Question 1

Let $\mathbf { r } ( t ) = - \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + 3 t \mathbf { k }$ Then the curvature K(t) is

Accepted Answer

A) 0 
B)  $\frac { 4 } { 13 }$ 
C)  $\frac { 4 } { 5 }$ 
D) 2 
E) 1 
A) 0 
B)  $\frac { 4 } { 13 }$ 
C)  $\frac { 4 } { 5 }$ 
D) 2 
E) 1

Question 2

The speed of a particle moving along the curve $\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } - t \mathbf { j } - t ^ { 2 } \mathbf { k }$ is

Accepted Answer

A)  $\sqrt { 5 }$ 
B)  $\frac { 1 } { \sqrt { 7 } }$ 
C)  $\sqrt { 8 t ^ { 2 } + 1 }$ 
D)  $\sqrt { 8 t ^ { 2 } - 1 }$ 
E)  $\sqrt { 13 }$ 
A)  $\sqrt { 5 }$ 
B)  $\frac { 1 } { \sqrt { 7 } }$ 
C)  $\sqrt { 8 t ^ { 2 } + 1 }$ 
D)  $\sqrt { 8 t ^ { 2 } - 1 }$ 
E)  $\sqrt { 13 }$

Question 3

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

Accepted Answer

A)  $\sqrt { ( \sin t + 2 ) ^ { 2 } + ( \cos t - 1 ) ^ { 2 } }$ 
B)  $\sqrt { ( \sin t - 1 ) ^ { 2 } + ( \cos t + 2 ) ^ { 2 } }$ 
C)  $\sqrt { ( \sin t + 1 ) ^ { 2 } + ( \cos t + 2 ) ^ { 2 } }$ 
D)  $\sqrt { ( \sin t - 1 ) ^ { 2 } + ( \cos t - 2 ) ^ { 2 } }$ 
E)  $\sqrt { ( \sin t + 1 ) ^ { 2 } + ( \cos t - 2 ) ^ { 2 } }$ 
A)  $\sqrt { ( \sin t + 2 ) ^ { 2 } + ( \cos t - 1 ) ^ { 2 } }$ 
B)  $\sqrt { ( \sin t - 1 ) ^ { 2 } + ( \cos t + 2 ) ^ { 2 } }$ 
C)  $\sqrt { ( \sin t + 1 ) ^ { 2 } + ( \cos t + 2 ) ^ { 2 } }$ 
D)  $\sqrt { ( \sin t - 1 ) ^ { 2 } + ( \cos t - 2 ) ^ { 2 } }$ 
E)  $\sqrt { ( \sin t + 1 ) ^ { 2 } + ( \cos t - 2 ) ^ { 2 } }$

Question 4

The normal component $a _ { \mathrm { N } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = e ^ { t } \mathbf { i } + e ^ { - t } \mathbf { j } + t \mathbf { k }$ is

Accepted Answer

A)  $\frac { 3 } { \sqrt { 2 } }$ 
B)  $\frac { \sqrt { 2 } } { 3 }$ 
C)  $\sqrt { 56 }$ 
D) 1 
E) 0 
A)  $\frac { 3 } { \sqrt { 2 } }$ 
B)  $\frac { \sqrt { 2 } } { 3 }$ 
C)  $\sqrt { 56 }$ 
D) 1 
E) 0

Question 5

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } - t \mathbf { j } - t ^ { 2 } \mathbf { k }$ 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 6

Which of the following equation implies that the motion of a planet lies in a plane?

