Unformatted text preview: blue MATH 32A Final Exam March 17, 2008 LAST NAME
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1 (20 pts) 5 (20 pts) 2 (20 pts) 6 (20 pts) 3 (20 pts) 7 (20 pts) 4 (20 pts) TOTAL 2 PROBLEM 1 (20 Points)
Let function f (x, y ) = x4 − x2 − y 2 .
(A) Find all critical points of f (x, y ) and classify each one
as a maximum, minimum, or saddle point. (B) Find the absolute maximum and minimum of f (x, y )
on the domain x2 + y 2 ≤ 4. Answer : 3 Solution: (A) First solve for the critical points:
1
fx = 4x3 − 2x = 2x(2x2 − 1) ⇒ x = 0, ± √
2
fy = −2y = 0 ⇒ y = 0 There are three critical points:
1
(0, 0)
( √ , 0),
2
The discriminant is 1
(− √ , 0)
2 2
D(x, y ) = fxx fyy − fxy = (12x2 − 2)(−2) − 0 = 4 − 24x2 We see that D(0, 0) = 4 > 0 so (0, 0) is a local min or max.
In fact, it is a local max since fx x(0, 0) = −2 < 0. We have
1
1
D(± √ , 0) = 4 − 24( ) = −8 < 0
2
2
Therefore the other two critical points are saddle points.
(B) We evaluate f (x, y ) at the critical points:
1
11
1
f (0, 0) = 0,
f (± √ , 0) = − − 0 = −
42
4
2
Next, we test f (x, y ) on the circle x2 + y 2 = 4. We ﬁnd
that on the circle
f (x, y ) = x4 − x2 − y 2 = x4 − 4 (for −2 ≤ x ≤ 2) Since x4 is positive, we may conclude immediately that the
maximum of f on the circle is
f (2, 0) = f (−2, 0) = 24 − 4 = 12 and the minimum value is f (0, 0) = −4.
Conclusion: the max of f (x, y ) on the disk x2 + y 2 ≤ 4 is
12 and the min is −4 (since the values at the critical points
lie between −4 and 12). 4 PROBLEM 2 (20 Points)
Suppose that f (x, y ) is a function of x and y , and that
x = u2 − v 2 , y = uv . Use the table of values ∂ f and ∂ f to
∂x
∂y
compute
∂f
∂f
and
∂ u (u,v)=(1,1)
∂ v (u,v)=(1,0)
x
0
1
0
1 Answer: Answer: y
0
0
1
1 ∂f
∂ u (u,v)=(1,1)
∂f
∂ v (u,v)=(1,0) ∂f
∂ x (x, y ) ∂f
∂ y (x, y ) 2
5
1
3 3
1
4
2 5 Solution: (A) When (u, v ) = (1, 1), we have
x = u2 − v 2 = 0, y = uv = 1 The table gives the values
∂f
∂f
= −1,
= −4
∂ x (x,y)=(0,1)
∂ y (x,y)=(0,1)
By the Chain Rule:
∂f ∂x ∂f ∂y
∂f
∂f
∂f
=
+
=
(2u) +
(v )
∂u
∂x ∂u ∂y ∂u
∂x
∂y
Evaluate at (u, v ) = (1, 1) and (x, y ) = (0, 1):
∂f
= (−1)(2) + (−4)(1) = −6
∂ u (u,v)=(1,1)
(B) When (u, v ) = (1, 0), we have
x = u2 − v 2 = 1, y = uv = 0 The table gives the values
∂f
∂f
= 5,
=1
∂ x (x,y)=(1,0)
∂ y (x,y)=(1,0)
By the Chain Rule:
∂f
∂f ∂x ∂f ∂y
∂f
∂f
=
+
=
(−2v ) +
(u)
∂v
∂x ∂v ∂y ∂v
∂x
∂y
Evaluate at (u, v ) = (1, 0) and (x, y ) = (0, 1):
∂f
= (5)(0) + (1)(1) = 1
∂ v (u,v)=(1,0) 6 PROBLEM 3 (20 Points)
Let r(t) = cos3 t, 0, sin3 t (A) Show that r (t)2 = 9 cos2 t sin2 t (B) Find a formula for the curvature κ(t) at time t. Answer : 7 Solution: We have
r (t) = −3 cos2 t sin t, 0, 3 sin2 t cos t
