Question 1

In the discussion of Kepler's Second Law of Planetary Motion, the expression $\frac { d A } { d t }$ can be expressed as

Accepted Answer

A)  $\frac { 1 } { 2 } \| \mathbf { r } ( t ) \times \mathbf { v } ( t ) \|$ 
B)  $\| \mathbf { r } ( t ) \times \mathbf { v } ( t ) \|$ 
C)  $\frac { 1 } { 2 } \| \mathbf { r } ( t ) \bullet \mathbf { v } ( t ) \|$ 
D)  $\| \mathbf { r } ( t ) \bullet \mathbf { v } ( t ) \|$ 
E)  $\mathbf { r } ( t ) \bullet \mathbf { v } ( t )$ 
A)  $\frac { 1 } { 2 } \| \mathbf { r } ( t ) \times \mathbf { v } ( t ) \|$ 
B)  $\| \mathbf { r } ( t ) \times \mathbf { v } ( t ) \|$ 
C)  $\frac { 1 } { 2 } \| \mathbf { r } ( t ) \bullet \mathbf { v } ( t ) \|$ 
D)  $\| \mathbf { r } ( t ) \bullet \mathbf { v } ( t ) \|$ 
E)  $\mathbf { r } ( t ) \bullet \mathbf { v } ( t )$

Question 2

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = \sin ( 2 t ) \mathbf { i } - \cos ( 2 t ) \mathbf { j } + 2 t \mathbf { k }$ is

Accepted Answer

A)  $\sqrt { 56 }$ 
B)  $\sqrt { 8 t ^ { 2 } + 1 }$ 
C) 0 
D)  $\sqrt { 65 }$ 
E)  $\frac { 8 t } { \sqrt { 8 t ^ { 2 } + 1 } }$ 
A)  $\sqrt { 56 }$ 
B)  $\sqrt { 8 t ^ { 2 } + 1 }$ 
C) 0 
D)  $\sqrt { 65 }$ 
E)  $\frac { 8 t } { \sqrt { 8 t ^ { 2 } + 1 } }$

Question 3

Let $\mathbf { r } ( t ) = \sin ( 2 t ) \mathbf { i } - \cos ( 2 t ) \mathbf { j } + 5 t \mathbf { k }$ Then the unit tangent vector $\mathbf { T } \left( \frac { \pi } { 4 } \right)$ is

Accepted Answer

A)  $\frac { 5 } { \sqrt { 29 } } \mathbf { j } + \frac { 2 } { \sqrt { 29 } } \mathbf { k }$ 
B)  $- \frac { 2 } { \sqrt { 29 } } \mathbf { j } - \frac { 5 } { \sqrt { 29 } } \mathbf { k }$ 
C)  $\frac { 2 } { \sqrt { 29 } } \mathbf { j } + \frac { 5 } { \sqrt { 29 } } \mathbf { k }$ 
D)  $\frac { 2 } { \sqrt { 29 } } \mathbf { j } - \frac { 5 } { \sqrt { 29 } } \mathbf { k }$ 
E)  $- \frac { 2 } { \sqrt { 29 } } \mathbf { j } + \frac { 5 } { \sqrt { 29 } } \mathbf { k }$ 
A)  $\frac { 5 } { \sqrt { 29 } } \mathbf { j } + \frac { 2 } { \sqrt { 29 } } \mathbf { k }$ 
B)  $- \frac { 2 } { \sqrt { 29 } } \mathbf { j } - \frac { 5 } { \sqrt { 29 } } \mathbf { k }$ 
C)  $\frac { 2 } { \sqrt { 29 } } \mathbf { j } + \frac { 5 } { \sqrt { 29 } } \mathbf { k }$ 
D)  $\frac { 2 } { \sqrt { 29 } } \mathbf { j } - \frac { 5 } { \sqrt { 29 } } \mathbf { k }$ 
E)  $- \frac { 2 } { \sqrt { 29 } } \mathbf { j } + \frac { 5 } { \sqrt { 29 } } \mathbf { k }$

Question 4

Let $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } - \sin ( 2 t ) \mathbf { j } + 6 t \mathbf { k }$ Then $\mathbf { r } ^ { \prime } \left( \frac { \pi } { 6 } \right)$ is

