Math233ExamIISol

7/28/2019 Math233ExamIISol

1/32

Math 233 - Exam II - Fall 2011October 19, 2011 - Renato Feres

NAME:

STUDENT ID NUMBER:

General instructions: This exam has 16 questions, each worth the same amount.Check that no pages are missing and notify your proctor if you detect any problemswith your copy of the exam. Mark your ID number on the six blank lines on the topof your answer card, using one line for each digit. Print your name on the top of thecard. Choose the answer that is closest to the solution and mark your answer cardwith a PENCIL by shading in the correct box. You may use a 3 5 card with notesand any calculator that does not have graphing functions. GOOD LUCK!

1. Solve the initial value problem

d2r

dt2= 2k

with initial conditions

r(0) = 20k anddr

dt

t=0

= i + 3j.

12.2 15

(A) r(t) = 2ti + tj + (10 t2) k(B) r(t) = ti tj + (20 + t2) k(C) r(t) = 3ti 2tj + (20 + t2) k(D) r(t) = 3ti tj + (40 t2) k(E) r(t) = ti j + (20 t2) k

(F) r(t) = ti + 3tj + (20 t2) k(G) r(t) = 4ti 2tj + (t2 10) k(H) r(t) = ti 3tj + (t2 20) k(I) r(t) = ti + (t2 20) k(J) (t2 20) k

1


2/32


3/32

2. Decide whether each of the following statements is always true (T), or sometimesfalse or simply meaningless (F): 11. 4 27, 31

33

a. (u v) w = u (v w)b. (u v) w = u (v w)c. (u v) w = u (v w)d. Ifu v = u w and u = 0, then v = we. (u v) v = 0

These are, respectively:

(A) F, F, F, F, F

(B) F, F, F, T, F(C) T, F, T, T, F

(D) T, T, T, F, F

(E) T, F, T, F, T

(F) F, T, F, T, F

(G) T, T, T, F, F,

(H) T, F, T, T, T

(I) T, F, F, F, T

(J) F, F, F, T, T

3


4/32

Solution: a. is always true as discussed in class; b. and c. are meaningless; d.is not always true: take, for example, v = u = i, and w = 0; d. is always trueas the cross product is always perpendicular to both factors.

4


5/32

3. Find a parametric equation for the tangent line to the curve 12.1 19

r(t) = sin t, t2 cos t, et

at the point with parameter value t = 0.

(A) x = 2t, y = 0, z = 1 t(B) x = t, y = 1, z = 1 t

(C) x = t, y = 1, z = 1 + t(D) x = t, y = 2, z = 1 + t

(E) x = t, y = 1, z = 1 + 2t(F) x = t, y =

1, z = 1 + 3t

(G) x = t, y = t, z = 1 + t(H) x = t, y = 1 t, z = 1 + t(I) x = t, y = 1 2t, z = 1 + t(J) x = 2t, y = 1 t, z = 1 + t

5


6/32

Solution:

We have r(0) = 0,1, 1 and r(0) = 1, 0, 1. Therefore, the vector equationof the tangent line is

s(t) = r(0) + r(0)t = 0,1, 1 + t1, 0, 1 = t,1, 1 + t.

6


7/32

4. Find the point on the curve

r(t) =5sin t, 5cos t, 12t

at a distance 26 units along the curve from the point (0, 5, 0) in the directionof increasing arc length. 12.3 9

(A (0, 5, 24)

(B) (0, 5, 12)

(C) (0, 5, 6)

(D) (0, 5, 4)

(E) (0, 5, 3)

(F) (5, 0, 24)

(G) (5, 0, 12)

(H) (5, 0, 6)

(I) (5, 0, 4)

(J) (5, 0, 3)

7


8/32

Solution: Notice that (0, 5, 0) = r(0). So we take the initial parameter to bet = 0. Also r(t) = 5cos t,5sin t, 12, so

|r(t)| = 25 + 144 = 13.

Thus the arc length parameter is

s(t) =

t0

|r()| d =t0

13 d = 13t.

Now s(t) = 26 implies t = 2 and the position vector of the point we want is

r(2) = 0, 5, 24.

8


9/32

5. Find the curvature for the curve

r(t) = et cos t, et sin t, 2at the point with parameter t = 1. 12.4 11

(A) e

2

(B)

2

(C) 2

2

(D) 1/

2

(E) 1/(2

2)

(F) 1/(e2

2)

(G) 1/(e2)(H) e/

2

(I) e2/

2

(J) e/(2

2)

9


10/32

Solution:

The first derivative of r(t) is

v(t) =

et(cos t sin t), et(sin t + cos t), 0

and its magnitude is

|v(t)| = et

(cos t sin t)2 + (sin t + cos t)2 = et

2.

Thus the unit tangent vector is

T(t) =1

2cos t sin t, sin t + cos t, 0 .