r (t) = 9 cos4 sin2 t + 9 sin4 t cos2 t
= 3 cos2 t sin2 t(cos2 t + sin2 t) = 3 cos t sin t r (t) = 6 cos t sin t − 3 cos3 t, 0, 6 cos2 t sin t − 3 sin3 t r (t) × r (t) = 0, 9 cos2 t sin2 t, 0 The curvature is
r (t) × r (t)
9 cos2 t sin2 t
1
κ(t) =
=
=
(t)3
3
r
3 cos t sin t
3 cos t sin t 8 PROBLEM 4 (20 Points)
Let P be the plane through the points
P1 = (1, 2, 0), P2 = (2, 0, 1), P3 = (0, 0, 2) (A) Find an equation for the plane P . Answer : (B) Find the coordinates of the point Q where the plane P
intersects the line through the two points
A = (1, 0, 0), Answer : B = (0, 2, 1) 9 Solution: (A) We have
P1 P2 = 2, 0, 1 − 1, 2, 0 = 1, −2, 1 P1 P3 = 0, 0, 2 − 1, 2, 0 = −1, −2, 2 We ﬁnd a normal to the plane by computing the cross product
P1 P2 × P1 P3 = 1, −2, 1 × −1, −2, 2 = −2, −3, −4 Since P1 = (1, 2, 0) lies on the plane, the equation of the
plane is
−2(x − 1) − 3(y − 2) − 4z = 0
or
2x + 3y + 4z = 8
(B) The line through (1, 0, 0) and (0, 2, 1) has the parametrization
r(t) = 1 − t, 2t, t
We solve for t in the equation
2(1−t)+3(2t)+4t = 8 ⇒ 2−2t+6t+4t = 8 ⇒ 8t = 6 ⇒ t = 3/4 133
3
Since r( 4 ) = 1 , 3 , 3 , we ﬁnd that the point Q = ( , , )
424
424
lines on the intersection of the plane and the line. 10 PROBLEM 5 (20 Points)
The gradient of pressure at the origin in r3 is
∇P = 4, 3, −1 (millibars per meter) A ﬂy carrying a pressure gauge travels in the positive direction along the y axis at 20 meters per second. How fast
is the pressure reading on the gauge changing when the ﬂy
passes through the origin? Give your answer in the correct
units. Answer : Solution: The velocity vector of the ﬂy is v = 0, 20, 0.
By the Chain Rule for Paths, the reading on the pressure
gauge is changing at a rate of
∇P · v = 4, 3, −1 · 0, 20, 0 = 60 millibars per second 11 PROBLEM 6 (20 Points)
Use the contour map to answer the following questions.
Show and explain the computation you made to obtain the
estimate.
(A) Estimate De f (P ) where e is the unit vector in the
direction of i + j and P is the point indicated on the map.
Answer: Answer: (B) Mark the point Q on the contour map where f (Q) = 400,
Answer: De f (P ) = ∂f
> 0,
∂x Q Figure 1 ∂f
=0
∂y Q 12 Solution: (A) The distance from P (which lies on the level
curve 400) to the level curve 200 along a segment in the i + j
direction (45o angle) is approximately 50 meters. Therefore
200 − 400
De f (P ) ≈
= −4
50
(B) The conditions ∂f
∂f
f (Q) = 400,
> 0,
=0
∂x Q
∂y Q
are satisﬁed at the point Q which is the leftmost point of
∂f
the level curve f (x, y ) = 400. Indeed,
> 0 because
∂x Q
f is increasing as you move horizontally to the right. And
since the tangent line to the level curve is vertical, we see
∂f
that
= 0.