Accepted Answer

A)  $\sqrt { 3 } \mathbf { i } - \mathbf { j } - 6 \mathbf { k }$ 
B)  $\sqrt { 3 } \mathbf { i } - \mathbf { j } + 6 \mathbf { k }$ 
C)  $- \sqrt { 3 } \mathbf { i } + \mathbf { j } + 6 \mathbf { k }$ 
D)  $- \sqrt { 3 } \mathbf { i } - \mathbf { j } + 6 \mathbf { k }$ 
E)  $- \sqrt { 3 } \mathbf { i } - \mathbf { j } - 6 \mathbf { k }$ 
A)  $\sqrt { 3 } \mathbf { i } - \mathbf { j } - 6 \mathbf { k }$ 
B)  $\sqrt { 3 } \mathbf { i } - \mathbf { j } + 6 \mathbf { k }$ 
C)  $- \sqrt { 3 } \mathbf { i } + \mathbf { j } + 6 \mathbf { k }$ 
D)  $- \sqrt { 3 } \mathbf { i } - \mathbf { j } + 6 \mathbf { k }$ 
E)  $- \sqrt { 3 } \mathbf { i } - \mathbf { j } - 6 \mathbf { k }$

Question 5

The domain of the vector function $\mathbf { v } ( t ) = e ^ { t } \mathbf { i } - \sqrt { 1 - t ^ { 2 } } \mathbf { j } + 8 \mathbf { k }$ is

Accepted Answer

A)  $[ - 1,1 )$ 
B)  $( - 1,1 )$ 
C)  $[ - 1,1 ]$ 
D)  $( - 1,1 ]$ 
E)  $( - \infty , \infty )$ 
A)  $[ - 1,1 )$ 
B)  $( - 1,1 )$ 
C)  $[ - 1,1 ]$ 
D)  $( - 1,1 ]$ 
E)  $( - \infty , \infty )$

Question 6

The solution $\mathbf { r } ( t )$ to the differential equation $\mathbf { r } ^ { \prime } ( t ) = 3 t ^ { 2 } \mathbf { i } - \frac { 1 } { t - 3 } \mathbf { j } + 4 \mathbf { k }$ with the condition $\mathbf { r } ( 4 ) = \mathbf { i } - \mathbf { j } + 3 \mathbf { k }$ is

Accepted Answer

A)  $- \left( t ^ { 3 } - 63 \right) \mathbf { i } - ( \ln | t - 3 | + 1 ) \mathbf { j } + ( 4 t - 13 ) \mathbf { k }$ 
B)  $\left( t ^ { 3 } - 63 \right) \mathbf { i } + ( \ln | t - 3 | + 1 ) \mathbf { j } - ( 4 t - 13 ) \mathbf { k }$ 
C)  $\left( t ^ { 3 } - 63 \right) \mathbf { i } - ( \ln | t - 3 | + 1 ) \mathbf { j } - ( 4 t - 13 ) \mathbf { k }$ 
D)  $\left( t ^ { 3 } - 63 \right) \mathbf { i } - ( \ln | t - 3 | + 1 ) \mathbf { j } + ( 4 t - 13 ) \mathbf { k }$ 
E)  $\left( t ^ { 3 } - 63 \right) \mathbf { i } + ( \ln | t - 3 | + 1 ) \mathbf { j } + ( 4 t - 13 ) \mathbf { k }$ 
A)  $- \left( t ^ { 3 } - 63 \right) \mathbf { i } - ( \ln | t - 3 | + 1 ) \mathbf { j } + ( 4 t - 13 ) \mathbf { k }$ 
B)  $\left( t ^ { 3 } - 63 \right) \mathbf { i } + ( \ln | t - 3 | + 1 ) \mathbf { j } - ( 4 t - 13 ) \mathbf { k }$ 
C)  $\left( t ^ { 3 } - 63 \right) \mathbf { i } - ( \ln | t - 3 | + 1 ) \mathbf { j } - ( 4 t - 13 ) \mathbf { k }$ 
D)  $\left( t ^ { 3 } - 63 \right) \mathbf { i } - ( \ln | t - 3 | + 1 ) \mathbf { j } + ( 4 t - 13 ) \mathbf { k }$ 
E)  $\left( t ^ { 3 } - 63 \right) \mathbf { i } + ( \ln | t - 3 | + 1 ) \mathbf { j } + ( 4 t - 13 ) \mathbf { k }$

