Sample Questions Math 122 Final Exam – Spring 2006

Sample Questions Math 122 Final Exam – Spring 2006
Seat in alternate seats in the portion of the room assigned to your section. Get your exam from your
instructor. When finished, hand your exam to your instructor. Instructors will not accept exams from
students in other sections.
Write your exam in pencil. Calculators are allowed; notes and texts are not.
Room Assignments
Parker 330 Strong, middle
Badawy 330 Strong, rear
Creese 3140 Wescoe, East rear
Brondi 3140 Wescoe, West rear
Fleissner 120 Snow
Mitchell 3140 Wescoe, East middle
Console 3140 Wescoe, West middle
Kummini 3140 Wescoe, East front
Carlson 3140 Wescoe, West front
Wagner 330 Strong, front
The final exam will consist of approximately 20 questions similar to those in this list. It is always important
to show your work. It is especially important when computing multiple integrals. You should show the
antiderivatives and the “ single ” integrals. The correct numerical value, by itself, will receive no credit.
1. Use implicit differentiation to find
2. Find
∂z
∂x
and
∂z
∂y
∂z
∂x
and
∂z
∂y
when ex + y 2 + z 3 = xyz.
when z = f (x4 − y 3 ).
3. Find an equation for the plane tangent to z = ex+2y at (ln 3, ln 5, 75).
4. Find the linear approximation L(x, y) of the function f (x, y) =
approximate
f (1.98, 1.01). Compute the difference quotients
5. Find the (total) differential of u =
p
4 + 2x + y 2 at (2, 1). Use it to
L(1.98, 1.01) − f (1.98, 1.01)
L(1.8, 1.1) − f (1.8, 1.1)
and
0.2
0.02
e2x sin y
.
ln z
6. Suppose that f (x, y) is a differentiable function of x and y, and g(u, v) = f (2u − 3v, −u + v 2 − 1). Use
the table of values to calculate gu (3, 2) and gv (3, 2). Show your work!
f g fx fy
(0,0) 5 2 1 -3
(3,2) 2 5 2 -4
8. At a certain time t0 , particle P is at (1, 2, 5) with velocity < 2, −1, 0 > and particle Q is at (1, 5, 9)
with velocity < 3, 1, −4 >. How fast is the distance |P Q| increasing at that instant?
9. Find the directional derivative Du f (3, 5) when f (x, y) = y 2 − x2 and u makes an angle of
positive x axis. Show your work!
π
5
with the
10. Find the directional derivative of f (x, y, z) = x + y 2 z + xz 2 at the point (2, 4, 1)
in the direction of v=< 3, −6, 2 >. Show your work!
11. Consider the hyperbolic paraboloid x − 9y 2 + 4z 2 = −24 at the point (8, 2, 1). Find a normal vector,
an equation for the tangent plane, and the symmetric equations for the normal line.
12. Use the second derivative test to classify the critical point at the origin. Show your work!
A. f (x, y) = 4x2 + xy − y 2 + 5y 4
B. f (x, y) = 3x2 + xy + 6y 4 + 2y 2
C. f (x, y) = x3 + x2 + 2xy + y 2
D. f (x, y) = cos x + ey − y
13. Find the critical points of f (x, y) = 2y 2 x − yx2 + 4xy. (Hint: factor y from fx (x, y) and factor x from
fy (x, y)). Classify the critical points. Show your work!
14. Use Lagrange multipliers to find the maximum and minimum values (and where they occur) of the
function f (x, y) = −6x + 2y
subject to the constraint
3x2 + y 2 = 16.
Show your work! In
particular, show the 3 equations in 3 unknowns.
15. Use Lagrange multipliers to find the positive numbers x, y, z which maximumize the value of the
function f (x, y, z) = xyz 2 subject to the constraint 2x + 4y + 6z = 48. Show your work!
[Hint: one approach is to solve the first three equations for xyz 2 /λ.]
√
16. Find the length of the curve r(t) =< et , 2 t, e−t >,
0 ≤ t ≤ ln 2.
|r0 (t) × r00 (t)|
at the point (1, 2, 0). Also
|r0 (t)|3
find the normal and tangential components of the acceleration at that point.
