Exam Two Review Two Solutions

Exam Two Review Two Solutions
Math 21a
1. The function
f (x, y) =
Spring, 2013
3x − 4y
.
x2 + y 2
satisfies which of the following partial differential equations.
(a) fxx = fyy
(b) fx = fyy
(c) fxx + fyy = 0
(d) fy = fxx
(e) None of the above.
Solution:
The correct answer is (c).
2. An astronaut is flying in a spacecraft along the path described by
r(t) = ht2 − t, 2 + t, −3/ti,
where t is given in hours.
(a) What is the velocity of the spacecraft when it reaches the point (6, 5, −1)?
Solution:
The spacecraft reaches the point (6, 5, −1), when t = 3.
h2t − 1, 1, 3/t2 i, the velocity at t = 3 is r0 (3) = h5, 1, 1/3i.
Since r0 (t) =
(b) What is the speed of the spacecraft when it reaches the point (6, 5, −1)?
√
235
0
Solution: The speed at t = 3 is |r (3)| =
.
3
(c) What is the acceleration of the spacecraft when it reaches the point (6, 5, −1)?
Solution: Since r00 (t) = h2, 0, −6/t3 i, the acceleration at t = 3 is r00 (3) = h2, 0, −2/9i.
(d) If the engines of the spacecraft are shut off when it reaches the point (6, 5, −1), where will the
spacecraft be 2 hours later?
Solution: The spacecraft will continue in a straight line (6, 5, −1) in the direction of the
velocity vector r0 (3) = h5, 1, 1/3i. Its position at time t is given by
p(t) = h6, 5, −1i + th5, 1, 1/3i
and p(2) = h16, 7, −1/3i.
3. Find r(t) if r0 (t) = (2t + 1)i + cos tj − et k and r(0) = −i + πj + 3k.
Solution:
Z
r0 (t)dt
Z
(2t + 1)i + cos tj − et kdt
r(t) =
=
= (t2 + t)i + sin tj − et k + C
Since r(0) = −k + C = −i + πj + 3k, we have r(t) = (t2 + t − 1)i + sin t + πj + (4 − et )k.
4. The intersection of the two surfaces x2 +
y2
2
= 1 and z 2 +
y2
2
= 1 consists of two curves.
(a) Parameterize each curve in the form r(t) = hx(t), y(t), z(t)i.
Math 21a
Exam Two Review Two Solutions
Spring, 2013
Solution: Both these surfaces are elliptical cylinders. The intersection curves, when translated
2
vertically into the xy-plane, each project onto the ellipse x2 + y2 = 1. This we can view as
2
(x)2 + √y2
= 1 and take the parameterization x = cos t and √y2 = sin t. What about z?
Well, from the equations it’s clear that z 2 = x2 , so z = x or z = −x (hence two curves). So
our parameterizations are
√
√
and
r(t) = hcos t, 2 sin t, − cos ti.
r(t) = hcos t, 2 sin t, cos ti
(b) Set up the integral for the arc length of one of the curves.
0
Solution:
√ We’ll work with the first one. Notice that r (t) = h− sin t,
0
|r (t)| = 2. Thus the arc length is given by the integral
Z
2π
2π
Z
0
|r (t)| dt =
√
√
2 cos t, − sin ti, so
2 dt.
0
0
(c) What is the arc length of this curve?
√
Solution: This integral is 2π 2.
5. Imagine the planet Earth as the unit sphere centered at the origin in three-dimensional space. An
asteroid is approaching from the point P (0, 4, 3) along the path
r(t) = h(4 − t) sin(t), (4 − t) cos(t), 3 − ti.
(a) When and where will it first hit the earth?
Solution: The asteroid will hit the earth when x2 + y 2 + z 2 = 1. In terms of t, this is when
(4−t)2 sin2 t+(4−t)2 cos2 t+(3−t)2 = 1
or
(4−t)2 +(3−t)2 = 1
or
2t2 −14t+24 = 0.
This factors to 2(t − 3)(t − 4) = 0, so t = 3 and t = 4. Thus the asteroid will first hit the earth
when t = 3 at r(1) = hsin(3), cos(3), 0i.
(b) What velocity will it have at impact?
Solution: The velocity will be r0 (3). We compute the derivative to be
r0 (t) = h(4 − t) cos t − sin t, (t − 4) sin t − cos t, −1i,
so r0 (3) = hcos(3) − sin(3),√− sin(3) − cos(3), −1i is the velocity. (The speed, by the way, will
be a nice simple |r0 (3)| = 3.)
