Chapter 11 Vector Functions and Curves

11.1 Vector Functions of One Variable

1) Let r(t) = 4t i + 3sin(t) j – 3cos(t) k be a differentiable vector function giving the position r of a particle at time t. Find the speed of the particle at time t = seconds.

A) 5.5 units/s

B) 5.0 units/s

C) 4.5 units/s

D) 4.0 units/s

E) 3.5 units/s

Answer: B

Diff: 1

2) Find the velocity at time t = -2 of a particle whose position at time t is given by r(t) = 3t i – j.

A) v(-2) = – 3i + 12j

B) v(-2) =- 3i – 12j

C) v(-2) =3i + 12j

D) v(-2) = 3i – 12j

E) v(-2) = 3i – 4j

Answer: D

Diff: 1

3) Find the acceleration at time t = -1 of a particle whose position at time t is given by r(t) = 4t i + j.

A) a(-1) = -2j

B) a(-1) = 2j

C) a(-1) = i – 2j

D) a(-1) = i + 2j

E) a(-1) = 2i + j

Answer: B

Diff: 1

4) A moving particle starts at an initial position (1, 0, 0) with initial velocity i – j + k. Its acceleration at time t is a(t) = 4t i + 6t j + k. Find its velocity and position at time t > 0.

A) v(t) = (2 + 1) i + (3 – 1) j + (t + 1) k; r(t) = ( + t + 1) i + ( – t) j + ( + t) k

B) v(t) = (2 + 1) i + (3 + 2) j + (t + 1) k; r(t) = ( + t + 1) i + ( + 2t) j + ( + t) k

C) v(t) = (2 + 1) i + (3 – 1) j + (t + 1) k; r(t) = ( + t + 1) i + ( – t) j + ( + t) k

D) v(t) = (2 + 1) i + (3 – 1) j + (t + 1) k; r(t) = ( + t + 1) i + ( – t) j – ( + t) k

E) v(t) = (2 – 1) i + (3 – 1) j – (t + 1) k; r(t) = ( + t + 1) i – ( – t) j + ( + t) k

Answer: A

Diff: 2

5) Find the velocity, speed, and acceleration at time t of a particle that has position function

r(t) = (2sin t) i + 6t j + (2cos t) k.

A) v(t) = (cos t) i + 6j – (2sin t) k; = ; a(t) = – (2sin t) i – (2cos t) k

B) v(t) = -(cos t) i + 6j + (2sin t) k; = ; a(t) = (2sin t) i + (2cos t) k

C) v(t) = (cos t) i + 6j – (2sin t) k; = 10; a(t) = – (2sin t) i – (2cos t) k

D) v(t) = (cos t) i + (2sin t) k; =2 ; a(t) = – (2sin t) i – (2cos t) k

E) v(t) = -(cos t) i + 6j + (2sin t) k; = 10; a(t) = (2sin t) i + (2cos t) k

Answer: A

Diff: 2

6) Find the velocity, speed, and acceleration at time t of a particle that has position function

r(t) = t i + j + k.

A) v(t) = i + j – k, + , a(t) = j + k

B) v(t) = i + j – k, – , a(t) = j + k

C) v(t) = i + j – k, + , a(t) = j – k

D) v(t) = j – k, , a(t) = j + k

E) v(t) = i + j + k, + , a(t) = j – k

Answer: A

Diff: 2

7) What kind of curve is r(t) = (sin t) i + (sin t) j + 2(cos t) k?

A) a circular helix

B) an oval plane curve that is, however, not an ellipse

C) an ellipse (but not a circle)

D) a circle

E) a parabola

Answer: D

Diff: 2

8) Describe the curve r(t) = (sin t) i + (cos t) j + k.

A) a helix wound around the cylinder + = 1

B) the circle (of radius 1) in which the plane z = intersects the sphere + + = 4

C) the circle (of radius ) in which the plane z = 1 intersects the sphere + + = 3

D) the circle (of radius ) in which the plane z = intersects the cylinder + = 1

E) a helix wound around the cylinder + = 3

Answer: B

Diff: 2

9) Find an equation of the line tangent to the parametric space curve

r(t) = ( + 3t + 1) i + (2 – 7t j + (4sin(t) -3)k at the point on the curve corresponding to t = 0.

A) r(u) = (1 + 3u) i + (2 – 7u) j + (- 3 + 4s) k, u ∈ R

B) 3x – 7y + 4z -23 = 0

C) x + y + z = 0

D) r(u) = (3 + u) i + (-7 + 2u) j + (4 – 3u) k, u ∈ R

E) x + 2y – 3z – 14 = 0

Answer: A

Diff: 2

10) A particle is moving to the right with constant speed 2 along the curve y = cos x in the xy-plane.

Find its velocity at the instant when it crosses the vertical line x = .

A) v = i – j

B) v = i – j

C) v = i – j

D) v = i – j

E) v = i + j

Answer: C

Diff: 2

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