Now,

dT

ds=

1

|v(t)|dT

dt=

1

et

2

1

2 sin t cos t, cos t sin t, 0

.

The vector in parenthesis on the right-hand side has unit length and the mul-tiplicative term in front of it is positive. So we can write

dT

ds=

1

et

2N.

Therefore,(t) =

1

et

2

which is equal to 1/(e

2) when t = 1.

10


11/32

6. Match the following surface equations: 11.6 1, 11, 97

a. x2 + y2 + 4z2 = 10

b. x2 + 4z2 = y2

c. x = z2 y2d. x2 + 2z2 = 8

with the surface descriptions:

1. hyperbolic paraboloid

2. cylinder with an elliptic cross-section

3. elliptical cone

4. ellipsoid

(A) a 1, b 2, c 3, d 4(B) a 4, b 3, c 2, d 1(C) a 1, b 4, c 3, d 2(D) a 2, b 3, c 4, d 1(E) a 3, b 2, c 1, d 4

(F) a 4, b 3, c 1, d 2

(G) a 4, b 1, c 3, d 2(H) a 4, b 2, c 3, d 1(I) a 3, b 1, c 2, d 4(J) a 4, b 1, c 2, d 3

11


12/32

Solution: This is readily deduced from the general form of these equations.

12


13/32

7. Find an equation for the circle of curvature of the curve r(t) = t, sin t at thepoint (/2, 1). 12.4 21

(A)

x 22 + (y + 5)2 = 1

(B)

x 2

2+ (y + 4)2 = 1

(C)

x 2

2+ (y + 3)2 = 1

(D)

x 2

2+ (y + 2)2 = 1

(E)

x 22

+ (y + 1)2 = 1

(F)

x 22

+ y2 = 5

(G) x 2

2+ y2 = 4

(H)

x 22 + y2 = 3

(I)

x 22

+ y2 = 2

(J)

x 2

2+ y2 = 1

13


14/32

Solution: The point (/2, 1) corresponds to the parameter, t = /2. The ve-locity of the curve is v(t) = 1, cos t, and the acceleration is a(t) = 0, sin t.At t = /2 we have

v = 1, 0 = i, a = 0,1 = j.

We can find by the formula

=|v a||v|3 = 1.

Thus the circle of curvature has radius R = 1 and center

OC = /2, 1 j = /2, 0

Therefore, the equation of the circle is

x

2

2+ y2 = 1.

14


15/32

8. Find the volume of the segment cut from the paraboloid z = x2 + y2 by the 11.6 47plane z = 1.

(In other words, find the volume of the region defined by x2 + y2 z 1.)

(A) 3

(B 2

(C)

(D) 32

(E) 2

(F) 2

2

(G) 2

3

(H) 32

2

(I)

22

3

(J) 1

15


16/32

Solution:

The cross-section of the region at level z = r2 is a disc of radius r, whose area

is A(z) = z. Thus the volume of the region is

V =

10

z dz =

2.

16


17/32

9. Find the velocity, speed, and acceleration at time t = of a particle in spacewhose position vector is r(t) = 2cos t, 3sin t, 4t. 12.1 11

(A) v = 0, 0, 1, v = 1, a = 2, 0, 0(B) v = 0, 5,5, v = 52, a = 2, 0, 0(C) v = 0,3, 4, v = 5, a = 0, 1, 0(D) v = 0, 4, 3, v = 5, a = 1, 0, 0(E) v = 0, 4,3, v = 5, a = 2, 0, 0(F) v = 0,3, 4, v = 5, a = 1, 0, 0(G) v = 0,3, 4, v = 5, a = 0, 2, 0(H) v =

0,

1, 1

, v =

2, a =

2, 0, 0

(I) v = 0,4, 4, v = 42, a = 1, 0, 0(J) v = 0,3, 4, v = 5, a = 2, 0, 0

17


18/32

Solution:

The velocity of the particle is

r() = 2sin t, 3cos t, 4|t= = 0,3, 4,

the acceleration is

r() = 2cos t,3sin t, 0|t= = 2, 0, 0,

the speed is|r()| = 9 + 16 = 5.

18


19/32

10. A projectile is fired with an initial speed of 100 m/sec at an angle of elevationof 45. How high overhead will be projectile be 1 km downrange? (Use theapproximation g = 9.8 m/sec2.) 12.2 21

(A) 5 m

(B) 9.8 m

(C) 10 m

(D) 20 m

(E) 25 m

(F) 30 m

(G) 40 m

(H) 50 m

(I) 100 m

(J) 150 m

19


20/32

Solution:

The parametric equation of the projectile in vector form is

r(t) = r0 + v0t 12

gt2j

where r0 = 0, 0 and v0 = 100cos45, sin45. Therefore,

x, y =

1002

t,100

2t 9.8

2t2

.

When x = 1000 meters downrange, t = 10

2, and the height is at that sametime is

y =100

210

2

9.8

2(10

2)2 = 1000

980 = 20.

