Math 235 - Review for Exam 1
1. Vectors
We think of a vector v as an arrow from the origin (0, 0, 0) to the point (x, y, z). We equate v with (x, y, z)
and often write v = (x, y, z) to indicate the direction of v even though we may apply this vector at some
other point. With vectors
v1 = (x1 , y1 , z1 ) v2 = (x2 , y2 , z2 )
we have the two basic operations:
addition: v1 + v2 = (x1 + x2 , y1 + y2 , z1 + z2 )
scalar multiplication av1 = (ax1 , ay1 , az1 )
If P and Q are points in R3 , the vector Q − P is the vector in the direction of the line segment from P
to Q.
The three standard basis vectors
i = (1, 0, 1), j = (0, 1, 0), k = (0, 0, 1)
are important since every vector v = (x, y, z) in R3 can be written as a linear combination
v = xi + yj + zk
of these three vectors.
2. Lines
If P is a point in R3 and v is a vector, the line passing through P and in the direction v is
`(t) = P + tv.
If P and Q are points in R3 , then the line passing through P and Q is given by
`(t) = P + t(Q − P ).
Here we are applying the definition of a line to the point P and the vector v = Q − P which is the vector
from P to Q.
3. Dot product
For two vectors u = (a1 , b1 , c1 ) and v = (a2 , b2 , c2 ), the dot product of u and v is
u · v = a1 a2 + b1 b2 + c1 c2 .
Note that (a, b, c) · (a, b, c) = a2 + b2 + c2 and is the square of the norm of a vector v = (a, b, c), i.e.,
p
√
kvk = a2 + b2 + c2 = v · v.
The norm kvk represents the length of v and for two points P and Q, the quantity
kQ − P k
is the distance between P and Q.
Using the law of cosines, one shows that for vectors u and v,
u · v = kukkvk cos θ,
where θ is the (acute) angle between u and v.
For two vectors u and v above, the vector
u·v
v
v·v
is the projection of u along v and moreover
u=
u·v
v+
v·v
u−
1
u·v v
v·v
2
can be written as sum of a vector in the direction of v and a vector orthogonal to v.
4. Cross Product
For two vectors u = (a1 , b1 , c1 ) and v = (a2 , b2 , c2 ), the

i
u × v = det a1
a2
vector
j
b1
b2

k
c1 
c2
is the cross product of u and v. The vector u × v turns out to be a vector which is perpendicular to both u
and v.
The triple product (u × v) · w, where w = (a3 , b3 , c3 ) is

a1 b1
det a2 b2
a3 b3
equal to

c1
c2 
c3
and the absolute value of this number the volume of the parallelpiped formed from the vectors u, v, w.
Unlike the dot product where u · v = v · u, the cross product satisfies
u × v = −v × u.
5. Planes
The plane passing through a point P and normal to a vector N is the set of vectors X = (x, y, z) satisfying
(X − P ) · N = 0.
This equation turns out to be an equation of the form
Ax + By + Cz + D = 0
and the vector (A, B, C) turns out to be the normal vector to the plane.
6. Surfaces
Be able to draw surfaces using ‘slices’ in various sorts of planes. There are several matching column
problems with surfaces in the old final exams.
7. Limits
Be able to discuss whether or not limits exist. For a limit to exist, the limit must be the same for all
paths approaching the point. For the limit not to exist, you just need to come up with two different paths
where the limit along these paths is different.
8. Partial derivatives
∂f
f (x0 + h, y0 ) − f (x0 , y0 )
(x0 , y0 ) = lim
h→0
∂x
h
∂f
f (x0 , y0 + k) − f (x0 , y0 )
(x0 , y0 ) = lim
k→0
∂y
k
The above two quantities represent the ‘partial derivatives’ of f with respect to x and y (respectively) and
are the rate of change of f with respect to the appropriate variable.
The plane tangent to f at (x0 , y0 ) is
z = f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )
= f (x0 , y0 ) + ∇f (x0 , y0 ) · (x − x0 , y − y0 ),
where ∇f = (fx , fy ) is the ‘gradient’ of f
3
9. Parametric curves
A curve if a function X : R → R3 , i.e., X(t) = (x(t), y(t), z(t)) traces out a curve in R3 as t runs through
the real numbers. The vector
X 0 (t) = (x0 (t), y 0 (t), z 0 (t))
is a vector tangent to the curve for every t and
`(t) = X(t0 ) + tX 0 (t0 )
is a tangent line to the curve at t0 .
We can think of X(t) is the position vector for a moving object, where t is time. The vector X 0 (t) will
be the velocity vector, kX 0 (t)k is the speed, and X 00 (t) is the acceleration vector. The quantity
Z b
kX 0 (t)kdt
a
is the length of the curve (or the distance traveled) from t = a to t = b.
10. The chain rule
If w = f (x(t), y(t), z(t)), then
dw
∂f dx ∂f dy ∂f dz
=
+
+
dt
∂x dt
∂y dt
∂z dt
If w = f (x(u, v), y(u, v), z(u, v), then
∂w
∂f ∂x ∂f ∂y
∂f ∂z
=
+
+
∂u
∂x ∂u ∂y ∂u
∂z ∂u
∂w
∂f ∂x ∂f ∂y ∂f ∂z
=
+
+
∂v
∂x ∂v
∂y ∂v
∂z ∂v
11. Directional derivatives
If A is a unit vector in R3 , P is a point, and f : R3 → R, then the quantity
∇f (P ) · A
is the rate of change of f at P in the direction A. This is also known as the directional derivative.