17. For the curve r(t) =< t2 , 2t, ln t >, find the curvature κ =
aT =
r0 (t) · r00 (t)
|r0 (t)|
aN =
|r0 (t) × r00 (t)|
|r0 (t)|
18. For the curve r(t) =< −1 + 3t, 6 + 4 cos t, π + 4 sin t >, find the unit tangent T(t) =
normal N(t) =
r0 (t)
, the unit
|r0 (t)|
T0 (t)
, and the binormal B(t) = T(t) × N(t).
|T0 (t)|
19. A batter hits a baseball 3.5 ft above the ground toward the center field fence, which is 370 ft from
where he hits the ball. The ball leaves the bat with speed 109 ft/sec at an angle of 48◦ above the horizontal.
Neglect air resistence, so that the only external force is gravity, g = 32 ft/sec2 . Write vector equations for
the acceleration, the velocity, and the position of the baseball. What is the height of the ball at the fence?
20. A particle moves through space with position function ~r(t) = 2t3 , 3t2 , 3t . Compute each of the
following functions and explain its physical interpretation in terms of the motion of the particle. (Some of
these have names). Make clear the difference between the first two. Make clear the difference between the
last two.
00
0 ~
r (t) ,
0
~r (t),
|~r(t)| ,
~r (t),
Z
s(t) =
t
~
r 0 (u) du
0
ZZ
xy 2 dA,
21. Calculate the double integral
R = {(x, y)| 3 ≤ x ≤ 5, 0 ≤ y ≤ 1}.
R
Z
3 Z y/3
y dx dy.
22. Evaluate the iterated integral
y2
0
ZZ
23. Express
f (x, y) dA as an iterated integral, when R is the triangle with vertices (1, 2), (4, 2), and
R
(4, 8). Now reverse the order of integration. (In other words, express it both as a dy dx integral and as a
dx dy integral).
ZZ
x dA,
24. Evaluate the double integral
when R is the region bounded by the curves y =
R
y2
x= .
8
x3
and
2
25. Find the volume under the surface z = x2 y and above the region in the xy-plane bounded by y = x2
and y = 4.
4Z 2
Z
26. Evaluate by reversing the order of integration:
0
√
3
ex dx dy. (Sketch the region. Display the new
y
iterated integral. Show intermediate steps.)
27. Use polar coordinates to find the volume under the paraboloid z = x2 + y 2 and above the annulus in
the xy-plane
4 ≤ x2 + y 2 ≤ 16. (Sketch the region. Display the iterated integral. Show intermediate steps.)
Z 1 Z √1−y2
1
28. Evaluate by converting to polar coordinates:
dx dy. (Sketch the region.
1 + x2 + y 2
0
0
Display the new iterated integral. Show intermediate steps.)
29. Consider the function f (x, y) =
xy 2
.
x2 + y 2
A. Find lim f (x, 0) = 0.
x→0
B. Convert to polar coordinates to find
C. Let g(x, y) =
xy 2
. Show that
x2 + 3y 2
Z
lim
(x,y)→(0,0)
lim
(x,y)→(0,0)
f (x, y) = 0.
g(x, y) = 0. What Theorem are you using?
1 Z x Z x+y
30. Evaluate the iterated integral
12z dz dy dx.
0
0
0
31. Find the Jacobian of the transformation
x = −2u2 + 2v 2
y = 2u2 + 2v 2 .
32. Suppose that at a certain instant t0 , a curve r(t) has r =< 2, 4, 0 >, T = 13 < 2, 2, 1 >, N = 13 <
−1, 2, −2 >, B = 31 < −2, 1, 2 >, and κ = 5. Give A. parametric equations for the tangent line, B. a scalar
equation of the normal plane, C. a scalar equation of the osculating plane, D. the center and radius of the
osculating circle.
7.
a.
b.
c.
d.
e.
f.
I
III
II
IV
V
VI
Write the letter of the surface and roman numeral of the contour map next to the function to which
they correspond.
