 2
√
√
⇔ 2(1 + 2) = 2(1 + 2).
Therefore our answers to Question 8 and 9 are the same.
10. Setting x = cosh(u) ⇒ dx = sinh(u)du, x = 1 ⇒ u = 0, and x = 2 ⇒ u =
arccosh(2). So
Z2
√
1
1
x2
−1
dx =
arcscosh(2)
Z
1
=
arccosh(2)
Z
1
du =
sinh(u) q
cosh2 (u) − 1
arccosh(2)
Z
sinh(u)
1
du
sinh(u)
0
du = arccosh(2).
0
11. We set x = sin(u) so that dx = cos(u)du, x = 0 ⇒ u = 0, and x =
This gives
1
π
Z2
Z6
2
x
sin2 (u)
√
p
cos(u) du.
dx =
1 − x2
1 − sin2 (u)
0
2
1
2
⇒ u = π6 .
0
2
Using 1 − sin (u) = cos (u) we get
1
Z2
0
MAS111
2
π
π
Z6
Z6
π6
1
1
1
u − sin(2u)
sin (u) du =
1 − cos(2u) du =
2
2
2
0
0
0
"
√ #
√ !
1 π 1 3
1 π
3
=
=
.
−
−
2 6 2 2
4 3
2
x
√
dx =
1 − x2
2
63
Spring 2016-17
12. The formula for the arc length of the graph of a function f (x) over an interval [a, b]
is
Zb p
arclength =
1 + (f ′ (x))2 dx.
a
In our case a = 0, b = 1, and
3
f (x) = x 2 ⇒ f ′ (x) =
3√
x.
2
So, the length of our curve is given by
"
3 #1
Z1 r
8
9 2
4 2
9
=
1+ x
.
arc length =
1 + x dx =
4
9 3
4
27
0
0
13
4
32
!
−1 .
13. The formula for the arc length of the graph of a function f (x) over an interval [a, b]
is
Zb p
arclength =
1 + (f ′ (x))2 dx.
a
In our case a = 0, b = 12 , and
f (x) = log(1 − x2 ) ⇒ f ′ (x) =
−2x
.
1 − x2
So, the length of our curve is given by
1
arc length =
Z2
0
=
Z
1
2
0
s
1
1
1+
−2x
1 − x2
(1 + x2
dx =
1 − x2
2
Z
0
1
2
dx =
Z2 s
0
x2 )2
4x2
(1 −
+
2
(1 − x )2
2
−1 +
dx =
1 − x2
Z
0
1
2
−1 +
dx =
Z2 s
0
(1 + x2 )2
dx
(1 − x2 )2
2
dx
(1 − x)(1 + x)
1
2
1
1
1
+
dx = [−x − log |1 − x| + log |1 + x|]02
1−x 1+x
0
1
3
1
1
= − − log
+ log
= log(3) − .
2
2
2
2
=
Z
−1 +
14. The formula for the arc length of the graph of a function f (x) over an interval [a, b]
is
Zb p
1 + (f ′ (x))2 dx.
arclength =
a
In our case a = 0, b = 1, and
f (x) = cosh(x) ⇒ f ′ (x) = sinh(x).
MAS111
64
Spring 2016-17
So the length of our curve is given by
Z1 q
0
2
1 + sinh (x) dx =
Z1 q
2
cosh (x) dx =
0
Z1
0
1
cosh(x) dx = sinh(x) = sinh(1).
0
15. The formula for the arc length of the graph of a function y(x) over an interval [a, b]
is
Zb p
1 + (y ′(x))2 dx.
arclength =
a
In our case a = 21 , b = 1, and
y(x) = x4 +
1
1
⇒ f ′ (x) = 4x3 −
.
2
32x
16x3
So, the length of our curve is given by
s
2
Z1
Z1 r
Z1 r
1
1
1
1
1
1 + 4x3 −
1 + 16x6 − +
16x6 + +
dx =
dx
dx =
3
6
16x
2 256x
2 256x6
1
2
1
2
1
2
=
Z1
1
=
2
s
1−
4x3
1
32
1
+
16x3
−
2
dx =
Z1
1
2
1
1
−
16 8
=
1
1
1
4
4x +
dx = x −
16x3
32x2 1
3
2
33
.
32
16. The formula for the arc length of the graph of a function y(x) over an interval [a, b]
is
Zb p
arclength =
1 + (y ′(x))2 dx.
a
In our case a = 1, b = 2, and
y(x) = x2 −
1
1
log(x) ⇒ f ′ (x) = 2x − .
8
8x
So, the length of our curve is given by
s
2
Z2
Z2 r
Z2 r
1
1
1
1
1
1 + 2x −
1 + 4x2 − +
4x2 + +
dx =
dx
dx =
2
8x
2 64x
2 64x2
1
1
1
s
2
2
2
Z
Z
2
1
1
1
2
2x +
dx = x + log(x) = 3 −
2x +
dx =
=
8x
8x
8
1
1
1
17. The formula for the arclength of the graph of a function f (x) over an interval [a, b]
is
Zb p
1 + (f ′ (x))2 dx.
arclength =
a
MAS111
65
Spring 2016-17
In our case a = π6 , b = π4 , and
f (x) = log(cos(x)) ⇒ −
sin(x)
cos(x)
by the chain rule. So, the length of our curve is given by
π
Z4
π
6
s
π
π
1+ −
sin(x)
cos(x)
2
Z4 s
Z4 s 2
2
cos (x) + sin2 (x)
sin (x)
dx =
dx
dx =
1+
cos2 (x)
cos2 (x)
π
6
π
6
π
=
π
Z4 s
π
6
Z4
1
dx =
cos2 (x)
1
dx.
cos(x)
π
6
Using the hint from Question 9, which tells us that
Z
sec(u) du = log | tan(u) + sec(u)| + C,
we have
π
4
π
6
arclength = [log | tan(u) + sec(u)|] = log |1 +
18. We have
f (x) = − log(cosec(x)) = − log
√
1 + √2 1
2
2| − log √ + √ = log √ 3
3
3 1
sin(x)
= log((sin(x)) .