Question 7

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = 2 t \mathbf { i } - \sin t \mathbf { j } + \cos t \mathbf { k }$ is

Accepted Answer

A)  $\sqrt { 56 }$ 
B)  $\sqrt { 65 }$ 
C) 0 
D)  $\sqrt { 73 }$ 
E)  $\sqrt { 78 }$ 
A)  $\sqrt { 56 }$ 
B)  $\sqrt { 65 }$ 
C) 0 
D)  $\sqrt { 73 }$ 
E)  $\sqrt { 78 }$

Question 8

The Geocentric Theory of Planetary Motion states that
​

Accepted Answer

A) The planets travel in circular orbits with the sun at the center. 
B) The sun, moon, and planets travel in circular orbits with Earth at the center. 
C) The planets travel in circular orbits with the moon at the center. 
D) The planets travel in elliptical orbits with the sun at a focus. 
E) The planets travel in elliptical orbits with Earth at a focus. 
A) The planets travel in circular orbits with the sun at the center. 
B) The sun, moon, and planets travel in circular orbits with Earth at the center. 
C) The planets travel in circular orbits with the moon at the center. 
D) The planets travel in elliptical orbits with the sun at a focus. 
E) The planets travel in elliptical orbits with Earth at a focus.

Question 9

The domain of the vector function $\mathbf { v } ( t ) = 2 ^ { t } \mathbf { i } - 5 t ^ { 2 } \mathbf { j } + \log _ { 2 } ( t - 4 ) \mathbf { k }$ is

Accepted Answer

A)  $( 4 , \infty )$ 
B)  $( - 4 , \infty )$ 
C)  $( - \infty , - 4 )$ 
D)  $( - \infty , - 4 ) \cup ( 4 , \infty )$ 
E)  $( - 4,4 )$ 
A)  $( 4 , \infty )$ 
B)  $( - 4 , \infty )$ 
C)  $( - \infty , - 4 )$ 
D)  $( - \infty , - 4 ) \cup ( 4 , \infty )$ 
E)  $( - 4,4 )$

Question 10

Let $\mathbf { a } ( t ) = - 32 \mathbf { k }$ be the acceleration of an object with velocity $\mathbf { v } ( t )$ . If $\mathbf { v } ( 0 ) = 0 \mathbf { i } + 0 \mathbf { j } + 0 \mathbf { k }$ , then $\mathbf { v } ( t )$ is

Accepted Answer

A)  $32 t \mathbf { k }$ 
B)  $- 32 t \mathbf { k }$ 
C)  $- 32 t \mathbf { i }$ 
D)  $- 32 t \mathbf { j }$ 
E)  $32 t \mathrm { j }$ 
A)  $32 t \mathbf { k }$ 
B)  $- 32 t \mathbf { k }$ 
C)  $- 32 t \mathbf { i }$ 
D)  $- 32 t \mathbf { j }$ 
E)  $32 t \mathrm { j }$

Question 11

The magnitude of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = 2 t \mathbf { i } - \cos ( 3 t ) \mathbf { j } + \sin ( 3 t ) \mathbf { k }$ is

Accepted Answer

A)  $2 \sqrt { 2 }$ 
B)  $\sqrt { 11 }$ 
C)  $4 \sqrt { 2 }$ 
D)  $\sqrt { 5 }$ 
E) 9 
A)  $2 \sqrt { 2 }$ 
B)  $\sqrt { 11 }$ 
C)  $4 \sqrt { 2 }$ 
D)  $\sqrt { 5 }$ 
E) 9

Question 12

Let $y = \sqrt { x }$ . Then the curvature K at (4, 2) is

Accepted Answer

A)  $\frac { 2 } { 25 \sqrt { 5 } }$ 
B)  $\frac { \sqrt { 2 } } { 11 }$ 
C)  $\frac { 2 } { 5 \sqrt { 5 } }$ 
D)  $\frac { 2 } { 17 \sqrt { 17 } }$ 
E)  $\frac { 1 } { 2 }$ 
A)  $\frac { 2 } { 25 \sqrt { 5 } }$ 
B)  $\frac { \sqrt { 2 } } { 11 }$ 
C)  $\frac { 2 } { 5 \sqrt { 5 } }$ 
D)  $\frac { 2 } { 17 \sqrt { 17 } }$ 
E)  $\frac { 1 } { 2 }$

