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Practice Exam with Answer - Calculus III | MATH 265, Exams of Advanced Calculus

Material Type: Exam; Class: CALCULUS III; Subject: MATHEMATICS; University: Iowa State University; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 09/02/2009

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Allen Math 265 B2 Practice Exam #1 Name:
1. Given that ~u =h−1,3i,~v =h2,1i, and ~w =h5,6iare three non-collinear vectors, find scalars aand bso
that ~w =a~u +b~v.
We need h−a,3ai+h2b, bi=h5,6i. This leads to the following system of equations:
a+ 2b= 5
3a+b= 6 a+ 2b= 5
6a2b=12 Adding the two equations together yields 7a=7. Thus
a= 1. Plugging back into a+ 2b= 5, we get 1+2b= 5 2b= 6. Therefore, b= 3.
2. Let ~u =h1,4,2i,~v =h−2,3,1i, and ~w =h4,1,1i. Perform the indicated operations.
(a) Compute the angle (in degrees) between ~u and ~v.
Use the formula ~u ·~v =
~u
~v
cos θ.~u ·~v = 1(2) + 4(3) + 2(1) = 2 + 12 + 2 = 12.
~u
=12+ 42+ 22=1+16+4 =21.
~v
=p(2)2+ 32+ 12=4+9+1=14.
Thus, 12 = 2114 cos θ12 = 76 cos θ26
7= cos θ. Therefore, θ45.5847.
(b) Find the volume of the parallelepiped determined by ~u,~v, and ~w.
We can use 2 methods.
~u ·(~v ×~w)
=
h1,4,2i · h−2,2,10i
from part (b). This equals
1(2) + 4(2) + 2(10)
=
2+820
=
14
= 14.
Another method we can use is to find
1 4 2
2 3 1
411
, which is
(31)(1) (2 4)(4) + (2 12)(2)
=
4(1) + 2(4) 10(2)
=
4 + 8 20
=
14
= 14.
3. Let ~u =h3,4iand ~v =h5,12i. We can write ~u =~m +~n, where ~m is parallel to ~v and ~n is perpendicular
to ~v. Find ~m and ~n.
~m = pr~v~u =~u ·~v
|~v|2~v.~u ·~v = 3(5) + 12(4) = 15 + 48 = 63. |~v|2= 52+ 122= 25 + 144 = 169.
Thus, ~m =63
169h5,12i=315
169,756
169.
Then ~n =~u ~m =h3,4i 315
169,756
169=517
169,676
169315
169,756
169=192
169,80
169.
1
pf3
pf4
pf5

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Allen – Math 265 B2 Practice Exam #1 Name:

  1. Given that ~u = 〈− 1 , 3 〉, ~v = 〈 2 , 1 〉, and w~ = 〈 5 , 6 〉 are three non-collinear vectors, find scalars a and b so

that w~ = a~u + b~v.

We need 〈−a, 3 a〉 + 〈 2 b, b〉 = 〈 5 , 6 〉. This leads to the following system of equations: { −a + 2b = 5

3 a + b = 6

−a + 2b = 5

− 6 a − 2 b = − 12

Adding the two equations together yields − 7 a = −7. Thus

a = 1. Plugging back into −a + 2b = 5, we get −1 + 2b = 5 ⇒ 2 b = 6. Therefore, b = 3.

  1. Let ~u = 〈 1 , 4 , 2 〉, ~v = 〈− 2 , 3 , 1 〉, and w~ = 〈 4 , − 1 , − 1 〉. Perform the indicated operations.

(a) Compute the angle (in degrees) between ~u and ~v.

Use the formula ~u · ~v =

~u

~v

cos θ. ~u · ~v = 1(−2) + 4(3) + 2(1) = −2 + 12 + 2 = 12. ∣ ∣ ~u

2

  • 4 2
  • 2 2 =

~v

2

  • 3 2
  • 1 2 =

Thus, 12 =

14 cos θ ⇒ 12 = 7

6 cos θ ⇒

= cos θ. Therefore, θ ≈ 45. 5847

◦ .

(b) Find the volume of the parallelepiped determined by ~u, ~v, and w~.

We can use 2 methods.