The formula for the arclength of the graph of a function f (x) over an interval [a, b]
is
Zb p
arclength =
1 + (f ′ (x))2 dx.
a
In our case a =
π
,
4
b=
π
,
3
and
f (x) = log(sin(x)) ⇒
cos(x)
sin(x)
by the chain rule. So, the length of our curve is given by
π
Z3
π
4
s
π
π
1+
cos(x)
sin(x)
2
dx =
Z3 s
1+
π
4
cos2 (x)
dx =
sin2 (x)
π
=
Z3 s
π
4
Z3 s
π
4
sin2 (x) + cos2 (x)
dx
sin2 (x)
π
1
dx =
sin (x)
2
Z3
π
4
1
dx.
sin(x)
This is a standard integral and we can write down the answer, or we can note that
sin(x)
1
sin(x)
=
.
=
2
sin(x)
1 − cos2 (x)
sin (x)
MAS111
66
Spring 2016-17
So, the arc length is given by
π
π
π
Z3
Z3
Z3
π
4
sin(x)
dx =
1 − cos2 (x)
sin(x)
dx =
(1 − cos(x))(1 + cos(x))
π
4
π
4
1
2
1
sin(x)
sin(x)
+ 2
dx.
1 − cos(x) 1 + cos(x)
Using the substitutions u = 1 − cos(x) and u = 1 + cos(x) we get that the arclength
equals
π3
"
1
1 + 2 dx ≥
x
Z∞
s
1
1 − cos(x)
1
log |1 − cos(x)| − log |1 + cos(x)| = log
2
2
1 + cos(x)
π
4
v
v
s
√
u
u
√ √
u1 − 3
u 1 − √12
(2 − 3)( 2 + 1)
2
t
t
√ − log
√ √
= log
= log
.
1 + √12
(2 + 3)( 2 − 1)
1 + 23
arclength =
# π3
π
4
19. The arc length of the curve is given by
Z∞ p
1 + (f ′ (x))2 dx =
Z∞ r
1
1
dx = ∞
1
where the inequality comes from the fact that x12 ≥ 1 for all x > 0 which implies
that
r
1
1 + 2 ≥ 1.
x
Hence the arclength of the graph of f (x) =
1
x
over the positive real axis is infinite.
20. The formula for the arc length of the graph of a function f (x) over an interval [a, b]
is
Zb p
1 + (f ′ (x))2 dx.
arclength =
a
In our case a = 0, b = 1, and
f (x) = x2 ⇒ f ′ (x) = 2x.
So, the length of our curve is given by
arclength =
Z1 √
1 + 4x2 dx.
0
The substitution x =
arcsin(1). i.e.,
arclength =
1
2
arcsin(1)
Z
0
MAS111
sinh(u) gives dx =
1
2
cosh(u), x = 0 ⇒ u = 0, x = 1 ⇒ u =
q
cosh(u) 1 + 4( 12 sinh(u))2 du =
67
arcsin(1)
Z
cosh2 (u)
0
Spring 2016-17
using cosh2 (u) − sinh2 (u) = 1. Using
cosh(u) =
eu + e−u
2
we get
arclength =
arcsin(1)
Z ex + e−x
2
0
2
du =
arcsin(1)
Z
e2u + e−2u + 2
du
4
0
arcsin(1)
Z
arcsin(1)
arcsin(1)
1 1
1 1
=
sinh(2u) + u
sinh(2u) + u
cosh(2u) + 1 du =
2 2
2 2
0
0
0
q
arcsin(1)
1
1
arcsin(1)
2
sinh(u) 1 + sinh (u) + u
=
= [sinh(u) cosh(u) + u]0
2
2
0
1 √
2 + arcsin(1) .
=
2
1
=
2
21. The formula for the area of a surface of revolution obtained by rotating the graph
of a function f (x) over an interval [a, b] is
surface area = 2π
Zb
p
|f (x)| 1 + (f ′ (x))2 dx.
a
In our case a = 0, b = 1,
f (x) =
1
⇒ f ′ (x) = 3x2 , and |f (x)| = f (x) = x3 as x > 0.
x
So, the area of our surface is given by
surface area = 2π
Z1
3
x
0
q
1+
(3x2 )2
dx = 2π
Z1
√
x3 1 + 9x4 dx
0
Setting u = 1 + 9x4 ⇒ du = 36x3 dx, and gives
π
surface area =
18
Z10
1
√
10
π √
π 2 3
2
10 10 − 1 .
u
=
u du =
18 3
27
1
22. The formula for the area of a surface of revolution obtained by rotating the graph
of a function f (x) over an interval [a, b] is
surface area = 2π
Zb
a
p
|f (x)| 1 + (f ′ (x))2 dx.
In our case a = 3, b = 8,
√
1
f (x) = 2 x ⇒ f ′ (x) = √ and |f (x)| = f (x) = cos(x) for x ∈ [3, 8].
x
MAS111
68
Spring 2016-17
So, the area of our surface is given by
surface area = 2π
Z8
√
2 x
3
=
r
1
1 + dx = 4π
x
Z8
√
3
2
x + 1 dx = 4π (x + 1) 2
3
3
8
3
8π
152π
(27 − 8) =
.
3
3
23. The formula for the area of a surface of revolution obtained by rotating the graph
of a function f (x) over an interval [a, b] is
surface area = 2π
Zb
a
p
|f (x)| 1 + (f ′ (x))2 dx.
In our case a = 0, b = 1,
f (x) = cosh(x) ⇒ f ′ (x) = sinh(x), and |f (x)| = f (x) = cosh(x) for all x.
So, the area of our surface is given by
surface area = 2π
Z1
cosh(x)
0
= 2π
π
=
2
Z1
0
Z1
q
1 + sinh (x) dx = 2π
0
cosh2 (x) dx = 2π
Z1
0
2x
e
−2x
+e
0
=
2
Z1
ex + e−x
2
2
cosh(x)
q
dx = 2π
cosh2 (x) dx
Z1
e2x + e−2x + 2
dx
4
0
1
π 1 2 1 −2
π 1 2x 1 −2x
+ 2 dx =
e − e
+ 2x =
e − e +2
2 2
2
2 2
2
0
π sinh(2)
+ π.