Question 13

Nicolaus Copernicus's Heliocentric Theory of Planetary Motion states that
​

Accepted Answer

A) The planets travel in circular orbits with the sun at the center. 
B) The planets travel in circular orbits with Earth at the center. 
C) The planets travel in circular orbits with the moon at the center. 
D) The planets travel in elliptical orbits with the sun at a focus. 
E) The planets travel in elliptical orbits with Earth at a focus. 
A) The planets travel in circular orbits with the sun at the center. 
B) The planets travel in circular orbits with Earth at the center. 
C) The planets travel in circular orbits with the moon at the center. 
D) The planets travel in elliptical orbits with the sun at a focus. 
E) The planets travel in elliptical orbits with Earth at a focus.

Question 14

The radius of curvature of $y = x ^ { 3 } - 2 x$ at (2, 4) is

Accepted Answer

A) 1 
B)  $\frac { 1 } { 2 \sqrt { 2 } }$ 
C)  $\frac { 2 } { 5 \sqrt { 5 } }$ 
D)  $\frac { 101 \sqrt { 101 } } { 12 }$ 
E)  $\frac { 7 } { 125 }$ 
A) 1 
B)  $\frac { 1 } { 2 \sqrt { 2 } }$ 
C)  $\frac { 2 } { 5 \sqrt { 5 } }$ 
D)  $\frac { 101 \sqrt { 101 } } { 12 }$ 
E)  $\frac { 7 } { 125 }$

Question 15

Let $\mathbf { a } ( t ) = - 32 \mathbf { k }$ be the acceleration of an object with velocity $\mathbf { v } ( t )$ . If $\mathbf { v } ( 0 ) = \mathbf { i } + 2 \mathbf { j } + 0 \mathbf { k }$ then the speed $\| \mathbf { v } ( t ) \|$ is

Accepted Answer

A)  $\sqrt { 1024 t ^ { 2 } + 5 }$ 
B)  $\sqrt { 1024 t ^ { 2 } - 5 }$ 
C)  $\sqrt { 1024 t ^ { 2 } + 25 }$ 
D)  $\sqrt { 1024 t ^ { 2 } - 25 }$ 
E)  $\sqrt { 1024 t ^ { 2 } + 125 }$ 
A)  $\sqrt { 1024 t ^ { 2 } + 5 }$ 
B)  $\sqrt { 1024 t ^ { 2 } - 5 }$ 
C)  $\sqrt { 1024 t ^ { 2 } + 25 }$ 
D)  $\sqrt { 1024 t ^ { 2 } - 25 }$ 
E)  $\sqrt { 1024 t ^ { 2 } + 125 }$

Question 16

Let $\mathbf { v } ( t ) = 5 t \mathbf { i } - 2 t ^ { 2 } \mathbf { j } - \mathbf { k }$ and $\mathbf { w } ( t ) = - 4 t \mathbf { i } + t ^ { 2 } \mathbf { j } - 3 \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$ is

Accepted Answer

A)  $- 14 t \mathbf { i } + 19 \mathbf { j } - 9 t ^ { 2 } \mathbf { k }$ 
B)  $14 t \mathbf { i } - 19 \mathbf { j } + 9 t ^ { 2 } \mathbf { k }$ 
C)  $14 t \mathbf { i } + 19 \mathbf { j } + 9 t ^ { 2 } \mathbf { k }$ 
D)  $14 t \mathbf { i } - 19 \mathbf { j } - 9 t ^ { 2 } \mathbf { k }$ 
E)  $14 t \mathbf { i } + 19 \mathbf { j } - 9 t ^ { 2 } \mathbf { k }$ 
A)  $- 14 t \mathbf { i } + 19 \mathbf { j } - 9 t ^ { 2 } \mathbf { k }$ 
B)  $14 t \mathbf { i } - 19 \mathbf { j } + 9 t ^ { 2 } \mathbf { k }$ 
C)  $14 t \mathbf { i } + 19 \mathbf { j } + 9 t ^ { 2 } \mathbf { k }$ 
D)  $14 t \mathbf { i } - 19 \mathbf { j } - 9 t ^ { 2 } \mathbf { k }$ 
E)  $14 t \mathbf { i } + 19 \mathbf { j } - 9 t ^ { 2 } \mathbf { k }$