~u · (~v × w~)

from part (b). This equals ∣ ∣ 1(−2) + 4(2) + 2(−10)

Another method we can use is to find

, which is

  1. Let ~u = 〈 3 , 4 〉 and ~v = 〈 5 , 12 〉. We can write ~u = m~ + ~n, where m~ is parallel to ~v and ~n is perpendicular

to ~v. Find m~ and ~n.

m ~ = pr~v~u =

~u · ~v

|~v| 2

~v. ~u · ~v = 3(5) + 12(4) = 15 + 48 = 63. |~v|

2 = 5

2

  • 12

2 = 25 + 144 = 169.

Thus, m~ =

Then ~n = ~u − m~ = 〈 3 , 4 〉 −

  1. Find the equation of the plane that passes through the points P = (1, 4 , 6), Q = (− 2 , 5 , −1), and

R = (1, − 1 , 1).

If ~u is the vector from Q to P, then ~u = 〈(1 + 2), (4 − 5), (6 + 1)〉 = 〈 3 , − 1 , 7 〉. Similarly, if ~v is the vector

from Q to R, then ~v = 〈(1 + 2), (− 1 − 5), (1 + 1)〉 = 〈 3 , − 6 , 2 〉. To compute a normal to the plane, we

compute ~u × ~v.

~u × ~v =

~i ~j ~k

= (−2 + 42)~i − (6 − 21)~j + (−18 + 3)

k = 40~i = 15~j − 15

k = 〈 40 , 15 , − 15 〉.

This vector has the same direction as 〈 8 , 3 , − 3 〉. So the equation of the plane has the form

8 x + 3y − 3 z = D. Using the point P, we get 8(1) + 3(4) − 3(6) = D ⇒ D = 8 + 12 − 18 = 2. Therefore,

the equation of the plane is 8x + 3y − 3 z = 2.

  1. Find the center and radius of the sphere 4x 2 + 4y 2 + 4z 2 − 24 x + 56y − 40 z − 197 = 0.

4 x 2

  • 4y 2
  • 4z 2 − 24 x + 56y − 40 z = 197 ⇒ x 2
  • y 2
  • z 2 − 6 x + 14y − 10 z =

⇒ (x 2 − 6 x) + (y 2

  • 14y) + (z 2 − 10 z) =

⇒ (x 2 − 6 x + 9) + (y 2

  • 14y + 49) + (z 2 − 10 z + 25) =

⇒ (x − 3) 2

  • (y + 7) 2
  • (z − 5) 2 =

⇒ (x − 3) 2

  • (y + 7) 2
  • (z − 5) 2 =

Center: (3, − 7 , 5). Radius:

  1. A particle P travels in the plane, and its vector position at time t is given by ~r(t) = (1 − t)~i +

1 − t

j.

(a) Find the Cartesian equation for the path of the particle.

From the vector equation, the parametric equations are given by x = 1 − t and y =

1 − t 2

. Solve

the first equation for t and get t = 1 − x. Plug into the second equation to get

y =

1 − (1 − x) 2 =

1 − (1 − 2 x + x 2 ) =

2 x − x 2

. Hence, the path of the particle is

y

2 = 2x − x

2 in Cartesian coordinates.

(b) Assuming that P starts moving at time t = 0, how far will it have traveled at t = 1/2 seconds?

L =

0

[x′(t)]^2 + [y′(t)]^2 dt. x ′ (t) = −1, and so [x ′ (t)] 2 = 1.

y ′ (t) =

(1 − t 2 ) − 1 / 2 (− 2 t) = −

t √ 1 − t 2

, and so [y ′ (t)] 2 =

t

2

1 − t 2

Hence, [x ′ (t)] 2

  • [y ′ (t)] 2 = 1 +

t

2

1 − t 2

1 − t

2

1 − t 2

t

2

1 − t 2

1 − t 2

. Therefore,

L =

0

1 − t 2

dt =

0

1 − t^2

dt =

[

sin

− 1 t

] 1 / 2

0

= sin

− 1 (1/2) − sin

− 1 (0) =

π

π

  1. A cannon sits atop a cliff that is 176 feet tall. It is aimed at an angle of 53. 1301

◦ above the horizontal,

and it shoots a projectile at the speed of 25 feet per second.