1. f (x) =
p
4 − y 2 − 8x2
2. g(x) = cos(xy)
2
3. h(x) = ex cos y
4. j(x) = xy 2 − x3
5. k(x) = cos(x) − sin(y)
6. l(x) = (y − sin(3x))3
Sample Solutions Math 122 Final Exam – Spring 2006
1. In problems like this and the next, we assume that x and y are independent.
∂z
∂z
∂z
yz − ex
∂
Method 1: First ∂x
: ex + 0 + 3z 2
= yz + xy . Then solve for
= 2
.
∂x
∂x
∂x
3z − xy
∂z
xz − 2y
By similar computation
= 2
.
∂y
3z − xy
ex − yz
∂z
Fx
= − 2
Method 2: Transform to F (x, y, z) = 0: ex + y 2 + z 3 − xyz = 0. Then
= −
and
∂x
Fz
3z − xy
2y − xz
Fy
∂z
=− 2
=−
∂y
Fz
3z − xy
2. Chain rule:
∂z
= f 0 (x4 − y 3 )4x3
∂x
∂z
= f 0 (x4 − y 3 )(−3y 2 ).
∂y
and
3.
z − 75 = 75(x − ln 3) + 150(y − ln 5). If you use the method of problem 11, you must first transform to
x+2y
e
− z = 0.
4. L(x, y) = 3 + 31 (x − 2) + 31 (y − 1). f (1.98, 1.01) ∼
= L(1.98, 1.01) = 2.996666. The first fraction equals
(2.966666 − 2.968164)/0.2 = −0.0074, and the second equals (2.996666 − 2.996681)/0.02 = −0.0007. These
computations illustrate that the ε’s in Definition 7, page 772, go to 0.
5. du =
2e2x sin y
dx +
ln z
e2x cos y
dy −
ln z
e2x sin y
dz.
z(ln z)2
6. First, if (u, v) = (3, 2), then (x, y) = (2 · 3 − 3 · 2, −3 + 22 − 1) = (0, 0). The Chain Rule says that
gu (3, 2) = fx (0, 0)xu (3, 2) + fy (0, 0)yu (3, 2) = 1 · 2 + (−3) · (−1) = 5.
Similarly, gv (3, 2) = 1 · (−3) + (−3) · 4 = −15. See 11,12 page 786.
8. Let x be the x-coordinate of Q minus the x-coordinate of P , etc. It is cleaner to work with the square of the
ds
dx
dy
dz
d
distance |P Q|, which satisfies s2 = x2 + y 2 + z 2 . Now dt
: 2s
= 2x
+ 2y
+ 2z . Substitute
dt
dt
dt
dt
ds
−20
d
=
= −2.
the given values after dt . The first equation gives s = 5, then solve the second equation:
dt
10
Also be prepared for problems like Exercises 32-35, page 787.
9. ∇f (x, y) =< −2x, 2y >;
1.024
∇f (3, 5) =< −6, 10 >;
Du f (3, 5) =< −6, 10 > · < cos π/5, sin π/5 >=
10. ∇f (x, y, z) =< 1 + z 2 , 2yz, y 2 + 2xz >; ∇f (2, 4, 1) =< 2, 8, 20 >;
1
Du f (2, 4, 1) =< 2, 8, 20 > · |v|
< 3, −6, 2 >= 71 (6 − 48 + 40) = − 27 .
11. ∇f (x, y, z) =< 1, −18y, 8z >. The gradient vector, ∇f (8, 2, 1) =< 1, −36, 8 >, is normal to the surface. An
equation for the tangent plane is 1(x − 8) − 36(y − 2) + 8(z − 1) = 0. The symmetric equations for the normal
x−8
y−2
z−1
=
=
line are
1
−36
8
1
12. A. D = 8 · (−2) − 1 · 1 = −17 < 0, saddle point.
B. D = 6 · 4 − 1 · 1 = 23 > 0, local extremum.
Then fxx = 6 > 0, local minimum.
C. D = 2 · 2 − 2 · 2 = 0, no conclusion.
D. D = −1 · 1 − 0 = −1, saddle point.
13. fx = 2y 2 − 2xy + 4y = y(2y − 2x + 4), fy = x(4y − x + 4), D(x, y) = −8xy − (4y − 2x + 4)2 .
fx = 0 = fy gives 2 · 2 = 4 cases.