Since
∇f (P ) · A = k∇f (P )k cos θ,
where θ is the angle between ∇f (P ) and A (A is a unit vector!), then
∇f (P ) direction where rate of change is the largest
−∇f (P ) direction where rate of change is the smallest.
12. Normal vector to a surface
If F (x, y, z) = 0 is a surface, then
∇F (P )
is a vector normal to this surface at the point P (assumed to be on the surface).
4
13. Some problems from old exams
1. Find the angle between a diagonal of a cube and one of its edges. [Hint: Use coordinates]
Solution. The diagonal is u = (1, 1, 1) and one of the edges is v = (0, 0, 1). Note that cos θ =
u·v
kukkvk
√
1
−1
(1/ 3).
= √ and so θ = cos
3
2. Consider the vectors a = (1, 1, −3), and b = (1, 0, 2). Find each of the following.
(a) A unit vector parallel to a.
(b) a unit vector perpendicular to a .
(c) the projection of a onto b.
1
−3
1
−1
a·b
−5
1
)b =
(1, 0, 2) = (−1, 0, −2)
Solution: (a) ( √ , √ , √ ) (b) ( √ , √ , 0) (other answers are possible) (c) (
11
11
11
2
2
b·b
5
3. Suppose u , v, and w are each vectors in R3 . State whether each expression is meaningful. If not, explain
why. If so, state whether the expression represents a scalar or a vector.
a) (v + w) × u
b) v · w + v × w
c) (u × u) × u
d) u · (v + u)
e) (u · v) × (u · w)
f) u · (v × w)
Solution: a) vector
b) makes no sense
c) vector
d) scalar
e) makes no sense
f) scalar
4. Consider the triangle with vertices A = (1, 2, 3), B = (−1, 4, −5), and C = (2, 2, 2).
(a)Find the lengths of the sides of the triangle.
(b) Find the projection of AB on AC.
(c) Find the angle between BA and BC.
(d) Compute the area of ABC.
(e) Find the equation of the plane containing the triangle ABC.
(f) Find the parametric equation of the line determined by intersecting the plane found in part e. with
the plane −x + 2y − 3z = 4.
√
√
√
Soln: (a) AB = (2, −2, 8); AC = (−1, 0, 1); BC = (3, −2, 7) kABk = 72; kACk = 2; kBCk = 62
AB · AC
6
(b)
AC = (−1, 0, 1) = (−3, 0, 3)
AC · AC
2
√ √
(c) BA · BC = kBAkkBCk cos (θ); 66 = 72 62 cos (θ); θ = cos−1 ( √ 66 ) ∼
= 8.9 degrees
72·62
√ √
1 kAB × BCk = 1 kABkkBCk sin (θ) = 1 72 62 sin (8.9) ∼ 5.2
(d) Area = 2
=
2
2
i
j
k −2
8 = −2i − 10j − 2k. One point on the plane is (1, 2, 3), so the plane is
(e) The normal to this plane is AB × AC, which is 2
−1
0
1 −2(x − 1) − 10(y − 2) − 2(z − 3) = 0
(f) The idea is that the cross product of the normal vectors to each plane will be normal to the normals, and hence will be on both planes.
Thus, this vector points in the direction of the line of intersection.
i
j
k −10
−2 = 34i − 4j − 14k. The point (0, 2, 0) is on both planes,
The line will be in the direction of (−2, −10, −2) × (−1, 2, −3) = −2
−1
2
−3 so x = 34t; y = 2 − 4t; z = −14t is the parametric equation of the line.
−
→
→
→
5. For a non-zero vector −
v , the direction angles α, β, and γ are the angles −
v makes with the vectors i ,
→
−
−
→
→
j , and k respectively. Find the direction angles for the vector −
v = (1, 2, 4).
Solution.
√
→
− − →
−
−1
−
( i ·→
v /k i kk→
v k) = cos
(1/ 21).
√
→
− →
→
−
−1
−
β = cos
(j ·−
v /k j kk→
v k) = cos
(2/ 21).
√
− − →
−
−1 →
−1
−
γ = cos
(k ·→
v /k k kk→
v k) = cos
(4/ 21).
α = cos
−1
−1
−
6. For a general vector →
v = (x, y, z), what does cos2 α + cos2 β + cos2 γ equal? Be sure to justify your
answer.
5
Solution. Note that
cos α =
→
−
→
−
v · i
x
= →
→
−
kvk
k−
vk
cos β =
→
−
→
−
v · k
z
= →
→
−
kvk
k−
vk
→
−
→
−
v · j
y
= →
→
−
kvk
k−
vk
cos γ =
x2 + y 2 + z 2
x2 + y 2 + z 2
Thus
2
2
2
cos α + cos β + cos γ =
kvk2
=
x2 + y 2 + z 2
= 1.
−
→
→
−
−
−
7. If the vectors →
a and b are parallel, what is →
a × b equal to? (Hint: Look at the formula for the cross
product.)