∂y Q 13 PROBLEM 7 (20 Points)
Find the coordinates of the point (a, b) lying on the surface
2x2 − 2xy + y 2 + 4x − 2y = 7 with the largest xcoordinate. Hint: use Lagrange multipliers. Answer : 14 Solution: We must ﬁnd the maximum value of the function f (x, y ) = x subject to the constraint
g (x, y ) = 2x2 − 2xy + y 2 + 4x − 2y − 7 = 0 The Lagrange equations are 1, 0 = λ4x − 2y + 4, −2x + 2y − 2 The equation 1 = λ(4x − 2y + 4) shows that λ = 0, and the
second equation yields
0 = −2x + 2y − 2 ⇒ y = x + 1 Substitute back in the constraint equation: 2x2 − 2xy + y 2 + 4x − 2y − 7 == 0 2x2 − 2x(x + 1) + (x + 1)2 + 4x − 2(x + 1) − 7 = 0 2x2 − 2x2 − 2x + x2 + 2x + 1 + 4x − 2x − 2 − 7 = 0
x2 + 2x − 8 = 0 (x + 4)(x − 2) = 0 Therefore, x = 2 (and y = x + 1 = 3) or x = −4 (and
y = x + 1 = −4). The maximum value of x occurs at the
point (2, 3). blue MULTIPLE CHOICE QUESTIONS
ENTER ALL ANSWERS ON YOUR SCANTRON
FORM 882ES (in pencil)
If you change any answers on the scantron form, please make sure to
erase completely and leave no smudges. 1 2 PROBLEM 1 (10 Points)
Find the equation of the tangent plane to the surface
x2 y + y 3 + z 4 = 3
at the point P = (1, 1, 1).
(A) 2x + 3y + 4z = 9
(B) 3x + 2y + 4z = 0
(C) 2x + 4y + 4z = 10
(D) 2x + 5y + 4z = 11
(E) 3x + 3y + 4z = 10 Solution: C . The gradient provides a normal vector:
∇F(1,1,1) = 2xy, x2 + 3y 2 , 4z 3 = 2, 4, 4 The equation of the plane is
or 2(x − 1) + 4(x − 1) + 4(x − 1) = 0
2x + 4y + 4z = 10 3 PROBLEM 2 (10 Points)
Let
f (x, y ) = x3 y 4 − x − y
Find a vector v pointing in a direction along which the directional
derivative of f (x, y ) at P = (2, −1) is equal zero.
(A) 33, 11
(B) 12, 4
(C) 12, −32
(D) 12, 32
(E) 11, −33
Solution: Both A and B are correct (and both answers will receive
full credit). We have
and ∇f = (3x2 y 4 − 1, 4x3 y 3 − 1 ∇f (2, −1) = (11, −33
The vector v = 33, 11 is perpendicular to ∇f (2, −1) and so the directional derivative in the direction of v = 33, 11 is zero. But
4
12, 4 = 33, 11
11
so 12, 4 is also correct. 4 PROBLEM 3 (10 Points)
Find the minimum value taken by f (x, y ) = xy − x2 y − 4xy 2 on the
hyperbola xy = 1.
