1
MATH 338
Solution Outlines for Assignment #4
Due on Friday, November 25, 2011
Problem 1: Suppose r is a root of the equation, and therefore satisfies br + c = r2 . Number
of roots on the left hand side is b and half of it would be 2b which if multiplied by itself gives
2
2
� 2b � . The number on the left hand side is c which if added to � 2b � and the square root is
�
2
extracted, gives � 2b � + c. Adding this to half the number of roots, namely 2b , gives the
�
2
familiar expression 2b + � 2b � + c for a root of the polynomial x2 − bx − c = 0.
As is shown in the figure below, if we locate K and N on RH so that KN = N M
and RN =N L, then the area of the rectangles BM RN and KN LT will be equal. Since
Area(ABRH) = c,the area of the region AM N LT KH is also c. As is clear from the picture
that AM LG is a square and
b 2
Area(AM LG) = Area(AM N LT KH)+Area(GHKT ) = Area(ABRH)+Area(GHKT ) = c+� � .
2
�
2
Since Area(AM LG) = (AG)2 , we obtain AG = � 2b � + c and as a result, x = AC = AG+GC =
�
2
� 2b � + c + 2b .
2
Problem 2:
To prove that the intersection points of the two conics xy = d and y 2 + dx − db = 0 satisfy
the cubic equation x3 + d = bx2 is easy! Simply substitute y = xd in the other equation to
2
2
get � xd � + dx − db = 0 which for the case x > 0 and d ≠ 0 can be multiplies by xd to give
x3 + d − bx2 = 0. To determine the condition on d and b for the conics to intersect in zero, one
or two points, one way is as follows: Note that the extremums are attained at x = 2b
3 , 0. Using
2b
the second derivative, it becomes clear that if b, d > 0, then x = 3 gives the local minimum
′′
and x = 0 gives the local maximum (since f ′′ ( 2b
3 ) = 2b > 0 and f (0) = −2b < 0) of the cubic.
If the conics intersect at two points x1 , x2 > 0 then x1 and x2 have to satisfy the original
cubic f (x) = x3 − bx2 + d. This means that the graph of f (x) intersects the positive x-axis
twice. Therefore, the local minimum m = f ( 2b
3 ) has to be negative and the local maximum
3
2
3
� − b � 2b
� + d < 0 ⇒ d < 4b
m = f (0) also has to be negative. In other words, � 2b
3
3
27 . Note that
we always have f (0) = d > 0. As a result, the conics intersect at two points if and only if
3
2b
d < 4b
27 . Similarly, to have one and zero intersection points, we need f ( 3 ) to be zero and
3
3
4b
positive, respectively. This implies d = 4b
27 and d > 27 , respectively. This is the same as
al-Tusi’s analysis.
A more geometric argument can be made as follows: The slope of the tangent line to the
parabola z 2 + dx − db = 0, which we denote by (x, z(x)) as we are only considering the first
d
quadrant, at the point (x0 , z(x0 )) is z ′ (x0 ) = − 2z(x
0)
(To do this, we can differentiate the equation of the parabola to get 2z(x)z ′ (x) + d = 0,
or use the techniques such as Descarte’s method as follows: We have z 2 (x) + dx − db = 0 ⇒
z 2 (x) = −dx + db. As is explained in problem 5 below, To find the slope of the normal line to
the parabola at the point x0 , we form the equality z 2 (x) + ν 2 − 2νx + x2 − n2 = (x − x0 )2 q(x),
where q(x) is a polynomial. In this case, as the left hand side is a second degree polynomial,
we have q(x) = 1. Equating the coefficients of all powers of x to zero, we get ν = x0 + d2 . Thus,
3
2z(x0 )
0)
the slope of the normal is z(x
ν−x0 =
d . The slope of the tangent would be the negative
d
inverse of that of normal’s which would be − 2z(x
.)