→
−
→
−
−
−
−
Solution. Suppose →
a = (a1 , a2 , a3 ). Since b is parallel to →
a , then b = c→
a = (ca1 , ca2 , ca3 ). From this we get
→
→
− 
−
→
−
i
j
k
→
−
→
−
→
−
→
−
→
−
→
−
→
−
a1
a2
a1
a3
a2
a3
→
−
− j det
+ k det
= 0 i +0j +0k
a × b = det  a1
a2
a3  = i det ca
ca1
ca2
ca1
ca3
ca3
2
ca1
ca2
ca3
8. Consider the triangle formed from the points A = (2, 0, 0), B = (0, 0, 6), C = (0, 5, 0). Find the projection
of the vector AB along the vector AC.
Solution. P1 = B − A = (−2, 0, 6), P2 = C − A = (−2, 5, 0). The projection of P1 along P2 is
P1 · P2
P2 · P2
P2 = (
−8 20
,
, 0).
29 29
9. For the points in the previous problem, find the angle between AB and AC.
Solution.
cos θ =
P1 · P2
kP1 kkP2 k
4
√
40 29
= √
which makes θ ≈ 1.4530 radians.
10. Compute the area of the triangle with vertices
A = (2, −3, 1) B = (1, −1, 2) C = (−1, 2, 3).
Solution: Let v1 = B − A = (−1, 2, 1) and v2 = C − A = (−3, 5, 2). The area of ABC is
1
2
kv1 × v2 k =
1
2
k(−1, −1, 1)k =
1√
2
3.
11. What is the equation of the plane which contains the triangle in part (a)?
Solution: The normal vector to this plane is N = v1 × v2 = (−1, −1, 1) and a point on the plane is P = (2, −3, 1). The equation of the plane is
then
((x, y, z) − P ) · N = 0
which turns out to be −x − y + z = 2.
12. Let P1 = (1, 0, 2), P2 = (2, 1, 3), and P3 = (1, 5, 4) be the vertices of a triangle.
(a) Find the lengths of the sides of this triangle.
(b) Compute the area of this triangle.
Solution.
(a)
`(P1 , P2 ) = kP2 − P1 k =
q
√
(2 − 1)2 + (1 − 0)2 + (3 − 2)2 = 3.
The other ones are very similar.
(b) Let A = P2 − P1 = (1, 1, 1) (the segment from P1 to P2 ) and B = P3 − P1 = (0, 5, 2) (the segment from P1 to P3 ). The area of the
triangle is then
1
1
1√
kA × Bk = k(−3, −2, 5)k =
38.
2
2
2
13. The plane passing through the points P1 , P2 , and P3 (as in the previous problem) intersects the plane
x − y − z = 2 in a line. Find the parametric form of this line.
6
Solution. The plane in the previous problem has normal direction (−3, −2, 5) and x − y − z = 2 has normal direction (1, −1, −1). The desired
line will have direction vector (−3, −2, 5) × (1, −1, −1) = (7, 2, 5). One checks that the point (5, −1, 4) lies on both planes and so the desired line
is
`(t) = (5, −1, 4) + t(7, 2, 5).
14. Find the equation of the plane which
(a) is perpendicular to the line
`(t) = (0, 7, 1) + t(−1, −2, 3)
and passes through the point (2, 4, −1).
(b) passes through the point (3, 2, −1) and contains the line
`(t) = (1, −1, 0) + t(3, 2, −2).
Solution. (a) A normal vector for this plane is the direction vector for the normal line which is (−1, −2, 3). The plane is then
((x, y, z) − (2, 4, −1)) · (−1, −2, 3) = 0.
(b) Three points in the plane are A = (3, 2, −1), B = `(0) = (1, −1, 0), and C = `(1) = (4, 1, −2). The vectors v1 = B − A = (−2, −3, 1) and
v2 = C − A = (1, −1, −1) are in the direction of the plane and so N = v1 × v2 = (4, −1, 5) is normal to the plane. Thus the plane is
((x, y, z) − (3, 2, −1)) · (4, −1, 5) = 0.
15. Find the equation of the plane consisting of all points (x, y, z) which are equidistant to the points
(−4, 2, 1) and (2, −4, 3).
Soln: We are looking for the set of points (x, y, z) so that
q
q
(x − 4)2 + (y − 2)2 + (z − 1)2 = (x − 2)2 + (y + 4)2 + (z − 3)2 .
Square both sides and multiply it all out. The squares will go away and you will be left with 12x−12y+4z = 8. Another way to solve this is by letting
P = (−4, 2, 1) and Q = (2, −4, 3). Geometric considerations say that Q − P = (6, −6, 2) is normal to the plane and M = (P + Q)/2 = (1, −1, 2)
is on the plane. Thus ((x, y, z) − (1, −1, 2)) · (6, −6, 2) = 0 is the equation of this plane.
16. The following planes
x + 2y + 2z = 4
2x + y − 3z = −5
intersect in a line. Find the parametric representation of that line.
Solution. The normal N1 to the first line is N1 = (1, 2, 2) and the normal N2 to the second is N2 = (2, 1, −3). A direction vector for the line is
V = N1 × N2 = (−8, 7, −3). Notice that the point P = (−2, 2, 1) lies on both the planes and so in on our line. Thus out line is
`(t) = (−2, 2, 1) + t(−8, 7, −3).