Question 17

The fact that $\mathbf { r } ( t ) \times \mathbf { v } ( t )$ is constant implies that

Accepted Answer

A) The speed of a planet around the sun is a constant. 
B) The motion of the planet lies in a plane. 
C) The speed of the sun is a constant. 
D) The force between the planet and the sun is a constant. 
E) The position of the planet is unrelated to the sun. 
A) The speed of a planet around the sun is a constant. 
B) The motion of the planet lies in a plane. 
C) The speed of the sun is a constant. 
D) The force between the planet and the sun is a constant. 
E) The position of the planet is unrelated to the sun.

Question 18

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = ( t + 2 ) \mathbf { i } + 2 t \mathbf { j } + 2 \mathbf { k }$ is

Accepted Answer

A)  $\sqrt { 56 }$ 
B)  $\sqrt { 8 t ^ { 2 } + 1 }$ 
C) 0 
D)  $\sqrt { 65 }$ 
E)  $\frac { 8 t } { \sqrt { 8 t ^ { 2 } + 1 } }$ 
A)  $\sqrt { 56 }$ 
B)  $\sqrt { 8 t ^ { 2 } + 1 }$ 
C) 0 
D)  $\sqrt { 65 }$ 
E)  $\frac { 8 t } { \sqrt { 8 t ^ { 2 } + 1 } }$

Question 19

The solution $\mathbf { r } ( t )$ to the differential equation $\mathbf { r } ^ { \prime } ( t ) = \sin t \mathbf { i } + \cos t \mathbf { j } + 3 \mathbf { k }$ with the condition $\mathbf { r } ( 0 ) = \mathbf { i } + \mathbf { j }$ is

Accepted Answer

A)  $( \cos t + 2 ) \mathbf { i } - ( \sin t + 1 ) \mathbf { j } + 3 t \mathbf { k }$ 
B)  $( - \cos t + 2 ) \mathbf { i } - ( \sin t + 1 ) \mathbf { j } + 3 t \mathbf { k }$ 
C)  $( \cos t + 2 ) \mathbf { i } + ( \sin t + 1 ) \mathbf { j } + 3 t \mathbf { k }$ 
D)  $( - \cos t + 2 ) \mathbf { i } + ( \sin t + 1 ) \mathbf { j } + 3 t \mathbf { k }$ 
E)  $( \cos t + 2 ) \mathbf { i } + ( \sin t - 1 ) \mathbf { j } + 3 t \mathbf { k }$ 
A)  $( \cos t + 2 ) \mathbf { i } - ( \sin t + 1 ) \mathbf { j } + 3 t \mathbf { k }$ 
B)  $( - \cos t + 2 ) \mathbf { i } - ( \sin t + 1 ) \mathbf { j } + 3 t \mathbf { k }$ 
C)  $( \cos t + 2 ) \mathbf { i } + ( \sin t + 1 ) \mathbf { j } + 3 t \mathbf { k }$ 
D)  $( - \cos t + 2 ) \mathbf { i } + ( \sin t + 1 ) \mathbf { j } + 3 t \mathbf { k }$ 
E)  $( \cos t + 2 ) \mathbf { i } + ( \sin t - 1 ) \mathbf { j } + 3 t \mathbf { k }$

Question 20

The solution $\mathbf { r } ( t )$ to the differential equation $\mathbf { r } ^ { \prime } ( t ) = t \mathbf { i } + e ^ { - t } \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$ with the condition $\mathbf { r } ( 1 ) = \mathbf { i } - \mathbf { j } + 3 \mathbf { k }$ is