6. If
x2 + y 2 z + xz 4 + xyz 7 = 0,
then which of the following is ∂z/∂x at the point (x, y, z) = (1, 0, −1).
(a)
3
4
(b) −
3
4
(c)
4
3
(d) −
4
3
(e) None of the above.
Exam Two Review Two Solutions
Math 21a
Solution:
Spring, 2013
Let
F (x, y, z) = x2 + y 2 z + xz 4 + xyz 7 .
Then
∂z
2x + z 4 + yz 7
Fx
= 2
=−
.
∂x
Fz
y + 4xz 3 + 7xyz 6
Therefore,
3
∂z = ,
∂x (x,y,z)=(1,0,−1) 4
and the correct answer is (a).
7. Fill in the boxes. No additional explanations required.
Answer
d
dt f (r(t))
Chain rule
=
· r0 (t)
∇f
√
h1, 1i/ 2
Directional derivative
Dh1,1i/√2 f (1, 1) = ∇f (1, 1) ·
Linearization of f (x, y) at (1, 1)
L(x, y) =
Equation of tangent line at (1, 1)
∇f (1, 1) · hx − 1, y − 1i =
0
Critical point (1, 1) of f
∇f (1, 1) =
h0, 0i or 0
Lagrange equations
∇f (x, y) =
+ ∇f (1, 1) · hx − 1, y − 1i
∇g(x, y), g(x, y) = c
f (1, 1)
λ
8. Let z = f (x, y) have the contour plot shown in Figure 1 below, where the horizontal axis is the x-axis
and the vertical axis is the y-axis. At which of the labelled points below is the directional derivative
in the direction u = h− √12 , √12 i the greatest?
(a) A
(b) B
(c) C
(d) D
Solution: At the point B, moving in the direction u (which I’ll call “northwest”) decreases the
function f . The same thing is happening at point D (moving from the level curve f = 4 toward the
level curve f = 2). At A we’re moving from f = 4 almost immediately back to f = 4. But at C,
we’re increasing quickly to f = 6 and f = 8. Thus the correct answer is (c).
9. Answer the following questions True or False:
(a) T F The directional derivative Dv f is a vector perpendicular to v.
Solution: This is False. The directional derivative Dv f is the scalar ∇f · v.
(b) T F Given a curve r(t) on a surface g(x, y, z) = 1, then
d
dt g(r(t))
= 0.
Solution: This is True. This is simply the chain rule applied to the equation g(r(t)) = 1.
(c) T F If f (x, y) has a local maximum at (0, 0), then it is possible that fxx (0, 0) > 0 and
fyy (0, 0) < 0.
Solution: This is False. This is saying that f is concave up in the x direction (in the xz-plane)
and concave down in the y direction (in the yz-plane). This cannot then be a local maximum.
Math 21a
Exam Two Review Two Solutions
Spring, 2013
4
2
2
D
8
6
0
4
-2
0
C
-6 -8
-4
2
B
-2
A
4
-4
-4
-2
0
2
4
Figure 1: Figure For Problem 8.
R1Ry
R1Rx
(d) T F Fubini’s theorem assures us that 0 0 f (x, y) dy dx = 0 0 f (x, y) dx dy.
This is False. The two regions described by the iterated integrals are not the same – one is
Region (iii) of Problem 6 and the other is Region (vi).
(e) T F If x + sin(xy) = 1, then
dy
dx
cos(xy)
= − 1+y
x cos(xy) .
Solution: This is True.
(f) T F The directional derivative Dv f (1, 1) is zero if v is a unit vector tangent to the level
curve of f which passes through the point (1, 1).
Solution: This is True. This is our most important fact about the gradient: it is perpendicular
to the tangent line (or plane) of the level curve (or surface).
10. Let f (x, y) = y 2 − x2 .
(a) Calculate the gradient of f , ∇f .
Solution: ∇f (x, y) = −2x i + 2y j or h−2x, 2yi.
(b) What is the directional derivative of f in the direction v = −3 i + 4 j at the point (1, 1)?
Solution: A unit vector in the direct of v is u = −(3/5) i + (4/5) j. Thus,
14
Du f = ∇f (1, 1) · u = h−2, 2i · − 35 , 45 = .
5
(c) Find the maximum and minimum values of f (x, y) = y 2 − x2 subject to the constraint
g(x, y) = x2 + 4y 2 = 4.
Math 21a
Exam Two Review Two Solutions
Solution:
Spring, 2013
The gradient of g is ∇g = h2x, 8yi. If ∇f = λ∇g, then
−2x = λ2x
2y = λ8y
2
x + 4y 2 = 4.