17. Find a unit vector which is parallel to both the planes
2x − y − z = 4 and x + y − 4z = 1.
Solution. The normals to the planes are N1 = (2, −1, −1) and N2 = (1, 1, −4). A little geometry says that a vector which is parallel to both the
planes is normal to both N1 and N2 . Such a vector is N1 × N2 = (5, 7, 3). The desired vector must be a unit vector which is
N 1 × N2
kN1 × N2 k
=
1
83
(5, 7, 3).
18. Find an equation for the plane that
(a) contains the points (1, 0, 1), (2, 1, 1), and (3, −1, 1).
(b) is perpendicular to the line l(t) = (1, 1, 0) + t(2, 1, 7), and contains the point (0, 0, 0).
(c) is tangent to the surface xz + yx − 1 = eyz at the point (2, 1, 0)
Solution: (a) z = 1 (notice that all the z-coordinates are equal to 1)
(b) 2x + y + 7z = 0
(c) Let g(x, y, z) = xz + yz − 1 − eyz = 0, then ∇g(x, y, z) = (z, x − zeyz , z + y − yeyz ) so ∇g(2, 1, 0) = (0, 0, 1) and the tangent plane is
z = 1.
7
19. Find the equation of the plane which contains the two parallel lines
`1 (t) = (1, 1, 1) + t(2, −1, 4) and `2 (t) = (0, 1, 0) + t(2, −1, 4).
Solution. Three points in the plane are
P1 = `1 (0) = (1, 1, 1)
P2 = `1 (1) = (3, 0, 5),
P3 = `2 (0) = (0, 1, 0).
So the problem is reduced to finding the plane containing these three points. Let
→
−
a = P2 − P1 = (2, −1, 4)
→
−
b = P3 − P1 = (−1, 0, −1).
The normal to this plane is N = a × b = (1, −2, −1). The desired plane is then
((x, y, z) − (1, 1, 1)) · (1, −2, −1) = 0.
20. Find the equation of the plane which contains the line `1 (t) = (2 − 3t, 2 + t, 2 + 2t) and is parallel to the
line `2 (t) = (1 + 2t, 1 − t, −1 + 4t).
Solution: The vector (−3, 1, 2) is the direction vector for `1 (and is parallel to the plane) and the vector (2, −1, 4) is the direction vector for `2
(and is also parallel to the plane). Thus a normal to this plane is (−3, 1, 2) × (2, −1, 4) = (6, 16, 1). A point on this plane is `1 (0) = (2, 2, 2). Thus
the equation of this plane is
((x, y, z) − (2, 2, 2)) · (6, 16, 1) = 0.
21. For an triangle with vertices A, B, C, the center of gravity is defined to be
1
G = (A + B + C).
3
Use vectors to show that the center of gravity of the triangle ABC is the same for the triangle whose vertices
are the midpoints of the sides of ABC.
Solution: The midpoints of the three sides are M1 = (A + B)/2, M2 = (B + C)/2, M3 = (A + C)/2. The center of gravity G0 of this new triangle
is
1
0
G = (M1 + M2 + M3 )
3
which turns out to be equal to
1
(A + B + C) = G.
3
22. Draw the curve X(t) = (a cos t, a sin t, t).
Solution: This turns out to be a helix with “radius” equal to a.
23. The helix X(t) = (cos t, sin t, t) intersects the curve Y (t) = (1 + t, t2 , t) at the point (1, 0, 0). Find the
angle of intersection of these curves.
Soln: The angle of intersection will be the angle between the tangent vectors. First notice that X(0) = Y (0) = (1, 0, 0). The two tangent vectors
are X 0 (0) = (0, 1, 1) and Y 0 (0) = (1, 0, 1). The angle between them is
cos
−1
(
X 0 (0) · Y 0 (0)
kX 0 (0)kkY 0 (0)k
= cos
−1
1
π
.
√ =
2
4
24. Show that the angle the tangent to X(t) makes with the z-axis is constant in t.
Solution: The tangent to X(t) is X 0 (t) = (−a sin t, a cos t, 1). The angle this vector makes with the z-axis (0, 0, 1) is then
arccos
X 0 (t) · (0, 0, 1)
kX 0 (t)kk(0, 0, 1)k
1
= arccos p
.
a2 + 1
→
−
→
−
25. The angular momentum L (t) and torque −
τ (t) of a moving particle of mass m and position vector →
r (t)
are given by the formulas
→
−
→
→
→
→
→
L (t) = m−
r (t) × −
v (t) −
τ (t) = m−
r (t) × −
a (t).
−
What is L0 (t) in terms of →
τ (t)? (You may assume that the usual product rule (from calculus I) works
for the cross product.)
8
Solution. L0 = mr × v 0 + mr 0 × v = mr × a + mv × v = τ + 0 = τ.
26. Let a particle moves in the plane so that its position vector at time t is
C(t) = (et cos t, et sin t).
Show that the angle between the position vector and the velocity vector is constant and compute this
constant.
Solution. A computation shows that the velocity vector
0
t
t
t
t
C (t) = (e cos t − e sin t, e cos t + e sin t)
√
and so after another computation and a trig formula C 0 (t) · C(t) = e2t . Also note that kC(t)k = et and kC 0 (t)k = 2et . Thus
cos θ =
C(t) · C 0 (t)
1
= √
2
kC(t)kkC 0 (t)k
and so θ = π/ which is constant in t.