(A) −4
(B) −3
(C) − 3
2 (D) −2
(E) − 1
2 Solution: The correct answer is B . The Lagrange condition is
This yields ∇f = y − 2xy − 4y 2 , x − x2 − 8xy = λy, x y − 2xy − 4y 2
x − x2 − 8xy
=
y
x
Notice that x and y are both nonzero if (x, y ) satisﬁes xy = 1, so it is
legitimate to divide by x and y . Thus we obtain
λ= 1 − 2x − 4y = 1 − x − 8y ⇒ 4y = x The constraint yields xy = 4y 2 = 1, so y = ± 1 and the critical points
2
are (2, 1 ) and (−2, − 1 ). We have
2
2
1
1
f (2, ) = xy − x2 y − 4xy 2 = 1 − 2 − 4( ) = −3
2
2
1
1
f (−2, − ) = 1 + 2 − 4(− ) = 5
2
2
We conclude that the minimum value is −3 5 PROBLEM 4 (10 Points)
Find a parametrization of the line through P = (1, 0, 5) that is perpendicular to the plane
12x − 6y − 24z = 7
(A) L(t) = 12t, −t, −24t
(B) L(t) = 5 − 24t, −t, 1 + 12t
(C) L(t) = 1 + 4t, −t, 5 − 12t
(D) L(t) = 1 + 2t, −t, 5 − 4t
(E) L(t) = 7 + 12t, −t, 24t
Solution: The correct answer is D . The vector n = 12, −6, −24
is normal to the plane and therefore it is a direction vector for the
line. We may also divide n by 6 and use v = 2, −1, −4 as a direction
vector. Therefore, the following is a parametrization of the desired line:
L(t) = 1, 0, 5 + t2, −1, −4 = 1 + 2t, −t, 5 − 4t 6 PROBLEM 5 (10 Points)
The ﬁgure below shows a cube. The point Q is located at the center
of the top face of the cube. Calculate the cosine of the angle between
→ → the vectors v = P R and w = P Q
Figure 1
(A)
(B) √ 2
2 1
4 1
(C) √
3
√
3
(D)
2
(E) 1
2 Solution: The correct answer is E . If we place the origin at the
vertex directly below P , then we have
11
P = (0, 0, 1), Q = ( , , 1), R = (1, 0, 0)
22
and
→
→
11
v = P R = 1, 0, −1,
w = P Q = , , 0
22
The cosine of the angle is
cos θ = 1
v·w
1
2
=√
=
v w
2
2 1/2 7 PROBLEM 6 (10 Points)
One moon of the Vulcan Planet has a circular orbit of radius 1000 km
and completes one revolution of its orbit is 50 hours. A second moon
completes its circular orbit in 400 hours. According to Kepler’s Third
Law, the radius of the second moon’s orbit is
(A) 250 km
(B) 125 km
(C) 500 km
(D) 2000 km
(E) 4000 km Solution: The correct answer is E . We have or 10003
r3
=
4002
502
r3 = 10003 Therefore 4002
= 10003 (82 )
502 r = 1000(22 ) = 4000 km 8 PROBLEM 7 (10 Points)
Calculate the directional derivative of f (x, y, z ) = xy 2 z 3 at P = (2, 1, 1)
in the direction of the vector v = 1, 2, 2.
3
(A)
2
(B) 8
3 (C) 4
(D) 7
(E) 8 Solution: The correct answer is D . We have
∇f (2, 1, 1) = y 2 z 3 , 2xyz 3 , 3xy 2 z 2 = 1, 4, 6 The unit vector in the direction of v is e = 1 , 2 , 2 and the direction
333
derivative is
122
122
1 8 12
∇f (2, 1, 1) · , , = 1, 4, 6 · , , = + +
=7
333
333
33
3 9 PROBLEM 8 (10 Points)
A particle moves to the right along the parabola y = x2 (units in
centimeters). At time t = 1 sec, the particle passes the origin, at which
time the velocity and acceleration vectors of the particle are
v = 4, 0, a = 3, 6 How fast is the particle’s speed changing at t = 1?
(A) 3 cm/s2
(B) 12 cm/s2
(C) 2 cm/s2
(D) 6 cm/s2
(E) 4 cm/s2 Solution: The correct answer is A . The speed is changing at a rate
of
v
4, 0
v = a ·
= 3, 6 ·
= 3 cm/s2
v
4 ...
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This note was uploaded on 10/26/2011 for the course MATH 32A taught by Professor Gangliu during the Spring '08 term at UCLA.
 Spring '08
 GANGliu
 Math

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