So the height is 20 meters.

20


21/32

11. Find the tangential and normal scalar components of the acceleration of a par-ticle that moves according to the function

r(t) = 2cos t, 2sin t, 3t .

12.5 1

(A) aT = 0, aN = 2t

(B) aT = 0, aN = t

(C) aT = 1, aN = 4(D) aT = 3, aN = 2

(E) aT = 2, aN = 2

(F) aT = 0, aN = 4(G) aT = 0, aN = 3

(H) aT = 0, aN = 2

(I) aT = 1, aN = 1

(J) aT = 1, aN = 2

21


22/32

Solution:

The velocity is

v(t) = 2sin t, 2cos t, 3and the acceleration is

a(t) = 2cos t,2sin t, 0.

The speed is|v(t)| = 4 + 9 =

13

so the tangential component of the acceleration is

aT =d|v(t)|

dt

= 0.

The normal component is

aN =|a|2 a2T =

4 0 = 2.

22


23/32

12. Find the distance from the point (2,3, 4) to the plane x + 2y + 2z = 13. 11.5 39(A)

1

(B) 0

(C) 1

(D) 2

(E) 3

(F) 4

(G) 5

(H) 6

(I) 7(J) 8

23


24/32

Solution:

The point Q = (13, 0, 0) is easily seen to lie on the plane, and n = 131, 2, 2 isa unit perpendicular vector to the plane. The distance from P = (2,3, 4) tothe plane is

D =QP n =

11,3, 4 131, 2, 2 =

11 6 + 83 = 3.

24


25/32

13. Find the unit binormal vector B and the torsion function of the curve

r(t) =cos t, sin t,

1

at the point with parameter t = /4. 12.5 7

(A) B = i, = 0

(B) B = k, = 1(C) B = k, = 1

(D) B = k, = 0(E) B = k, = 0

(F) B = sin ti + cos tj, = 1/t

(G) B = cos ti + sin tj, = 0

(H) B = sin ti cos tj, = t(I) B = sin tj + cos tk, = t

(J) B = sin ti + cos tk, = 0

25


26/32

Solution:

This is a plane curve that lies on the plane z = 1. Therefore, B = k and = 0. An easy inspection gives B = k.

26


27/32

14. Find the point in which the line

x = 1

t, y = 3t, z = 1 + t

meets the plane2x y + 3z = 6.

11.5 51

(A)1

2 ,12 ,

32

(B)

12

,12

, 32

(C) (0, 0, 2)

(D) (3, 0, 0)

(E) (1,

1, 1)

(F) (2,2, 0)(G)

32 ,

32 ,12

(H)

35 ,35 , 15

(I)

32

,32

, 12

(J)

37 ,37 , 17

27


28/32

Solution:

The parameter t for the point we want satisfies

2(1 t) 3t + 3(1 + t) = 6.

This gives t = 1/2. Substituting into the parametric equation for the linegives

x =3

2, y = 3

2, z =

1

2.

28


29/32

15. Match the function expressions

I. f(x, y) = cos1(y

x2)

II. f(x, y) =

(x2 4)(y2 9)III. f(x, y) =

y x 2

with their domains among the sketched regions. 13.1 5, 9, 11

(A) I f, II d, I II a(B) I a II d, I II c(C) I e, I I d, I II b(D) I f, II c, I II a(E) I a, I I d, I II f

(F) I d, I I e, I II b(G) I b, I I d, I II c(H) I f, II a, I II e(I) I d, I I c, I II a(J) I b, I I c, I II f

29


30/32

Solution:

The domain of cos1(yx2) is determined by the inequalities 1 yx2 1,which are equivalent to x

2

1 y x2

+ 1. This is the region between twoparabolas, f.

The domain of

(x2 4)(x2 9) consists of points such that

x2 4 and y2 9, or x2 4 and y2 9,

which corresponds to d.

The domain of

y x 2 is the set of points such that y x 2 0, ory 2 + x. This region is shown in a.

30


31/32

16. The limits 13.2 15, 21

I.

lim(x,y)(1,1)

xy y 2x + 2x 1

II.

lim(x,y)(0,0)

sin(x2 + y2)

x2 + y2

are, respectively:

(A) 0 and 0

(B) 1/2 and 1/2

(C) 2 and 1(D) 2 and 1(E) 1 and 0

(F) 1 and 0(G) 0 and 1

(H) 1 and 1(I) 1 and 1(J) 1 and 2

31


32/32

Solution:

For the first limit notice that the numerator is (x 1)(y 2) so that for x = 1the quotient is equal to y 2. This function is continuous, therefore the limitis 1 2 = 1.For the second limit, notice that lim(x,y)(0,0) (x

2 + y2) = 0 and recall that thelimit of sin

as 0 is 1. Therefore, the second limit is 1.

Math233ExamIISol

Documents

Transcript of Math233ExamIISol