27. Commander Skywalker’s ship moves along the path X(t) = (et cos t, et sin t), 0 ≤ t < ∞.
(a) Find the distance he travels for 0 ≤ t ≤ 1.
(b) At time t = 2π, Commander Skywalker turns off his engines (causing him to fly in a line). Find the
equation of that line.
(c) The planet Xardon is located at the point (e2π + 20e2π , 20e2π ). Will Commander Skywalker (with his
engines off) make it to Xardon?
Solution. (a)
Z 1
0
Z 1
kX (t)kdt =
dist =
t
t
t
t
k(−e sin t + e cos t, e cos t + e sin t)kdt
0
0
A computation shows this is equal to
Z 1√
√
t
2e dt = 2(e − 1).
0
(b) Point on the (tangent) line is X(2π) = (e2π , 0) and the direction vector is the tangent vector X 0 (2π) = (e2π , e2π ). The tangent line is
then
2π
`(t) = (e
2π
, 0) + t(e
2π
,e
).
(c) Notice that `(20) is the exact location of Xardon. He will make it! May the force be with you.
28. Sketch a graph of x + y 2 − z 2 = 0, explaining the different steps you take in making this drawing.
Soln: I want to see level curves for some choice of variables. If you choose level curves for z, then the level curves will be parabolas. If you choose
level curves for x, then the level curves will be hyperbolas. In any case, you should compute at least 4 level curves to get the idea of what is going
on, then try to graph those on the xyz axes. The pictures are very difficult to draw, but I wanted to see your approach.
29. Draw x2 − y 2 − z 2 = 1
Solution: The surface intersects the plane z = 0 in the curve x2 − y 2 = 1 which is a hyperbola opening the x-axis with vertex x = 1, x = −1.
The surface intersects the plane y = 0 in the curve x2 − z 2 = which is a hyperbola opening along x-axis with vertex x = 1, x = −1.
The surface intersects the planes x = c > 1 in the circles y 2 + z 2 = c2 − 1 and so the resulting surface is a hyperboloid of two sheets opening
along the x-axis with vertex at x = 1, x = −1.
30. Draw the surface y 2 − x2 − z 2 = 1. Explain in detail what you are doing.
Solution. In the xy plane (z = 0) the trace is y 2 − x2 = 1 which is a hyperbola with its vertices at y = ±1. The trace in the yz plane (x = 0) is
y 2 − z 2 = 1 which is also a hyperbola with its vertices at y = ±1. In the planes y = ±1, ±2, ±3, ... the traces are circles. In total, the surface is a
hyperboloid of two sheets.
31. Draw the graph of x2 − y 2 + z 2 = −1. Be sure to explain what you are doing.
Solution: In the xy-plane, the surface traces out the curve y 2 − x2 = 1 which is a hyperbola with vertex at y = ±1 and opens along the y axis.
In the yz-plane, the surface traces out the curve y 2 − z 2 = 1 which is hyperbola with vertex at y = ±1 and opening along the y-axis. One can
check that in the planes y = ±2, ±3, ±4, ..., the trace is a circle. Thus the surface is a hyperboloid of 2 sheets with vertex at y = ±1 and opening
along the y-axis.
9
32. (a) Sketch the region bounded by the surfaces
p
z = x2 + y 2 and z = 2 − x2 − y 2 .
(b) Show that the set of points X = (x, y, z) such that (X − (1, 2, 3)) · (X − (4, 5, 6)) = 0 is a sphere and
find its center and radius.
Soln: (a) The first surface is a cone with its vertex at the origin. The second surface is a paraboloid (upside down) with its pear at (0, 0, 2).
Together, they form an ‘ice cream’ cone region.
(b) The above equation gives us
(x − 1)(x − 4) + (y − 2)(y − 5) + (z − 3)(z − 6) = 0.
Multiply all this out to get
2
2
2
(x − 5x + 4) + (y − 7y + 10) + (z − 9z + 18) = 0.
Complete the square three times to get
2
2
(x − 5/2) + (y − 7/2) + (z − 9/2)
2
= 27/4.
p
27/4.
This is a sphere whose center is (5/2, 7/2, 9/2) and radius is
33. Draw the surface x2 − y 2 = −1. Be sure to explain what you are doing.
Solution: In the xy-plane, the trace is y 2 − x2 = 1 which is a hyperbola with vertex at y = ±1 and opens along the y axis. Since there is no z in
the formula, the trace will be the same in every plane z = c and so the surface will be a ‘wall’ constructed by extending the hyperbola y 2 − x2 = 1
in the various planes z = c.
34. At what point does the tangent plane to the surface z = x2 + 2y 3 at (1, 1, 3) meet the z-axis?
Soln: The tangent plane is z − 3 = 2(x − 1) + 6(y − 1). When x = y = 0, we get z = −5, and that is the point where the tangent plane meets the
z-axis.
35. Compute the following limit or explain why it does not exist.
sin(xy)
(x,y)→(0,0) x2 + y 2
lim
Solution. This limit does not exist. Along the line y = 0 (the x-axis), the limit is clearly equal to zero. Along the line y = x, the limit is
sin(x2 )
lim
=
2x2
x→0
1
2
.