2
24. The formula for the area of a surface of revolution obtained by rotating the graph
of a function f (x) over an interval [a, b] is
surface area = 2π
Zb
a
p
|f (x)| 1 + (f ′ (x))2 dx.
In our case a = 0, b = π4 ,
h πi
f (x) = cos(x) ⇒ f ′ (x) = sin(x), and |f (x)| = f (x) = cos(x) for x ∈ 0,
.
4
So, the area of our surface is given by
π
surface area = 2π
Z4
0
MAS111
69
cos(x)
q
1 + sin2 (x) dx.
Spring 2016-17
Setting u = sin(x) ⇒ du = cos(x)dx, and gives
√1
surface area = 2π
Z 2√
1 + u2 du.
0
Setting u = sinh(v) ⇒ du = cosh(v)dv, and gives
arcsinh √1
surface area = 2π
arcsinh √1
2
Z
0
arcsinh √1
2
Z
= 2π
Z
q
cosh(v) 1 + sinh2 (v) dv = 2π
2
cosh2 (v) dv
0
arcsinh √1
1
(cosh(2v) + 1) dv = π
2
Z
0
2
cosh(2v) + 1 dv
0
arcsinh √1
2
1
sinh(2v) + v
2
0
1
1
1
+ arcsinh √ .
sinh 2 arcsinh √
=π
2
2
2
=π
This expression can be simplified by expressing sinh(2v) in terms of sinh(v), but we
leave the answer as shown.
25. The formula for the area of a surface of revolution obtained by rotating the graph
of a function f (x) over an interval [a, b] is
surface area = 2π
Zb
a
p
|f (x)| 1 + (f ′ (x))2 dx.
In our case a = 1, b = 2,
f (x) =
and |f (x)| = f (x) =
x3
12
+
1
x
as
x3 1
x2
1
+
⇒ f ′ (x) =
− 2,
12 x
4
x
x3
12
+
1
x
≥ 0 for x ∈ [1, 2].
So the area of our surface is given by
s
2
4
s
2
2
Z2 3
Z2 4
x
x −4
x
x + 12
1
1
surface area = 2π
+
− 2
dx = 2π
dx
1+
1+
12 x
4
x
12x
4x2
1
1
s
r
2
2
Z2 4
Z
4
8
4
x + 12 x + 8x + 16
x + 12
x4 + 4
= 2π
dx = 2π
dx
12x
16x4
12x
4x2
1
=
π
24
1
Z2
(x4 + 12)(x4 + 4)
π
dx =
3
x
24
Z2
2
48
π x6
24
35
2
x + 16x + 3 dx =
+ 8x − 2 = π.
x
24 6
x 1 16
1
π
=
24
1
MAS111
Z2
x8 16x4 48
+ 3 + 3 dx
x3
x
x
1
5
70
Spring 2016-17
26. The formula for the area of a surface of revolution obtained by rotating the graph
of a function f (x) over an interval [a, b] is
surface area = 2π
Zb
p
|f (x)| 1 + (f ′ (x))2 dx.
a
In our case a = 0, b = 1, and
f (x) = ex ⇒ f ′ (x) = ex , and |f (x)| = f (x) = ex .
So, the area of our surface of revolution is given by
surface area = 2π
Z1
√
ex 1 + e2x dx.
0
Setting u = ex ⇒ du = ex dx, and gives
surface area = 2π
Ze √
1 + u2 du.
1
Setting u = sinh(v) ⇒ du = cosh(v)dv, and gives
surface area = 2π
arcsinh(e)
Z
arcsinh(1)
= 2π
q
cosh(v) 1 + sinh2 (v) dv = 2π
arcsinh(e)
Z
arcsinh(e)
Z
cosh2 (v) dv
arcsinh(1)
ev + e−v
2
arcsinh(1)
2
dv =
π
2
arcsinh(e)
Z
e2v + e−2v + 2 dv
arcsinh(1)
arcsinh(e)
arcsinh(e)
π e2v − e−2v
(ev − e−v ) (ev + e−v ) v
=
+v
=π
+
2
2
2×2
2 arcsinh(1)
arcsinh(1)
arcsinh(e)
q
h
v iarcsinh(e)
v
2
= π sinh(v) cosh(v) +
= π sinh(v) 1 + sinh (v) +
2 arcsinh(1)
2 arcsinh(1)
√
√
1
1
.
=π
1 + e2 + arcsinh(e) − 2 −
2
2
27. The formula for the area of a surface of revolution obtained by rotating the graph
of a function f (x) over an interval [a, b] is
surface area = 2π
Zb
a
p
|f (x)| 1 + (f ′ (x))2 dx.
In our case a = 1, b = ∞, and
f (x) =
MAS111
1
1
1
⇒ f ′ (x) = − 2 , and |f (x)| = f (x) = as x > 0.
x
x
x
71
Spring 2016-17
So, the area of our surface is given by
s
2
Z∞
Z∞
1
1
1
surface area = 2π
dx.
dx ≥ 2π
1+ − 2
x
x
x
1
1
q
2
as 1 + − x12 > 1 for x ≥ 1. Hence the area of our surface of revolution is
surface area ≥ 2π
Z∞
1
dx = 2π [log(x)]∞
1 = ∞.
x
1
28. The equation of the circle is
x2 + (y − 2)2 = 1.
This implies that
√
y = 2 ± 1 − x2
√
√
with y = 2 + 1 − x2 and y = 2 − 1 − x2 corresponding to the upper and lower
semicircles. The total surface area of the doughnut is the sum of the areas of the
surfaces obtained by rotating the upper and lower semicircles about the x-axis.
Thus, using the formula for the area of the surface of revolution, we get
s
Z1
2
√
√
d
2 + 1 − x2
surface area = 2π |2 + 1 − x2 | 1 +
dx
dx
−1
s
Z1
2
√
√
d
2
2
2− 1−x
dx.
+2π |2 − 1 − x | 1 +
dx
−1
√
√
√
Now, 2 ± 1 − x2 > 0 on [−1, 1] so |2 ± 1 − x2 | = 2 ± 1 − x2 in our integrands.