Case 1: y = 0 and x = 0 yield critical point (0, 0); D(0, 0) = −16 saddlepoint.
Case 2: y = 0 and 4y − x + 4 = 0 yield critical point (4, 0); D(4, 0) = −16 saddlepoint.
Case 3: 2y − 2x + 4 = 0 and x = 0 yield critical point (0, −2); D(0, −2) = −16 saddlepoint.
Case 4: 2y − 2x + 4 = 0 and 4y − x + 4 = 0 yield critical point ( 34 , − 23 ); D( 34 , − 23 ) = 48
9 local extremum.
Then fxx = 34 local minimum.
Sometimes students create new cases. For example x = 0 and 4y − x + 4 = 0 give the point (0, −1). But
fx (0, −1) = −2, not 0.
14. The three equations are −6 = λ · 6x; 2 = λ · 2y; and 3x2 + y 2 = 16. The first two equations give
− x1 = λ = y1 , and then y = −x. Substituting into the constraint gives x2 = 4; so x = ±2 and y = ∓2.
f (−2, 2) = 16 is the maximum value, and f (2, −2) = −16 is the minimum value.
Students sometimes give extra points. For example x = 2 and y = 2 satisfies 3x2 + y 2 = 16, but does not satisfy
y = −x.
15. ∇f = λ∇g gives three equations: yz 2 = 2λ; xz 2 = 4λ, and 2xyz = 6λ. One approach is to multiply the
first equation by x/λ, the second by y/λ, and the third by z/2λ. All the left sides are xyz/λ, so 2x = 4y = 3z.
Substituting into the constraint gives 2x + 2x + 2(2x) = 48, hence x = 6, y = 3, and z = 4. Maximum value is
6 · 3 · 42 = 288. (Warning: there are several places where a factor of 2 is easily forgotten).
√
r0 · r0 = e2t + 2 + e−2t ;
|r0 | = et + e−t ;
16. r0 =< et , 2, −e−t >;
ln 2
Z ln 2
1
3
et + e−t dt = et − e−t = (2 − ) − (1 − 1) =
s=
2
2
0
0
17. First the derivatives: r(t) =< t2 , 2t, ln t > r0 (t) =< 2t, 2, 1/t > r00 (t) =< 2, 0, −1/t2 >. We are asked
about the point r(t) =< 1, 2, 0 > so t = 1, and then substituting, r0 =< 2, 2, 1 > and r00 =< 2, 0, −1 >.
We continue computing. |r0 | = 3; r0 × r00 =< −2, 4, −4 >; |r0 × r00 | = 6; and finally κ = 2/9.
r0 · r00
|r0 × r00 |
aT is the length of the projection of r00 onto T, or
=
1.
By
formula,
a
=
= 2. Alternatively,
N
|r0 |
|r0 |
the Pythagorean Theorem gives r00 · r00 = aT 2 + aN 2 , then aN = 2. It is not necessary, but we can check that
r00 = 1T + 2N.
18. r0 · r0 = 9 + 16 sin2 t + 16 cos2 t = 25;
T = 15 < 3, −4 sin t, 4 cos t >;
T0 = 51 < 0, −4 cos t, −4 sin t >;
N =< 0, − cos t, − sin t >;
B = 15 < 4, 3 sin t, −3 cos t >
19. a =< 0, −32 >; v =< 72.9, 81.0 − 32t >; r =< 72.9t, 3.5 + 81.0t − 16t2 >; The ball is at the fence
when t = 370/72.9 = 5.1, then the height is 0.4. It hits the fence, but does not go over, assuming that the fence is
at least 5 inches high.
2 , 6t, 3 is the velocity of the particle at time t.
20. ~r 0 (t) = 6t
√
|~r 0 (t)| = 36t4 + 36t2 + 9 = 6t2 + 3 is the particle’s speed at time t.
The velocity is a vector; it has magnitude and direction. The speed, a scalar, is the magnitude of the velocity.
~r 00 (t) = h12t,
√ 6, 0i is the particle’s acceleration at time t.
|~r(t)| = 4t6 + 9t4 + 9t2 is the particle’s displacement at time t, that is, the straight line distance from the
origin, (where
it started at t =R 0), to where it is at time t.