Note the last equality follows from L’Hospital’s rule. Since the limit is two different values from two distinct paths, the limit fails to exist.
36. Compute the following limits (or explain why they don’t exist).
(a)
lim
(x,y)→(0,0)
(b)
lim
(x,y)→(0,0)
sin(x + 2y)
.
x + 2y
sin(2x) − 2x + y
.
x3 + y
Solution: (a) Let u = x + 2y and notice that as (x, y) → (0, 0), u → 0. Thus the above limit is equal to
lim
sin u
u→0
u
= 1.
(b) Along the path x = 0 (the y-axis), the above limit is equal to
lim
y→0
y
y
= 1.
Along the path y = 0 (the x-axis), the limit is
lim
x→0
sin(2x) − 2x
x3
which, after several applications of L’Hopital’s rule, is equal to −4/3. Since the limits along two different paths are different, the limit does not
exist.
10
37. Compute the following limit (or explain why it does not exist).
x4 − y 4
(x,y)→(0,0) x2 + y 2
lim
Solution:
lim
(x,y)→(0,0)
x4 − y 4
x2 + y 2
=
(x2 − y 2 )(x2 + y 2 )
lim
x2 + y 2
(x,y)→(0,0)
2
38. Find a normal to the surface z = f (x, y) = e−x
−y 2
=
lim
2
x −y
2
=0
(x,y)→(0,0)
at the point (1, 1).
2
2
Solution: If F = e−x −y − z = 0, then point on this surface is (1, 1, e−2 ). The normal at this point is
−x2 −y 2
∇F = (−2xe
−2
evaluated at (1, 1, e
−2
). A calculation shows this equals (−2e
−2
, −2e
−x2 −y 2
, −2ye
, −1)
, −1).
39. Explain why the direction of the gradient ∇f is the direction along which f is increasing the fastest.
Soln: If we take unit vectors u to determine the directional derivative, then the largest possible value of ∇f · u = k∇f kkuk cos (θ) = k∇f k cos (θ)
is when θ = 0 (since cos is maximized at 1 when the angle is 0). This implies that the direction of maximal increase is in the same direction as
the gradient as claimed.
40. What the largest the directional derivative of f (from the previous problem) can be at the point (1, 1)?
Solution: The largest directional derivative is
−2
k∇f (1, 1)k = k(−2e
−2
, −2e
)k =
√ −2
8e
.
1
41. Consider the function f (x, y, z) = p
. At the point (1, 2, 2) compute the directional deriva+ y2 + z2
tive in the direction in which f decreases most rapidly.
x2
Solution. f decreases the most in the direction −∇f (1, 2, 2). A computation shows that
∇f = −
1
(x2 + y 2 + z 2 )3/2
(x, y, z)
and so - ∇f (1, 2, 2) = (1/27, 2/27, 2/27). This is the direction in which f decreases most rapidly. The actual directional derivative is −k∇f k =
−1/9.
42. Find the maximum rate of change of f (x, y, z) = log(xy 2 z 3 ) at (1, −2, 3) and the direction it occurs.
Soln: The direction the maximum rate of change takes place is ∇f (1, −2, 3). A computation shows that ∇f = (1/x, 2/y, 3/z) and so ∇f (1, −2, 3) =
√
(1, −1, 1). The maximum rate of change is k∇f (1, −2, 3)k = 3.
43. Let f (x, y) = 1 + y + xexy . Compute the directional derivative of f at (0, 0) in the direction (2, −4).
√
Solution. The direction vector is A = (2, −4) but this is, unfortunately, not unit. The unit direction vector is A/kAk = (2, −4)/ 20. The
gradient turns out to be
xy
xy
2 xy
∇f = (xye
+ e ,1 + x e )
which, evaluated at (0, 0) is ∇f (0, 0) = (1, 1). The directional derivative is thus
1
−2
∇f (1, 1) · A/kAk = (1, 1) · √ (2, −4) = √ .
20
20
44. Show that if (a, b, c) is a point on the surface xyz = 1 (i.e., abc = 1), the product of the x-,y-, and
z-intercepts of the tangent plane to this surface at (a, b, c) always comes out to be the same number. What
is this number?
Soln: Let G = xyz − 1 = 0. The normal to this surface at (a, b, c) is ∇G(a, b, c). A computation shows that ∇G = (yz, xz, xy) and so
∇G(a, b, c) = (bc, ac, ab). The tangent plane is
((x, y, z) − (a, b, c)) · (bc, ac, ab) = 0.
Multiplying all this out and using the fact that abc = 1, we get
xbc + yac + zab = 3.
The x-intercept is 3/bc, the y-intercept is 3/ac, and the z-intercept is 3/ab. The product is (using abc = 1 one more time!) 27.
11
45. Compute the tangent plane to the surface z = sin(3x + y 2 ) at x = 0, y =
p
π/2.
Solution. Let F (x, y, z) = sin(3x + y 2 ) − z and note that
2
2
∇F = (3 cos(3x + y ), 2y cos(3x + y ), −1)
p
and so ∇F (0, π/2, 1) = (0, 0, −1). This vector is normal to our surface and this is the normal to the desired tangent plane. Our plane is
q
((x, y, z) − (0, π/2, 1)) · (0, 0, −1) = 0
which reduces to just z = 1.