Moreover,
2
2 2
√
√
d
x2
x
d
2
2
2+ 1−x
2− 1−x
=
=
= −√
.
dx
1 − x2
dx
1 − x2
So, the surface area is given by
surface area = 8π
Z1 r
−1
x2
1+
dx = 8π
1 − x2
Z1
−1
√
1
dx.
1 − x2
Setting x = sin(u) ⇒ dx = cos(u)du, and gives
surface area = 8π
π
π
π
Z2
Z2
Z2
− π2
MAS111
cos(u)
p
du = 8π
1 − sin2 (u)
− π2
72
cos(u)
du = 8π
cos(u)
du = 8π 2 .
− π2
Spring 2016-17
29. The formula for the area of a surface of revolution obtained by rotating the graph
of a function f (x) over an interval [a, b] is
surface area = 2π
Zb
a
p
|f (x)| 1 + (f ′ (x))2 dx.
In our case a = 0, b = ∞, and
f (x) = e−x ⇒ f ′ (x) = −e−x , and |f (x)| = f (x) = e−x as f (x) > 0.
So, the area of our surface of revolution is given by
surface area = 2π
Z∞
√
e−x 1 + e−2x dx.
0
Setting u = e−x ⇒ du = −e−x dx, and gives
surface area = −2π
Z0 √
1 + u2 du = 2π
1
Z1 √
1 + u2 du.
0
Setting u = sinh(v) ⇒ du = cosh(v)dv, and gives
surface area = 2π
arcsinh(1)
Z
= 2π
arcsinh(1)
Z
cosh(v)
0
0
q
2
1 + sinh (v) dv = 2π
arcsinh(1)
Z
cosh2 (v) dv
0
arcsinh(1)
1
sinh(2v)
(cosh(2v) + 1) dv = π
+v
2
2
0
= π [sinh(v) cosh(v) + v]arcsinh(1)
0
arcsinh(1)
q
√
2
= π sinh(v) 1 + sinh (v) + v
=π
2 + arcsinh(1) .
0
We finish by showing that a = arcsinh(1) = log(1 +
sinh(a) = 1 ⇔
√
2). Note that
ea − e−a
= 1.
2
We now solve for a:
ea − e−a
1
= 1 ⇔ ea − a = 2 ⇔ e2a − 1 = 2ea ⇔ (ea )2 − 2ea − 1 = 0
2
e
which is a quadratic in ea . i.e.,
√
√
√
2± 4+4
a
= 1 ± 2 ⇒ a = log(1 + 2)
e =
2
as ea > 0. So, the area of the surface of revolution is
√
√
π( 2 + log(1 + 2))
as claimed.
MAS111
73
Spring 2016-17
30. The formula for the volume of a solid of revolution obtained by rotating the area
under the graph of a function f (x) about the x-axis over an interval [a, b] is
volume = π
Zb
(f (x))2 dx.
a
In our case a = 0, b = 4, and so the volume of revolution is
volume = π
Z4
0
2 2
x
dx = π
Z4
x4 dx =
0
π 5 4 1024π
x 0=
.
5
5
31. The formula for the volume of a solid of revolution obtained by rotating the area
under the graph of a function f (x) about the x-axis over an interval [a, b] is
volume = π
Zb
(f (x))2 dx.
a
In our case a = 0, b = 1, and so the volume of revolution is
volume = π
Z1
x 2
(e ) dx = π
Z1
e2x dx =
0
0
π 2x 1 (e2 − 1)π
e 0=
.
2
2
32. The formula for the volume of a solid of revolution obtained by rotating the area
under the graph of a function f (x) about the x-axis over an interval [a, b] is
volume = π
Zb
(f (x))2 dx.
a
In our case a = −1, b = 1, and so the volume of revolution is
volume = π
Z1
cosh2 (x) dx = π
−1
Z1 −1
1
π 1 2x 1 −2x
e − e
+ 2x
=
4 2
2
−1
ex + e−x
2
2
Z1
e2x + e−2x + 2
dx
4
−1
π 2
1
−2
−2
2
=
e −e − e −e +8 = π
sinh(2) + 1 .
8
2
dx = π
33. At x = h we need f (x) = r i.e., mh = r ⇒ m = hr .
The formula for the volume of a solid of revolution obtained by rotating the area
under the graph of a function f (x) about the x-axis over an interval [a, b] is
volume = π
Zb
(f (x))2 dx.
a
In our case a = 0, b = h, and so the volume of revolution is
volume = π
Zh 0
MAS111
Zh
r 2
πr 2 h πr 2 h
πr 2
x dx = 2
x2 dx = 2 x3 0 =
.
h
h
3h
3
0
74
Spring 2016-17
34. The formula for the volume of a solid of revolution obtained by rotating the area
under the graph of a function f (x) about the x-axis over an interval [a, b] is
volume = π
Zb
(f (x))2 dx.
a
In our case a = 1, b = ∞, and so the volume of revolution is
∞
Z∞
1
1
1
=π
lim −
volume = π
+ 1 = π.
dx = π −
x→∞
x2
x 1
x
1
35. The curves intersect when x = x2 i.e., when x = 0 and x = 1. On the interval [0, 1],
the graph of y = x2 is smaller than the graph of y = x. So, the bounded region is
the set
{(x, y) ∈ R2 | 0 ≤ x ≤ 1, x2 ≤ y ≤ x}.
In our case, we calculate the volume of the solid of revolution by finding the volume
of the solid obtained by rotating y = x about the x-axis and subtracting the volume
of the solid obtained by rotating y = x2 about the x-axis.
The formula for the volume of a solid of revolution obtained by rotating the area
under the graph of a function y(x) about the x-axis over an interval [a, b] is
volume = π
Zb
(y(x))2 dx.
a
In our case, a = 0, b = 1, and so the volume of revolution is
volume = π
Z1
0
2
x dx − π
Z1
x2
0
2
dx =
hπ
3
x3 −
π 5 i1 2π
=
x
.