By the last equation, x and y cannot both be zero. If x 6= 0, then λ = −1 and y = 0.
Thus, we have critical points at (±2, 0). If y 6= 0, then λ = 1/4 and x = 0. In this case we
have critical points at (0, ±1). Therefore, f (2, 0) = f (−2, 0) = −4 is the minimum value and
f (1, 0) = f (−1, 0) = 1 is the maximum value.
11.
(a) Locate and classify all the critical points of
f (x, y) = 3y − y 3 − 3x2 y.
Solution:
The critical points of f (x, y) are those points where ∇f = 0. Since ∇f =
2
h−6xy, 3 − 3y − 3x2 i, the critical points are where xy = 0 and x2 + y 2 = 1. This gives us four
points: (0, ±1) and (±1, 0).
We can compute fxx = −6y, fxy = fyx = −6x, and fyy = −6y. Thus D = 36y 2 − 36x2 , and we
get a small chart:
Crit. Pt.
(0, 1)
(0, −1)
(1, 0)
(−1, 0)
D
36
36
−36
−36
fxx
−6
6
0
0
Classification
Maximum
Minimum
Saddle
Saddle
f (x, y)
2
−2
0
0
(b) Where on the parameterized surface
D
E
2
2
r(x, y) = hu, v, wi = xy 3 , x2 , 3y2
is the function g(u, v, w) = u − v − w extremal? To investigate this, find all the critical points
2
2
of the function f (x, y) = xy 3 − x2 − 3y2 . For each critical point, specify whether it is a local
maximum, a local minimum, or a saddle point and show how you know.
Solution: The critical points of f (x, y) are again those points where ∇f = 0. In this case
∇f = hy 3 − x, 3xy 2 − 3yi, so the critical points are where x = y 3 and 3y(xy − 1) = 0. Thus
either y = 0 (so x = 0 as well) or xy = 1, in which case x4 = 1. This gives us three points:
(0, 0), (1, 1), and (−1, −1).
We compute fxx = −1, fxy = fyx = 3y 2 , and fyy = 6xy − 3. Thus D = 3 − 6xy − 9y 4 , and we
again get a small chart:
Crit. Pt.
(0, 0)
(1, 1)
(−1, −1)
D
3
−12
−12
fxx
−1
−1
−1
Classification
Maximum
Saddle
Saddle
f (x, y)
0
−1
−1
12. Find an equation of the tangent plane to the surface x2 y + exz + yz = 3 at the point (0, 1, 2).
Solution:
Let F (x, y, z) = x2 y + exz + yz. Then F (x, y, z) = 3 is a level surface for F . Since
∇F (x, y, z) = h2xy + zexz , x2 + z, xexz + yi,
Math 21a
Exam Two Review Two Solutions
Spring, 2013
∇F (0, 1, 2) = h2, 2, 1i is a normal vector to the surface at (0, 1, 2). Therefore, the tangent plane is
given by
h2, 2, 1i · hx − 0, y − 1, z − 2i = 0
or
2x + 2(y − 1) + (z − 2) = 0.
13.
(a) Use the technique of linear approximation to estimate f ( π2 + 0.1, 2.9) for
f (x, y) = 10 sin(x) − 5y 2 + 8
Solution:
1/3
.
Near the point ( π2 , 3), we have the linear approximation
f (x, y) ≈ L(x, y) = f ( π2 , 3) + fx ( π2 , 3)(x − π2 ) + fy ( π2 , 3)(y − 3).
Since
−2/3
10 cos(x) =
10 cos(x)
3(10 sin(x) − 5y 2 + 8)2/3
−2/3
(−10y) = −
10y
,
3(10 sin(x) − 5y 2 + 8)2/3
fx =
1
3
10 sin(x) − 5y 2 + 8
fy =
1
3
10 sin(x) − 5y 2 + 8
and
we get f ( π2 , 3) = −3, fx ( π2 , 3) = 0, and fy ( π2 , 3) = − 10
9 . Thus
f (x, y) ≈ −3 + 0(x − π2 ) −
and so
f ( π2 + 0.1, 2.9) ≈ −3 + 0(0.1) −
10
(y − 3)
9
10
26
(−0.1) = .
9
9
(b) Find the unit vector at ( π2 , 3) in the direction where this function increases fastest.
Solution: This function increases fastest in the ∇f direction. Since ∇f ( π2 , 3) = h0, − 10
9 i, the
unit vector in this direction is h0, −1i.

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