46. Find the tangent plane and normal line to z =
x
at the point x = 2 and y = 1.
y
Solution. Let F (x, y, z) = x/y − z and notice that ∇F (x, y, z) = (1/y, −x/y 2 , −1). The vector ∇F (2, 1, 2) = (1, −2, −1) is normal to the surface
at (2, 1, 2) and so the tangent plane is
((x, y, z) − (2, 1, 2)) · (1, −2, −1) = 0
and the normal line is
`(t) = (2, 1, 2) + t(1, −2, −1).
47. At what point does the tangent plane to the surface
exy
1
=
1 + z2
2
at (0, 2, 1) meet the z-axis?
Solution: First let’s find the equation of the tangent plane to this surface at (0, 2, 1). Let
F (x, y, z) =
exy
1 + z2
−
1
2
and note that ∇F (0, 2, 1) is a normal to this surface at (0, 2, 1). A computation shows that
∇F (x, y, z) = (
yexy
1 + z2
,
xexy
1 + z2
,
−2zeey
(1 + z 2 )2
)
and so ∇F (0, 2, 1) = (1, 0, −1/2). The tangent plane is then
((x, y, z) − (0, 2, 1)) · (1, 0, −1/2) = 0
or equivalently 2x − z + 1 = 0. This meets the z-axis when x = y = 0 which occurs when z = 1. Thus the tangent plane meets the surface at the
point (0, 0, 1).
48. Suppose that a duck is swimming in the circle x = cos (t), y = sin (t) and that the water temperature is
given by the formula T = xy 3 + yx3 . Find dT
dt , the rate of change in the temperature the duck might feel:
(a) by the chain rule; (b) by expressing T in terms of t and differentiating.
Soln: We first find dT
by writing the formula T = cos (t) sin (t)(sin2 (t) + cos2 (t)) = cos (t) sin (t) and differentiating, yielding dT
= − sin2 (t) +
dt
dt
cos2 (t).
The second approach is to calculate the gradient of f , ∇f = (y 3 + 3x2 y, 3xy 2 + x3 ), evaluate that at (cos (t), sin (t)), yielding (sin3 (t) +
3 cos2 (t) sin (t), cos3 (t) + 3 sin2 (t) cos (t)), and dot that with the derivative of the path, namely (− sin (t), cos (t)) using the chain rule. This gives
the same answer as the first part.
49. Suppose f is a differentiable function of x and y and
g(u, v) = f (eu + sin v, eu + cos v).
Use the table below to compute gu (0, 0).
(0, 0)
(1, 2)
f
3
6
g
6
3
fx
4
2
fy
8
5
Soln: By the chain rule, gu = fx xu + fy yu . When u = v = 0, then x = 1 and y = 2. Thus
gu (0, 0) = fx (1, 2) · 1 + fy (1, 2) · 1 = 2 + 5 = 7.
12
50. For the function F (x, y) = xy and x = x(t), y = y(t), one is given the following information: x(1) =
1/2, x0 (1) = 1/3, y(1) = 2, y 0 (1) = −3. Compute dF/dt at t = 1.
Solution. By the chain rule,
Ft = Fx xt + Fy yt = yxt + xyt
which at t = 1 becomes 2(1/3) + 1/2(−3) = −5/6.
51. Given h(x, y, z) = f (x2 + y, zx, zy) and ∇f (1, 0, 0) = (3, 1, −1) compute
Solution:
∂h
∂x
∂h
∂x
at the point (0, 1, 0).
= ∇f (1, 0, 0) · (2x, z, 0)|(0,1,0) = (3, 1, −1) · (0, 0, 0) = 0.
52. Let z = f (x, y), where fx (4, 4) = 7, fy (4, 4) = 9, x = 2e3t + t2 − t + 2, y = 5e3t + 3t − 1. Find dz/dt for
t = 0.
Solution: By the chain rule dz/dt = fx x0 (t) + fy y 0 (t). Notice that x(0) = y(0) = 4 and so when t = 0 we get
dz/dt|t=0 = fx (4, 4) · 5 + fy (4, 4) · 18 = 7 · 5 + 9 · 18 = 197.
53. If z = xy 3 + x2 y 2 + f (x2 , y 2 ), what is zx (0, 2)?
Solution. Let u = x2 , v = y 2 . By the chain rule,
3
2
zx = y + 2xy + fu ux + fv vx
and so when x = 0 and y = 2 we get
zx (0, 2) = 8 + 0 + fu (0, 4)0 + fv (0, 4)0 = 8.
54. Let u = xyf (1/x + 1/y). Find a function G(x, y) such that x2 ux − y 2 uy = G(x, y)u.
Solution:
0
ux = xyf (1/x + 1/y)(
0
uy = xyf (1/x + 1/y)(
2
2
2
−1
x2
−1
y2
) + yf (1/x + 1/y)
) + xf (1/x + 1/y).
2
x ux − y uy = x yf (1/x + 1/y) − y xf (1/x + 1/y) = xyf (1/x + 1/y)(x − y).
Thus the desired function G(x, y) = x − y.
55. Let w = f (y − x − t, z − y + t). Show that
∂w
∂w ∂w ∂w
+2
+
+
= 0.