Accepted Answer

A)  $- \left( e ^ { t } - e \right) \mathbf { i } - \left( \frac { 3 } { 2 } t ^ { 2 } - \frac { 5 } { 2 } \right) \mathbf { j } + t ^ { 3 } \mathbf { k }$ 
B)  $- \left( e ^ { t } - e \right) \mathbf { i } + \left( \frac { 3 } { 2 } t ^ { 2 } - \frac { 5 } { 2 } \right) \mathbf { j } + t ^ { 3 } \mathbf { k }$ 
C)  $\left( e ^ { t } - e \right) \mathbf { i } - \left( \frac { 3 } { 2 } t ^ { 2 } - \frac { 5 } { 2 } \right) \mathbf { j } + t ^ { 3 } \mathbf { k }$ 
D)  $\left( e ^ { t } - e \right) \mathbf { i } + \left( \frac { 3 } { 2 } t ^ { 2 } - \frac { 5 } { 2 } \right) \mathbf { j } + t ^ { 3 } \mathbf { k }$ 
E)  $\left( e ^ { t } - e \right) \mathbf { i } - \left( \frac { 3 } { 2 } t ^ { 2 } - \frac { 5 } { 2 } \right) \mathbf { j } - t ^ { 3 } \mathbf { k }$ 
A)  $- \left( e ^ { t } - e \right) \mathbf { i } - \left( \frac { 3 } { 2 } t ^ { 2 } - \frac { 5 } { 2 } \right) \mathbf { j } + t ^ { 3 } \mathbf { k }$ 
B)  $- \left( e ^ { t } - e \right) \mathbf { i } + \left( \frac { 3 } { 2 } t ^ { 2 } - \frac { 5 } { 2 } \right) \mathbf { j } + t ^ { 3 } \mathbf { k }$ 
C)  $\left( e ^ { t } - e \right) \mathbf { i } - \left( \frac { 3 } { 2 } t ^ { 2 } - \frac { 5 } { 2 } \right) \mathbf { j } + t ^ { 3 } \mathbf { k }$ 
D)  $\left( e ^ { t } - e \right) \mathbf { i } + \left( \frac { 3 } { 2 } t ^ { 2 } - \frac { 5 } { 2 } \right) \mathbf { j } + t ^ { 3 } \mathbf { k }$ 
E)  $\left( e ^ { t } - e \right) \mathbf { i } - \left( \frac { 3 } { 2 } t ^ { 2 } - \frac { 5 } { 2 } \right) \mathbf { j } - t ^ { 3 } \mathbf { k }$

In the discussion of Kepler's Second Law of Planetary Motion, the expression $\frac { d A } { d t }$ can be expressed as

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = \sin ( 2 t ) \mathbf { i } - \cos ( 2 t ) \mathbf { j } + 2 t \mathbf { k }$ is

Let $\mathbf { r } ( t ) = \sin ( 2 t ) \mathbf { i } - \cos ( 2 t ) \mathbf { j } + 5 t \mathbf { k }$ Then the unit tangent vector $\mathbf { T } \left( \frac { \pi } { 4 } \right)$ is

Let $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } - \sin ( 2 t ) \mathbf { j } + 6 t \mathbf { k }$ Then $\mathbf { r } ^ { \prime } \left( \frac { \pi } { 6 } \right)$ is

The domain of the vector function $\mathbf { v } ( t ) = e ^ { t } \mathbf { i } - \sqrt { 1 - t ^ { 2 } } \mathbf { j } + 8 \mathbf { k }$ is

The solution $\mathbf { r } ( t )$ to the differential equation $\mathbf { r } ^ { \prime } ( t ) = 3 t ^ { 2 } \mathbf { i } - \frac { 1 } { t - 3 } \mathbf { j } + 4 \mathbf { k }$ with the condition $\mathbf { r } ( 4 ) = \mathbf { i } - \mathbf { j } + 3 \mathbf { k }$ is

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = 2 t \mathbf { i } - \sin t \mathbf { j } + \cos t \mathbf { k }$ is

The Geocentric Theory of Planetary Motion states that

The domain of the vector function $\mathbf { v } ( t ) = 2 ^ { t } \mathbf { i } - 5 t ^ { 2 } \mathbf { j } + \log _ { 2 } ( t - 4 ) \mathbf { k }$ is

Let $\mathbf { a } ( t ) = - 32 \mathbf { k }$ be the acceleration of an object with velocity $\mathbf { v } ( t )$ . If $\mathbf { v } ( 0 ) = 0 \mathbf { i } + 0 \mathbf { j } + 0 \mathbf { k }$ , then $\mathbf { v } ( t )$ is