0)
Similarly, for the hyperbola, the slope of the tangent line at a point (x0 , y(x0 )) is y ′ (x0 ) =
0)
− y(x
x0 . It is intuitively clear that for any value of b and d, there exists a point x0 > 0(as shown
in the figures below) such that y ′ (x0 ) = z ′ (x0 ) (This can be shown in many different ways, e.g.,
in today language by, putting together the facts that z ′ (x) and y ′ (x) are continuous functions
and limx→∞ y ′ (x) = 0, limx→0+ y ′ (x) = −∞, limx→b− z ′ (x) = −∞, z ′ (0) is finite negative and
then applying intermediate value theorem on a suitable closed interval of (0, ∞) (e.g., [�, b−�]
for small enough �) in order to obtain the existence of the point x0 . Using the fact that
z ′ (x), y ′ (x) are decreasing and increasing respectively, on (0, ∞), we obtain the uniqueness
of x0 for any value of b and d).
Now that we know the existence of uniqueness of x0 , we can easily locate x0 , y(x0 ) and
z(x0 ) for given values of b and d as follows. By our definition of x0 , we have y ′ (x0 ) =
x2
0)
d
z ′ (x0 ), which implies − y(x
= − 2z(x
. Since y(x0 )x0 = d, we obtain z(x0 ) = 20 . Since
x0
0)
z 2 (x0 ) + dx0 − bd = 0, we have
root of the polynomial
x40
4
+ dx0 − bd = 0. This means that x0 has to be the only positive
x4
+ dx − bd,
(1)
4
and f (x) locates the point x0 in (0, ∞). It is easy to show that f (x) has only one positive
root for any positive values of b and d (Since f (x) is increasing being a linear combination
of x4 and x with positive coefficients which are increasing on (0, ∞)). We can even try to
solve this equation using Ferrari’s method by adding 2ax2 + a2 to both sides of the equation
and express x0 explicitly in terms of a, b and d. But, the cubic equation that the parameter
a has to satisfies is as hard to solve as it is for our original cubic(Give it a try!).
As we discussed earlier and is intuitively clear from the figures below, the conics have
zero, one or two intersection points if and only if y(x0 ) > z(x0 ) (when (x0 , y(x0 )) is outside
the parabola), y(x0 ) = z(x0 ) (when (x0 , y(x0 )) is on the parabola) or y(x0 ) < z(x0 ) (when
x2
x2
(x0 , y(x0 )) is inside the parabola) respectively, which are equivalent to xd0 > 20 , xd0 = 20 or
d
x0
<
x20
2 .
f (x) =
This implies the following:
zero intersection point → x0 < (2d) 3 ,
1
one intersection point → x0 = (2d) 3 ,
1
two intersection points → x0 > (2d) 3 .
1
Had we derived x0 earlier, using Ferrari’s method, in terms of b and d, the above inequalities would have provided us with the desired conditions on b and d. But, as we said earlier,
Ferrari’s method needs the cubic for the parameter a to be solved which is no easier than
our original cubic. We can use the fact that x0 is a root of the polynomial f (x) in (1). This
polynomial is increasing on (0, ∞) as was discussed earlier. This means that in the case of
1
1
zero intersection point (in which x0 < (2d) 3 )) we have f (x0 ) < f ((2d) 3 ). Since f (x0 ) = 0, we
have:
4
(2d) 3
4b3
1
1
0 < f ((2d) 3 ) =
+ d(2d) 3 − bd ⇒
< d.
4
27
4
Similarly, in the case of one intersection point (in which x0 = (2d) 3 )), we obtain f (x0 ) =
1
3
f ((2d) 3 )) which, together with f (x0 ) = 0, gives 4b
27 = d. Finally, the case of two intersection
3
points is equivalent to the condition 4b
27 > d.