(a) Find the initial velocity vector, ~v(0) of the projectile.

We are given that

~v(0)

= 25, and the angle between ~v(0) and the ground is 53. 1301

. Let x be

the horizontal component of the velocity and y be the vertical component of the velocity. Then,

x = 25 cos(53.1301) = 15, and y = 25 sin(53.1301) = 20. Therefore, ~v(0) = 15

i + 20

j.

(b) Using part(a), the fact that ~a(t) = − 32

j, and assuming ~r(0) = 176

j, find the velocity, ~v(t), and the

position, ~r(t), of the projectile for any time t.

Taking the antiderivative of ~a(t) yields ~v(t) = C 1 ~i + (− 32 t + C 2 )~j. At t = 0, v(0) = C 1 ~i + C 2 ~j. But

from part (a), this should be 15~i+20~j. Hence, C 1 = 15, C 2 = 20, and thus ~v(t) = 15~i+(− 32 t+20)~j.

Now, we antidifferentiate again to get ~r(t) = (15t + C 3 )~i + (− 16 t 2

  • 20t + C 4 )~j. At t = 0,

~r(0) = C 3 ~i + C 4 ~j. But this should be 176~j. Therefore, C 3 = 0, C 4 = 176, and thus

~r(t) = 15t~i + (− 16 t 2

  • 20t + 176)~j.

(c) What is the maximum height of the projectile?

To answer this question, we set the vertical component of velocity equal to zero. So, − 32 t + 20 =

0 ⇒ 32 t = 20 ⇒ t =

= 0.625 seconds. Next we plug this time in for the vertical component

of position and get −16 (5/8)

2

  • 20 (5/8) + 176 = −

Therefore, the maximum height is

= 182.25 feet.

(d) At what time does the projectile hit the ground?

This happens when the vertical component of the position is zero. That is, when

− 16 t

2

  • 20t + 176 = 0 ⇒ 4 t

2 − 5 t − 44 = 0. You can use the quadratic formula to find t, but luckily

4 t 2 − 5 t − 44 factors into (4t + 11)(t − 4). Thus, 4t + 11 = 0 or t − 4 = 0, which implies t = −

or t = 4. Since t = −

is preposterous, we must have t = 4. So the projectile hits the ground 4

seconds after it is shot.

(e) With what speed does the projectile come smashing into the basin below the cliff?

First, we need to find the velocity at t = 4. ~v(4) = 15

i + (−32(4) + 20)

j = 15

i − 108

j. Then, the

speed is

~v(4)

2

  • 108 2 =

11 , 889 ≈ 109 .037 feet per second.

  1. Let ~r(t) = t~i + t

j +

t

k.

(a) Find T~ (1).

~v(t) = ~i + 2t~j + 2t

k. Then ~v(1) = ~i + 2~j + 2~k. Therefore,

T^ ~ (1) =

i + 2

j + 2

k √ 1 + 4 + 4

i + 2

j + 2

k √ 9

~i +

~j +

~k =

(b) Find N~ (1).

~a(t) = 2~j + 4t~k. Then ~a(1) = 2~j + 4~k. Therefore,

aT (1) = T~ (1) · ~a(1) =

Note that |~a(1)|

2 = 2 2

  • 4 2 = 20. Using the Pythagorean Identity |~a(1)|

2 = a 2 T (1) + a 2 N (1), we have

aN (1) =

2

Then, since ~a(1) = aT (1) T~ (1) + aN (1) N~ (1), we can compute N (1) as follows:

N^ ~ (1) =

~a(1) − aT (1) T~ (1)

aN (1)

(c) Find the curvature, κ at t = 1.

κ(1) =

∣~v(1) × ~a(1)

~v(1)

3

. But aN (1) =

∣~v(1) × ~a(1)

~v(1)

. Thus, we can use our calculation for aN (1) in

part (b) to get κ =

aN (1) ∣ ∣ ~v(1)

2

2

  • 2 2

(d) Find the binormal B~(1).

B(1) =

T (1) ×

N (1) =

i

j

k

1 / 3 2 / 3 2 / 3

− 2 / 3 − 1 / 3 2 / 3