5
15
0
36. The curves intersect when x2 = 1 i.e., when x = ±1. On the interval [−1, 1], the
graph of y = x2 is smaller than the graph of y = 1. So, the bounded region is the
set
{(x, y) ∈ R2 | − 1 ≤ x ≤ 1, x2 ≤ y ≤ 1}.
In our case, we calculate the volume of the solid of revolution by finding the volume
of the solid obtained by rotating y = 1 about the x-axis and subtracting the volume
of the solid obtained by rotating y = x2 about the x-axis.
The formula for the volume of a solid of revolution obtained by rotating the area
under the graph of a function y(x) about the x-axis over an interval [a, b] is
volume = π
Zb
(y(x))2 dx.
a
In our case a = −1, b = 1, and so the volume of revolution is
volume = π
Z1
−1
MAS111
dx − π
Z1
−1
x2
2
dx = 2π −
75
hπ
5
x5
i1
−1
= 2π −
8π
2π
=
.
5
5
Spring 2016-17
37. The equation of the circle is
x2 + (y − 2)2 = 1.
This implies that
√
y = 2 ± 1 − x2
√
√
with y = 2 + 1 − x2 and y = 2 − 1 − x2 corresponding to the upper and lower
semicircles. The total volume of the doughnut is the volume of the solid obtained
by rotating the region below the upper semicircle about the x-axis and subtracting
the volume of the solid obtained by rotating the region below the lower semicircle.
Thus, using the formula for the volume of the solid of revolution, we get
volume = π
Z1 √
1+
−1
x2
2
dx−π
Z1 √
1−
−1
x2
2
dx = 2π
Z1
−1
2π 3
x
x dx =
3
2
1
−1
=
4π
.
3
38. There is no paradox.
Firstly, the amount of paint needed to paint a surface is not the same as the surface
area of the surface; a 3m × 3m wall has surface area 9m2 and requires 9 × ǫ cubic
metres to paint the wall (i.e., the surface area times the uniform (very thin) thickness
ǫ of the layer of paint applied).
As we tend to ∞, the distance from x1 to − x1 gets smaller than any number ǫ
(including the width of the paint required to inside of the horn). So, it is not
possible to paint the inside of Gabriel’s horn with a uniform thickness of paint!
If we paint the outside of the horn then the paint we are using on the outside
intersects the paint inside the horn along a surface which has zero volume. As there
is no containment, we can not expect the latter to be finite.
It is not surprising that Gabriel’s horn has infinite surface area8 , and so we need
an infinite amount of paint (ǫ × (surface area)) to cover the outside of the horn
uniformly9 .
Finally, if we are allowed to vary the thickness of paint, then we can paint the inside
1
with a finite amount of paint (let the vertical thickness of paint be 2x
). Can you
paint the outside of the horn with a finite amount of paint if we are allowed to vary
the thickness of the paint?
Chapter 11
1. We evaluate this integral directly:
Zy=1
Zy=1 2 x=1
Zy=1 Zx=1
1
1 y 2 y=1 1
x dy =
= .
y dy = . y. xy dx dy =
2 x=0
2
2 2 y=0 4
y=0 x=0
y=0
y=0
8
you’ve already shown this in Question 27, but it should be intuitively clear that it should be infinite
(why?).
9
If it seems peculiar to you that it may require a different amount of paint to cover the inside than
the outside, then you should calculate the amount of paint it requires to coat the inside and outside of a
sphere of radius 1 with a layer of thickness ǫ
MAS111
76
Spring 2016-17
2. To do this, we graph the lines x = 0, x = 1, y = x, y = 1 + x in the coordinate
plane, shown below.
x=0
y =1+x
2
y=x
1
x
−2
1
−1
2
x=1
The only region bounded by these lines is highlighted in grey below:
x=0
y =1+x
2
y=x
1
x
−2
1
−1
2
x=1
3. (a) When we integrate first with respect to x, we first find the area of a slice
parallel to the x-axis. The picture below tells us that we are integrating from
the curve described the equation y = x2 to the curve described by the equation
√
y = x. Namely, we are integrating from x = − y (note that the value of x is
negative) to x = y.
y
x=−
√
x=y
1
y
x
1
−1
Now, y varies from 0 to 1. So,
ZZ
f (x, y) dA =
Zy=1 Zx=y
f (x, y) dx dy.
√
y=0 x=− y
A
(b) When we integrate first with respect to y, we first find the area of a slice
parallel to the y-axis. The picture below tells us that we are integrating from
the curve described the equation y = x2 to the curve described by the equation
y = 1 if −1 ≤ x ≤ 0, and from the equation y = x to the curve described by
the equation y = 1 if 0 ≤ x ≤ 1.
MAS111
77
Spring 2016-17
y
y=x
1
y = x3
x
1
−1
So,
ZZ
f (x, y) dA =
Zx=0 Zy=1
x=−1 y=x2
A
Zx=1 Zy=1
f (x, y) dy dx.
f (x, y) dy dx +
x=0 y=x
4. (a) We note that the region D is bounded by the lines x = 1, x = 3, and y = ±2
so that the domain D looks like:
y
x=1
x=2
y=2
2
D
1
x
1
2
−1
y = −2
−2
So, we have
ZZ
D
Zy=2 Zx=3
Zx=3 Zy=2
x2 − y 2 dy dx
x2 − y 2dx dy =
f (x, y)dA =
x=1 y=−2
y=−2 x=1
(b) We note that the region D is bounded by the lines y = −1, y = 2x + 1
), and y = −2x + 1 (⇔ x = 1−y
). The region is shown in the
(⇔ x = y−1
2
2
diagram below:
y
y = 2x + 1
(0, 1)
x
D
(−1, −1)
(1, −1)
y = −1
y = −2x + 1
This means that:
ZZ
f (x, y)dA =
1−y
2
x + y dx dy
y=−1 x= y−1
D
MAS111
Zy=1 x=
Z
78
2
Spring 2016-17
If we integrate first with respect to y, then the upper limits of integration for
x change at x = 0;
ZZ
f (x, y)dA =
Zx=0
Z
Zx=1 y=1−2x
x + y dy dx
x + y dy dx +
y=2x+1
Z
x=−1 y=−1
D
x=0 y=−1
(c) The region D is the first quadrant of the unit circle; the region
√ is bounded by
2
the
plines x = 0, y = 0, and the curve described by y = + 1 − x (⇔ x =
+ 1 − y 2).