∂x
∂y
∂z
∂t
Solution. Let u = y − x − t and v = z − y + t. By the chain rule we get
wx = wu ux + wv vx = wu (−1) + wv (0) = −wu
wy = wu uy + wv vy = wu (1) + wv (−1) = wu − wv
wz = wu uz + wv vz = wu (0) + wv (1) = wv
wt = wu ut + wv vt = wu (−1) + wv (1) = wv − wu
Now use the above to show
wx + 2wy + wz + wt = 0.
56. Let w = f (x, y) and x = 2u + v and y = u − v. Show that
5
∂2w
∂2w
∂2w ∂2w
∂2w
+2
+2 2 =
+
.
2
∂x
∂x∂y
∂y
∂u2
∂v 2
13
Solution. A calculation using the chain rules yields
wu = wx xu + wy yu = 2wx + wy
wuu
=
(2wx )u + (wy )u
=
(wxx xu + wxy yu ) + (wyx xu + wyy yu
=
4wxx + 4wxy + wyy
In a similar way,
wv = wx xv + wy yv = wx − wy
wvv
=
(wx )v − (wy )v
=
(wxx xv + wxy yv ) − (wyx xv + wyy yv )
=
wxx − 2wxy + wyy
From here a simple calculation shows that
5wxx + 2wxy + 2wyy = wuu + wvv .
57. Let A and B be vectors in 3-space for which A.B = 0 and kAk = kBk = 1. Show that the curve
F (t) = A sin t + B cos t
lies on a sphere.
Solution.
F (t) · F (t)
=
(A cos t + B sin t) · (A cos t + B sin t)
=
A · A cos t + 2A · B cos t sin t + B · B sin t
=
kAk cos t + kBk sin t
=
cos t + sin
2
2
2
2
2
2
2
2
=1
Thus F (t) lies on the sphere of radius equal to one.
58. A function
p U (x, y) is ‘harmonic’ if Uxx + Uyy = 0. Which of the following three functions are harmonic
(a) U = log x2 + y 2 (b) U = ex cos(y) (c) U = cos(xy)?
Solution. (a) harmonic (b) harmonic (c) not harmonic.
59. If U and V are harmonic functions and ∇U ⊥ ∇V , show that U V is also harmonic.
Solution.
(U V )x = Ux V + Vx U
(U V )xx = (Ux V + Vx U )x = Uxx V + 2Ux Vx + U Vxx
(U V )y = Uy V + Vy U
(U V )yy = (Uy V + Vy U )y = Uyy V + 2Uy Vy + U Vyy
Thus
(U V )xx + (U V )yy = V (Uxx + Uyy ) + U (Vxx + Vyy ) + 2∇U · ∇V = 0.
60. Let u = f (x, y) and x = r cos θ, y = r sin θ. Show that
(
∂u 2
∂u
∂u
1 ∂u
) + ( )2 = ( )2 + 2 ( )2 .
∂x
∂y
∂r
r ∂θ
Solutions. By the chain rule,
ur = ux xr + uy yr = ux cos θ + uy sin θ
uθ = ux xθ + uy yθ = ux (−r sin θ) + uy (r cos θ).
A little algebra shows that the quantity
(
∂u 2
1 ∂u 2
) +
(
)
∂r
r 2 ∂θ
evaluates to
(
∂u 2
∂u 2
) +(
) .
∂x
∂y
61. Suppose that g(u, v) = f (eu + cos v, eu + sin v). Find gu (0, 0) given the following table of values:
f (0, 0) = π
g(0, 0) = e
fx (0, 0) = 1
fy (0, 0) = 2
fx (1, 2) = 3
fy (1, 2) = 4
fx (2, 1) = 5
fy (2, 1) = 6
(Hint: You need not use all the values on the table)
f (2, 1) = e
f (1, 2) = π 2
14
Solution. Let x = eu + cos v and y = eu + sin v. By the chain rule,
gu = fx xu + fy yu .
When u = 0 then x = 2. When v = 0, then y = 1. Thus
gu (0, 0) = fx (2, 1)xu (0, 0) + fy (2, 1)yu (0, 0).
Note that xu = eu and yu = eu . Thus xu (0, 0) = 1 and yu (0, 0) = 1. From the table, fx (2, 1) = 5 and fy (2, 1) = 6. Thus
gu (0, 0) = 5 · 1 + 6 · 1 = 11.
62. Let u(x, t) satisfy Burger’s equation ut = uxx + u2x and set w(x, t) = eu(x,t) . Show that w satisfies the
heat equation wt = wxx .
Solution. wt = eu(x,t) ut and wx = eu(x,t) ux . Using the product rule,
u(x,t)
wxx = e
u(x,t)
uxx + ux e
u(x,t)
ux = e
2 u(x,t)
uxx + ux e
.
Thus
u(x,t)
wt = e
u(x,t)
ut = e
2
(uxx + ux ) = wxx .
63. Given w = f (x, y), x = eu cos v, y = eu sin v. What is e−2u (wu2 + wv2 ) equal to?
Solution. By the chain rule,
u
w u = wx e
wv = −wx e
u
sin v
u
cos v.
cos v + wy e
u
sin v + wy e
A computation shows that
2
(wu ) + (wv )
and so
−2u
e
2
2
=
2
2 2u
wx e
2
2 2u
+ wy e
2
(wu + wv ) = wx + wy