The magnitude of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = 2 t \mathbf { i } - \cos ( 3 t ) \mathbf { j } + \sin ( 3 t ) \mathbf { k }$ is

Let $y = \sqrt { x }$ . Then the curvature K at (4, 2) is

Nicolaus Copernicus's Heliocentric Theory of Planetary Motion states that

The radius of curvature of $y = x ^ { 3 } - 2 x$ at (2, 4) is

Let $\mathbf { a } ( t ) = - 32 \mathbf { k }$ be the acceleration of an object with velocity $\mathbf { v } ( t )$ . If $\mathbf { v } ( 0 ) = \mathbf { i } + 2 \mathbf { j } + 0 \mathbf { k }$ then the speed $\| \mathbf { v } ( t ) \|$ is

Let $\mathbf { v } ( t ) = 5 t \mathbf { i } - 2 t ^ { 2 } \mathbf { j } - \mathbf { k }$ and $\mathbf { w } ( t ) = - 4 t \mathbf { i } + t ^ { 2 } \mathbf { j } - 3 \mathbf { k }$ Then $\frac { d } { d t } ( \mathbf { v } ( t ) \times \mathbf { w } ( t ) )$ is

The fact that $\mathbf { r } ( t ) \times \mathbf { v } ( t )$ is constant implies that

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = ( t + 2 ) \mathbf { i } + 2 t \mathbf { j } + 2 \mathbf { k }$ is

The solution $\mathbf { r } ( t )$ to the differential equation $\mathbf { r } ^ { \prime } ( t ) = \sin t \mathbf { i } + \cos t \mathbf { j } + 3 \mathbf { k }$ with the condition $\mathbf { r } ( 0 ) = \mathbf { i } + \mathbf { j }$ is

The solution $\mathbf { r } ( t )$ to the differential equation $\mathbf { r } ^ { \prime } ( t ) = t \mathbf { i } + e ^ { - t } \mathbf { j } - \frac { 1 } { t ^ { 2 } } \mathbf { k }$ with the condition $\mathbf { r } ( 1 ) = \mathbf { i } - \mathbf { j } + 3 \mathbf { k }$ is

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Exam 12: Vector Functions

In the discussion of Kepler's Second Law of Planetary Motion, the expression dAdt\frac { d A } { d t }dtdA​ can be expressed as

The tangential component aT(t)a _ { \mathrm { T } } ( t )aT​(t) of the acceleration of a particle moving along the curve r(t)=sin⁡(2t)i−cos⁡(2t)j+2tk\mathbf { r } ( t ) = \sin ( 2 t ) \mathbf { i } - \cos ( 2 t ) \mathbf { j } + 2 t \mathbf { k }r(t)=sin(2t)i−cos(2t)j+2tk is

Let r(t)=sin⁡(2t)i−cos⁡(2t)j+5tk\mathbf { r } ( t ) = \sin ( 2 t ) \mathbf { i } - \cos ( 2 t ) \mathbf { j } + 5 t \mathbf { k }r(t)=sin(2t)i−cos(2t)j+5tk Then the unit tangent vector T(π4)\mathbf { T } \left( \frac { \pi } { 4 } \right)T(4π​) is

Let r(t)=cos⁡(2t)i−sin⁡(2t)j+6tk\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } - \sin ( 2 t ) \mathbf { j } + 6 t \mathbf { k }r(t)=cos(2t)i−sin(2t)j+6tk Then r′(π6)\mathbf { r } ^ { \prime } \left( \frac { \pi } { 6 } \right)r′(6π​) is

The domain of the vector function v(t)=eti−1−t2j+8k\mathbf { v } ( t ) = e ^ { t } \mathbf { i } - \sqrt { 1 - t ^ { 2 } } \mathbf { j } + 8 \mathbf { k }v(t)=eti−1−t2​j+8k is

The tangential component aT(t)a _ { \mathrm { T } } ( t )aT​(t) of the acceleration of a particle moving along the curve r(t)=2ti−sin⁡tj+cos⁡tk\mathbf { r } ( t ) = 2 t \mathbf { i } - \sin t \mathbf { j } + \cos t \mathbf { k }r(t)=2ti−sintj+costk is