The region is shown in the diagram below:
y
1
x2 + y 2 = 1
D
x
1
So,
√
√
2
Zy=1 x= Z 1−y
Zx=1 y=Z 1−x2
xy dx dy.
xy dy dx =
f (x, y)dA =
ZZ
x=0
y=0
y=0
x=0
D
5. We now directly calculate the integrals from Question 4:
(a)
Zy=2 3
Zy=2 Zx=3
Zy=2
x=3
x
x2 − y 2 dx dy =
9 − 3y 2 − ( 31 − y 2)dy
− xy 2 dy =
x=1
3
y=−2
y=−2 x=1
=
Zy=2
y=−2
26
3
− 2y 2dy =
y=−2
Zx=3 Zy=2
x=1 y=−2
26y 2y 3 y=2
= 24.
−
3
3 y=−2
Zx=3
Zx=3
3 y=2
y
x2 − y 2dy dx =
x2 y − dx =
4x2 −
3 y=−2
x=1
16
dx
3
x=1
3
16x x=3
4x
= 24.
−
=
3
3 x=1
(b)
Zy=1 x=
Z
1−y
2
x + y dx dy =
y=−1 x= y−1
MAS111
2
Zy=1
y=−1
Zy=1
x= 1−y
x2
2
2
+ yx y−1 dy =
y − y 2 dy = − .
x= 2
2
3
79
y=−1
Spring 2016-17
Zx=0
y=2x+1
Z
x + y dy dx +
x=−1 y=−1
Zx=0
=
x=−1
Zx=0
=
x=−1
Z
Z y=1−2x
x + y dy dx
y=−1
x=0
y2
xy +
2
y=2x+1
dx +
y=−1
2
4x + 4x dx
x=−1
Z
0dx =
x=0
x=−1
(c)
x=−1
Z y2
xy +
2
x=0
y=1−2x
dx
y=−1
x=−1
Z
4x3
2
2 =− .
+ 2x 3
3
x=0
√ 2
Zy=1 x= Z 1−y
Zy=1 2 x=√1−y2
yx xy dx dy =
dy
2 x=0
y=0
y=0
x=0
=
y=1
2
Zy=1
1
y4
y
2
1
= .
−
y(1 − y )dy = 2
2
4 y=0 8
1
2
y=0
√
Zx=1 y=Z 1−x2
Zx=1 2 y=√1−x2
xy xy dy dx =
dx
2 y=0
x=0
y=0
x=0
=
1
2
x=1
Zx=1
1
1 x2 x4
2
= .
−
x(1 − x )dy =
2 2
4 x=0 8
x=0
RR
6. The area is given by
dA where D is the region bounded by the line y = 1, and
D
the curves described by y = x1/2 , and x = − sin(πy). The three curves intersect at
the point (0, 0), (0, 1), and (1, 1). The domain is shown below:
y
y = x1/2
y=1
(0, 1)
(1, 1)
x = − sin(πy)
D
x
(0, 0)
This gives
area =
Zy=1
x=y
Z 2
y=0 x=− sin(πy)
=
MAS111
3
Zy=1 x=y2
x
dx dy =
x=− sin(πy)
y=0
Zy=1
y 2 + sin(πy)dy
dy =
y=0
cos(πy) y=1 1
1
1
1 2
y
= − (− − ) = + .
−
y=0
3
π
3
π π
3 π
80
Spring 2016-17
7. (a)
ZZ
2
Z1 Z2
2
(x − 3y ) dx dy =
S
=
0
1
Z1
0
2
2
(x − 3y ) dx dy =
7
− 3y 2
3
Z1 0
x=2
x3
2
− 3y x
dy
3
x=1
7
dy = y − y 3
3
1
0
=
7
4
−1= .
3
3
(b)
ZZ
1
dx dy =
(x + y 2)2
Z2 Z1
1
S
1
dy dx =
(x + y 2 )2
0
Z1 Z2
0
1
dx dy.
(x + y 2 )2
1
The second integral is easier.
Z1 Z2
0
Z1 2
Z1 1
1
dy =
−
dy
1 + y2 2 + y2
1
0
0
1
1
1
π
1
−1
−1 y
= tan y − √ tan √
= − √ tan−1 √ .
4
2
2 0
2
2
1
dx dy =
(x + y 2)2
1
1
−
x + y2
8.
ZZ
D
x=1−y
Z1 3
Z1 Z1−y
x
2
2
2
2
2
− 3y x
dy
(x − 3y ) dx dy =
(x − 3y ) dx dy =
3
x=0
0
0
Z1
0
(1 − y)3
2
− 3y (1 − y) dy
=
3
0
1
(1 − y)4
1
3 4
3
= −
=− .
−y + y
12
4
6
0
Now let’s do it the other way round:
ZZ
D
Z1 Z1−x
Z1
2
y=1−x
(x2 − 3y 2 ) dx dy =
(x2 − 3y 2) dy dx =
x y − y 3 y=0 dx
0
=
Z1
0
0
0
x2 (1 − x) − (1 − x)3 dx
x4 x3 (1 − x)4
= − +
+
4
3
4
9.
1
0
1
=− .
6
The region is given in the following picture:
MAS111
81
Spring 2016-17
y=2
(1, 2)
x= y−1
x= 3−y
y=1
(0, 1)
(2, 1)
From the diagram we see that it will be more efficient to make x the inner variable
(between y − 1 and 3 − y) and y the outer one. The other way round, we would
need to split the integral as a sum of two integrals.