1
Problem 3: According to Ferrari’s method, we add −2bx2 + b2 to both sides of the
equation and make a complete square on both sides of this equation as follows:
2
2
√
2
2
2
x −2bx +b = (x −b) = 10x −4x−8−2bx +b = � 10 − 2bx − √
� −8+b −� √
� .
10 − 2b
10 − 2b
(2)
Thus, in order to have a complete square on both sides of the above equation we put
4
2
2
2
2
2
b2 − 8 +
2
2
4
= 0 ⇒ b3 − 5b2 − 8b + 38 = 0.
10 − 2b
(3)
Let, bi be a root of the above cubic equation. If 10 − 2bi ≥ 0, which means bi ≤ 5 then (2)
2
implies −8 + b2i − � √10−2b
� = 0 and we can take the square root from both sides of (2) to
i
obtain:
�
�
2
2
x2 − 8 = ±( 10 − 2bi x − √
) ⇒ x2 ∓ 10 − 2bi x − 8 ± √
=0
10 − 2bi
10 − 2bi
�
�
√
√
2
2
± 10 − 2bi + 10 − 2bi − 4(−8 ± √10−2b
)
± 10 − 2bi − 10 − 2bi − 4(−8 ± √10−2b
)
i
i
⇒x1,2 =
, x3,4 =
.
2
2
2
These are all the solutions to the original equation when b = bi . We have the same expression
for other roots of (3) as long as bi ≤ 5
Problem 4: Since in our map (and on the sphere) one centimeter of the great circle
corresponds to one degree the radius of the sphere is 180
π as is stated in the problem. But
the length scales of the circles on the sphere parallel to the equator in nonzero longitudes
φ is different from how they are presented in the map (horizontal lines) by a factor cos φ.
Also, as is stated in the book the distance between the equator and a parallel circle to it
5
in longitude φ is D(φ) times the radius of the sphere as defined in the statement of the
problem and the distance between two such circle in longitudes φ1 and φ2 is D(φ1 ) − D(φ2 )
times the radius of the circle. Thus, in the map, the distance from the equator to the 10th
π
π
parallel circle to the equator which would be in longitude 10 degrees(10 180
= 18
radians),
180
π
is π D( 18 ) ≈ 10.05 centimeters. Similarly, in our map the distance between the equator
π
and the circle parallel to the equator in the longitude φ = 30○ = π6 rad is 180
π D( 6 ) ≈ 31.47
centimeters. Also, The distances between the equator and the parallel circles and longitudes
φ = 5○ , 15○ , 25○ is approximately 5.01, 15.17 and 25.83 respectively.
Problem 5: We first try to find the equation of the normal line at (x0 , x30 ) via the
tangent circle and the slope of the tangent line would be the negative inverse of the slope of
the normal. We first equate (x3 )2 + ν 2 − 2νx + x2 − n2 and (x − x0 )2 q(x). Therefore, q(x) has
to be polynomial of degree four, say x4 + ax3 + bx2 + cx + d. The result would be
(x3 )2 + ν 2 − 2νx + x2 − n2 =x6 + (a − 2x0 )x5 + (x20 − 2x0 a + b)x4 + (ax20 − 2bx0 + c)x3
+(bx20 − 2cx0 + d)x2 + (cx20 − 2dx0 )x + dx20 .
Now we make all the coefficients of xk vanish which results in the following system of equations:
a − 2x0
2
x0 − 2x0 a + b
ax20 − 2bx0 + c
bx20 − 2cx0 + d
cx20 − 2dx0
dx20
=0
=0
=0
=1
= −2ν
= ν 2 − n2 .
Solving the first five equation gives a = 2x0 , b = 3x20 , c = 4x30 , d = 1 + 5x40 and ν = x0 + 3x50 . This
−x30
(x0 )
1
is enough to find the slope of the normal line at (x0 , x20 ) which is 0−f
ν−x0 = x0 +3x50 −x0 = − 3x20 .
Thus, the slope of the tangent line at the point (x0 , x30 ) is 3x20 .