The Geocentric Theory of Planetary Motion states that ​

The domain of the vector function v(t)=2ti−5t2j+log⁡2(t−4)k\mathbf { v } ( t ) = 2 ^ { t } \mathbf { i } - 5 t ^ { 2 } \mathbf { j } + \log _ { 2 } ( t - 4 ) \mathbf { k }v(t)=2ti−5t2j+log2​(t−4)k is

Let a(t)=−32k\mathbf { a } ( t ) = - 32 \mathbf { k }a(t)=−32k be the acceleration of an object with velocity v(t)\mathbf { v } ( t )v(t) . If v(0)=0i+0j+0k\mathbf { v } ( 0 ) = 0 \mathbf { i } + 0 \mathbf { j } + 0 \mathbf { k }v(0)=0i+0j+0k , then v(t)\mathbf { v } ( t )v(t) is

The magnitude of the acceleration of a particle moving along the curve r(t)=2ti−cos⁡(3t)j+sin⁡(3t)k\mathbf { r } ( t ) = 2 t \mathbf { i } - \cos ( 3 t ) \mathbf { j } + \sin ( 3 t ) \mathbf { k }r(t)=2ti−cos(3t)j+sin(3t)k is

Let y=xy = \sqrt { x }y=x​ . Then the curvature K at (4, 2) is

Nicolaus Copernicus's Heliocentric Theory of Planetary Motion states that ​

The radius of curvature of y=x3−2xy = x ^ { 3 } - 2 xy=x3−2x at (2, 4) is

The fact that r(t)×v(t)\mathbf { r } ( t ) \times \mathbf { v } ( t )r(t)×v(t) is constant implies that

The tangential component aT(t)a _ { \mathrm { T } } ( t )aT​(t) of the acceleration of a particle moving along the curve r(t)=(t+2)i+2tj+2k\mathbf { r } ( t ) = ( t + 2 ) \mathbf { i } + 2 t \mathbf { j } + 2 \mathbf { k }r(t)=(t+2)i+2tj+2k is

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In the discussion of Kepler's Second Law of Planetary Motion, the expression $\frac { d A } { d t }$ can be expressed as

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = \sin ( 2 t ) \mathbf { i } - \cos ( 2 t ) \mathbf { j } + 2 t \mathbf { k }$ is

Let $\mathbf { r } ( t ) = \sin ( 2 t ) \mathbf { i } - \cos ( 2 t ) \mathbf { j } + 5 t \mathbf { k }$ Then the unit tangent vector $\mathbf { T } \left( \frac { \pi } { 4 } \right)$ is

Let $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } - \sin ( 2 t ) \mathbf { j } + 6 t \mathbf { k }$ Then $\mathbf { r } ^ { \prime } \left( \frac { \pi } { 6 } \right)$ is

The domain of the vector function $\mathbf { v } ( t ) = e ^ { t } \mathbf { i } - \sqrt { 1 - t ^ { 2 } } \mathbf { j } + 8 \mathbf { k }$ is

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = 2 t \mathbf { i } - \sin t \mathbf { j } + \cos t \mathbf { k }$ is

The Geocentric Theory of Planetary Motion states that

The domain of the vector function $\mathbf { v } ( t ) = 2 ^ { t } \mathbf { i } - 5 t ^ { 2 } \mathbf { j } + \log _ { 2 } ( t - 4 ) \mathbf { k }$ is

Let $\mathbf { a } ( t ) = - 32 \mathbf { k }$ be the acceleration of an object with velocity $\mathbf { v } ( t )$ . If $\mathbf { v } ( 0 ) = 0 \mathbf { i } + 0 \mathbf { j } + 0 \mathbf { k }$ , then $\mathbf { v } ( t )$ is

The magnitude of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = 2 t \mathbf { i } - \cos ( 3 t ) \mathbf { j } + \sin ( 3 t ) \mathbf { k }$ is

Let $y = \sqrt { x }$ . Then the curvature K at (4, 2) is

Nicolaus Copernicus's Heliocentric Theory of Planetary Motion states that

The radius of curvature of $y = x ^ { 3 } - 2 x$ at (2, 4) is

The fact that $\mathbf { r } ( t ) \times \mathbf { v } ( t )$ is constant implies that

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = ( t + 2 ) \mathbf { i } + 2 t \mathbf { j } + 2 \mathbf { k }$ is