ZZ
2x + y + 2
dx dy =
x+y+1
Z2 Z3−y
2x + y + 2
dx dy
x+y+1
1 y−1
R
Z2 Z3−y
2−
=
1 y−1
y
x+y+1
dx dy
(by long division of polynomials in x)
=
Z2
[2x − y ln(x + y + 1)]3−y
y−1 dy
Z2
(8 − 4y − y ln 4 + y ln 2y) dy
Z2
(8 − 4y − 2y ln 2 + y ln 2 + y ln y) dy
1
=
1
=
1
2 Z2
1
1 2
1 2
2
2
= 8y − 2y − (ln 2)y + y ln y −
y dy
2
2
2y
1
1
(integrating by parts)
2
1 2
1 2
1
2
2
= 8y − 2y − (ln 2)y + y ln y − y
2
2
4
1
3
5 + 2 ln 2
3
.
= 8 − 6 − ln 2 + 2 ln 2 − =
2
4
4
10.
The region is given by the following:
MAS111
82
Spring 2016-17
x=1
y=1
D
x=
√
y, i.e., y = x2
√
Z1 Z y
ZZ
3
(3x−x3 )
e
dx dy =
e(3x−x ) dx dy,
0
0
D
where the region D is indicated in the diagram.
So
√
Z1 Z1
Z1
Z1 Z y
3
(3x−x3 )
(3x−x3 )
e
dx dy =
e
dy dx = (1 − x2 )e(3x−x ) dx
0
0
0 x2
=
0
1 (3x−x3 )
e
3
1
0
1
= (e2 − 1).
3
11.
Za Za
0
x
x
p
x2 + y 2
dy dx =
Za Zy
0
0
=
Za
0
x
p
hp
√
x2 + y 2
x2 + y 2
= ( 2 − 1)
iy
0
2 a
y
2
0
dx dy
Z2 √
dx dy = ( 2 − 1)y dy
0
1 √
= ( 2 − 1)a2 .
2
y=a
x=a
x=y
x=0
0
MAS111
83
Spring 2016-17
12.
Z1 Z1
0 y2
√
Z1 Z x √
√
9
y 1 − x6 dx dy =
y 9 1 − x6 dy dx
=
0
0
Z1
0
1 10
y
10
√x
√
0
1 − x6 dx
Z1
1 5√
x 1 − x6 dx
10
0
1
1
1
6 3/2
= .
= − (1 − x )
90
90
0
=
x = y2
y=1
y=
√
x
x=1
y=0
13.
ZZ
√
D
4x3
y
dx dy =
− 3x + 1
Z1
1/2
√ 2 1
Zx − 4
0
Z1 =
1/2
=
Z1
1/2
√
4x3
y
dy dx
− 3x + 1
2
2
√
√
4x3
y
− 3x + 1
x2 −
1
4
2 4x3 − 3x + 1
√x2 − 41
dx
0
dx
√
i1
2
1 h√ 3
.
=
4x − 3x + 1
=
12
12
1/2
(Note that
MAS111
d
(4x3
dx
− 3x + 1) = 12x2 − 3 = 12(x2 − 14 )).
84
Spring 2016-17
x2 − y 2 =
1
4
(y =
q
x2 − 14 )
x=1
14. In this example we express
respect to y.
RR
2
D
ex dA as an iterated integral, integrating first with
y
y=x
(1, 1)
D
(0, 0)
x
(1, 0)
The line between (0, 0) and (1, 1) is described by the line y = x. This means:
ZZ
D
Zx=1
Zx=1
Zx=1 Zy=x
y=x
2
2
2
xex dx.
ex dy dx =
ex y dx =
e dA =
x2
y=0
x=0 y=0
x=0
x=0
To evaluate the last integral we set u = x2 ⇒ du = 2x dx; when x = 0 we have
u = 0, and when x = 1 we have u = 1. This substitution means
ZZ
1
e dA =
2
x2
Zu=1
e−1
eu du =
.
2
u=0
D
Note: it is not possible to integrate first with respect to x in this example.
15. (a) Here is the region of integration:
MAS111
85
Spring 2016-17
x2 + y 2 = 4
x2 + y 2 = 1
y=x
We put x = r cos θ, y = r sin θ, so that dx dy = r dr dθ. We see that r goes from
r = 1 to r = 2, and θ from θ = π/4 to θ = π/2. So the region of integration
becomes a rectangle
R = {(r, θ) | 1 ≤ r ≤ 2, π/4 ≤ θ ≤ π/2},
so the integral becomes
ZZ
θ=π/2
Z Zr=2
r2
2
er r dr dθ
e dA =
θ=π/4 r=1
R
θ=π/2
Z
=
θ=π/4
 r=2

Z
 rer2 dr  dθ
r=1
θ=π/2
Z "
=
θ=π/4
θ=π/2
Z
=
2
er
2
#r=2
dθ
r=1
e4 − e
dθ
2
θ=π/4
=
π(e4 − e)
.
8
(b) We are integrating over the ellipse:
x2
9
+
y2
4
=
π
2
0
MAS111
86
Spring 2016-17
We put x = 3u, y = 2v, so that the region D is transformed to the disk
C = {(u, v) | u2 + v 2 ≤ π/2}. The function becomes sin(u2v 2 ). Finally,
∂(x, y) 3 0
= 6,
=
∂(u, v) 0 2
so that dx dy = 6 du dv. Then the integral becomes
ZZ
ZZ
y2
x2
sin( 9 + 4 ) dA = 6
sin(u2 + v 2 ) dA.
D
C
Now we move to polar coordinates; put u = r cos θ, v = r sin θ, so that du dv =
r dr dθ. The disk C is given by r 2 ≤ π/2, and 0 ≤ θ ≤ 2π. So
ZZ
ZZ
y2
x2
sin(u2 + v 2 ) dA
sin( 9 + 4 ) dA = 6
C
D
=6
√
θ=2π
Z r=Z π/2
sin(r 2 ) r dr dθ
θ=0
=6
=6

√
Z π/2


2

 dθ
r
sin(r
)
dr


θ=2π
Z
θ=0
=6
r=0

r=0
θ=2π
Z θ=0
θ=2π
Z
r=
cos(r 2 )
−
2
r=√π/2
dθ
r=0
1
dθ
2
θ=0
=6×
1
× 2π = 6π.