Rt 0
s(t) = 0 |~r (u)| du = 0t 6u2 + 3 du = 2t3 + 3t is the distance along its path from the origin the particle to
where it is at time t.
The last two are different because the particle is not traveling in a straight line.
2
5Z 1
Z
5
Z
2
xy dy dx =
21.
3
0
3
3 Z y/3
Z
Z
y2
0
y/3
3
y dx dy =
22.
y 3 1
x dx =
3 0
0
yx
y2
1 1
· (25 − 9) =
3 2
3
y 2 /3 − y 3 dy =
(y 3 /9 − y 4 /4) =
8
3
−69/4
0
8Z 4
f (x, y) dx dy
and
y/2
2
2
1
3
0
Z
f (x, y) dy dx
23.
x dx =
3
Z
1 x2 5
·
=
3 2 3
5
Z
dy =
4 Z 2x
Z
1
3
√
√
Z 2
Z 2 Z 8x
√
8x
y2
24.
x=
xy x dy dx =
becomes y = 8x.
dx =
8
0
0
x3 /2
x3 /2
Z 2√
√ 2 5/2
2
x4
32 16
1
16
8x3/2 −
8 · x − x5 =
dx =
−
=
.
2
5
10
5
5
5
0
0
Z
2
Z
4
25.
−2 x2
2
1
8
( x3 − x7 ) =
3
14
−2
26. x =
√
2
1 2 2 4
1
(8x2 − x6 ) dx =
x y dx =
2
−2 2
−2
x2
64 128
−64 −128
512
( −
)−(
−
)=
3
14
3
14
21
Z
x2 y dy dx =
y becomes y =
x2 .
2
Z
2 Z x2
2
x3
e
0
2
Z
x3
e
dy dx =
4 Z 2π
Z
2
4
r rdθdr =
27. x + y = r ; dA = rdθdr.
2
0
2
x2
· y dx =
2
Z
0
2π
r θ dr =
Z
3
0
2 x3
x e
0
0
0
Z
2
2
Z
4
dx =
3
r dr =
2π
1 x3 2 1 8
e
= (e − 1)
3 0 3
2
r 4 4
2π · =
4 2
120π
28. The region is the portion of the unit disk in the first quadrant, so the new integral is
Z 1 Z π/2
Z
Z
1
π 1 r dr
π 2 du
π
r dθ dr =
dθ =
= ln 2.
2
2
1+r
2 0 1+r
4 1 u
4
0
0
x · 02
0
0
= lim 2 = 0 (not ).
x→0
x→0 x2 + 02
x→0 x
0
r3 cos θ sin2 θ
2
B. f (x, y) becomes
= r cos θ sin θ, which goes to 0 as r goes to 0+ .
r2
C. 0 ≤ |g(x, y)| ≤ |f (x, y)|. The limit at (0, 0) of the left and right functions is 0. By the Squeeze Theorem, the
limit of the middle function is 0.
29. A. lim f (x, 0) = lim
1 Z x Z x+y
Z
0
Z
1
0
31.
Z
1Z x
12z dz dy dx =
30.
0
0
0
x
6x y + 6xy + 2y dx =
2
2
3
Z
0
1
0
3
x+y
6z 2 1Z x
14 1 7
= x4 =
4
2
0
0
6x2 + 12xy + 6y 2 dy dx =
0
14x dx =
0
0
Z
dy dx =
∂(x, y)
= −4u · 4v − 4v · 4u = −32uv
∂(u, v)
32. a) r0 + 3sT gives x = 2 + 2s, y = 4 + 2s, and z = s (not unique);
b) The normal plane is perpendicular to T, (not N); so an equation is 2(x − 2) + 2(y − 4) + z = 0;
c) The osculating plane is perpendicular to B, so an equation is −2(x − 2) + (y − 4) + 2z = 0;
1
1
1
29 62
2
d) radius is = ; center is r0 + N =< , , − >
κ
5
κ
15 15 15
7. 1 b III;
2 c V;
3 a II
4eI
5 f VI
6 d IV
3