2
16. The region D is described by the inequalities r < 1, 0 ≤ θ ≤ π. Using the change
of variables x = r cos θ, y = r sin θ, we have x2 + y 2 = r 2 , and we find
ZZ
D
MAS111
Z1 Zπ
r 2 (cos θ + sin θ)2
r dθ dr
1 + r4
0 0
 π

 1
Z
Z
3
r
dr   cos2 θ + sin2 θ + 2 sin θ cos θ dθ
=
1 + r4
0
 0

1 Zπ
1
=
ln(1 + r 4 )  1 + sin(2θ) dθ
4
0
0
π
1
π
1
=
ln 2 θ − cos(2θ) = ln 2.
4
2
4
0
(x + y)2
dx dy =
1 + (x2 + y 2 )2
87
Spring 2016-17
1
1
17. z =
p
R2 − x2 − y 2 , so
s
1+
∂z
∂x
∂z 2
∂x
=√
+
−x
R2 −x2 −y 2
∂z 2
∂y
=
−x
z
and similarly
r
x2 y 2
= 1+ 2 + 2 =
z
z
R
.
=p
2
R − x2 − y 2
r
∂z
∂y
=
−y
.
z
So
z 2 + x2 + y 2
z2
p
The area of the graph of z = R2 − x2 − y 2 lying above D is an eighth of the
surface area of the sphere. Therefore the surface area of the sphere is
8
ZZ
D
Zπ/2ZR
R
r dr dθ
R2 − r 2
0 0
iR
√
πh
= 8 −R R2 − r 2 = 4πR2 .
2
0
R
p
dx dy = 8
R2 − x2 − y 2
θ=
√
π
2
r=R
θ=0
18.
MAS111
R
∂(u, v) 2x
2y = 2(x + y 2)e−x .
=
∂(x, y) −ye−x e−x 88
Spring 2016-17
The region D corresponds to 1 ≤ u ≤ 2, 0 ≤ v ≤ 1. Hence
ZZ
ZZ
∂(u, v) (x + y 2)e−x
1
dx dy
dx dy =
x2 + y 2 + 1
2(x2 + y 2 + 1) ∂(x, y) D
=
D
Z1
0
=
Z1
Z2
1
du dv
2(u + 1)
1
1
[ln(u + 1)]21 dv =
2
0
Z1
1 3
1 3
ln dv = ln .
2 2
2 2
0
y = ex (v = 1)
x2 + y 2 = 2 (u = 2)
y = 0 (v = 0)
x2 + y 2 = 1 (u = 1)
19. The region D corresponds to the region R of the (u, v)-plane described by the
inequalities 1 ≤ u ≤ 2 and 1 ≤ v ≤ 2.
y = x2
y2 = x
y 2 = 2x
2y = x2
∂(u, v) 2x/y −x2 /y 2
= 3.
=
∂(x, y) −y 2 /x2 2y/x MAS111
89
Spring 2016-17
So
∂(x,y)
∂(u,v)
= 13 . Hence
ZZ
ln(xy) dx dy =
D
ZZ
1
ln(uv) du dv
3
R
1
=
3
=
1
3
Z2 Z2
1
1
Z2
[u ln u − u + u ln v]21 dv
Z2
(2 ln 2 + ln v − 1) dv
1
1
=
3
1
=
(ln u + ln v) du dv
2
1
[(2 ln 2 − 1)v + v ln v − v]21 = (2 ln 2 − 1).
3
3
20. The triangle D in the (x, y)-plane corresponds to the triangle T in the (u, v)-plane
bounded by the lines v − u = 1 (a.k.a. y = 1/2), u = 0 (a.k.a. x = y) and v = 2u + 2
(a.k.a. 3y = x + 2).
3y = x + 2
v = 2u + 2
T
D
y=
x=y
1
2
v =u+1
u=0
−1
∂(x,y) 1
So ∂(u,v)
= 2 . Then
MAS111
∂(u, v) 1 −1
=
= 2.
∂(x, y) 1 1 90
Spring 2016-17
ZZ
x2 −y 2
(x − y)(2x − 2y + 1)e
dx dy =
D
ZZ
1
u(2u + 1)euv du dv
2
T
1
=
2
Z0 2u+2
Z
u(2u + 1)euv dv du
−1 1+u
1
=
2
Z0
[(2u + 1)euv ]2u+2
u+1 du
Z0
(2u + 1)(e2(u
−1
1
=
2
2 +u)
− eu
2 +u
) du
−1
0
1 1 2(u2 +u)
u2 +u
=
= 0.
e
−e
2 2
−1
21. The region D corresponds to the region of the (u, v)-plane described by the inequalities 1 ≤ u ≤ 2 and 6 ≤ v ≤ 8.
xy = 2
x2
+
y2
=8
x2 + y 2 = 6
So
MAS111
D
xy = 1
∂(u, v) y 2x
=
= 2(y 2 − x2 ).
x
2y
∂(x, y)
∂(u, v) 2
2
∂(x, y) = 2(x − y ),
91
Spring 2016-17
(since x > y on D). Hence du dv = 2(x2 − y 2 ) dx dy. Then
ZZ
x2 − y 2
1
dx dy =
2
2
2
(x + xy + y )
2
D
ZZ
R
Z8
1
1
du dv =
2
(v + u)
2
Z8 Z2
6
1
du dv
(v + u)2
1
Z8
2
1
−1
−1
1
dv =
dv
+
v+u 1
2
v+2 v+1
6
6
8
1
1
v+1
= (ln(9/10) − ln(7/8))
=
ln
2
v+2 6 2
1
72
36
1
= ln
.
= ln
2
70
2
35
1
=
2
22. We set x(u, v) = u3 and y(u, v) = v2 ; the Jacobian of the transformation (u, v) 7→
(x(u, v), y(u, v)) is 16 . Moreover, the image of the region bounded by 9x2 + 4y 2 = 1
is the unit disc D bounded by u2 + v 2 = 1. This means:
ZZ
1
area =
du dv.
6
D
Changing to polar coordinates, we get
1
area =
6
ZZ
1
r dr dθ =
6
r dr dθ =
π
.
6
θ=0 r=0
D
MAS111
θ=2π
Z Zr=1
92
Spring 2016-17