Accepted Answer

A)  $\mathbf { r } ( t ) \times \mathbf { v } ( t ) = 0$ 
B)  $\frac { d } { d t } ( \mathbf { r } ( t ) \times \mathbf { v } ( t ) ) = 0$ 
C)  $\mathbf { r } ( t ) \bullet \mathbf { v } ( t ) = 0$ 
D)  $\frac { d } { d t } ( \mathbf { r } ( t ) \cdot \mathbf { v } ( t ) ) = 0$ 
E)  $\frac { d } { d t } ( \mathbf { r } ( t ) \bullet \mathbf { v } ( t ) ) = 1$ 
A)  $\mathbf { r } ( t ) \times \mathbf { v } ( t ) = 0$ 
B)  $\frac { d } { d t } ( \mathbf { r } ( t ) \times \mathbf { v } ( t ) ) = 0$ 
C)  $\mathbf { r } ( t ) \bullet \mathbf { v } ( t ) = 0$ 
D)  $\frac { d } { d t } ( \mathbf { r } ( t ) \cdot \mathbf { v } ( t ) ) = 0$ 
E)  $\frac { d } { d t } ( \mathbf { r } ( t ) \bullet \mathbf { v } ( t ) ) = 1$

Question 7

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } + \left( 2 t ^ { 2 } - 1 \right) \mathbf { j } - 3 t ^ { 2 } \mathbf { k }$ is

Accepted Answer

A)  $2 \sqrt { 14 }$ 
B)  $\sqrt { 58 }$ 
C)  $\sqrt { 65 }$ 
D)  $\sqrt { 73 }$ 
E)  $\sqrt { 78 }$ 
A)  $2 \sqrt { 14 }$ 
B)  $\sqrt { 58 }$ 
C)  $\sqrt { 65 }$ 
D)  $\sqrt { 73 }$ 
E)  $\sqrt { 78 }$

Question 8

The radius of curvature of $\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k }$ at the point corresponding to $t = \frac { \pi } { 2 }$ is

Accepted Answer

A)  $\frac { 2 } { 17 \sqrt { 17 } }$ 
B)  $\frac { 1 } { 2 \sqrt { 2 } }$ 
C) 2 
D)  $\frac { 101 \sqrt { 101 } } { 12 }$ 
E)  $\frac { 7 } { 125 }$ 
A)  $\frac { 2 } { 17 \sqrt { 17 } }$ 
B)  $\frac { 1 } { 2 \sqrt { 2 } }$ 
C) 2 
D)  $\frac { 101 \sqrt { 101 } } { 12 }$ 
E)  $\frac { 7 } { 125 }$

Question 9

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

Accepted Answer

A)  $\sqrt { ( \sin t - 2 ) ^ { 2 } + ( \cos t - 1 ) ^ { 2 } }$ 
B)  $\sqrt { ( \sin t - 1 ) ^ { 2 } + ( \cos t - 2 ) ^ { 2 } }$ 
C)  $\sqrt { ( \sin t - 1 ) ^ { 2 } + ( \cos t + 2 ) ^ { 2 } }$ 
D)  $\sqrt { ( \sin t + 1 ) ^ { 2 } + ( \cos t + 2 ) ^ { 2 } }$ 
E)  $\sqrt { ( \sin t + 1 ) ^ { 2 } + ( \cos t - 2 ) ^ { 2 } }$ 
A)  $\sqrt { ( \sin t - 2 ) ^ { 2 } + ( \cos t - 1 ) ^ { 2 } }$ 
B)  $\sqrt { ( \sin t - 1 ) ^ { 2 } + ( \cos t - 2 ) ^ { 2 } }$ 
C)  $\sqrt { ( \sin t - 1 ) ^ { 2 } + ( \cos t + 2 ) ^ { 2 } }$ 
D)  $\sqrt { ( \sin t + 1 ) ^ { 2 } + ( \cos t + 2 ) ^ { 2 } }$ 
E)  $\sqrt { ( \sin t + 1 ) ^ { 2 } + ( \cos t - 2 ) ^ { 2 } }$

Question 10

Let $\mathbf { r } ( t ) = - \sin ( 2 t ) \mathbf { i } + 2 t \mathbf { j } - \cos ( 2 t ) \mathbf { k }$ Then the unit normal vector $\mathbf { n } \left( \frac { \pi } { 6 } \right)$ is

Accepted Answer

A)  $\frac { 1 } { 2 } \mathbf { i } + \frac { \sqrt { 3 } } { 2 } \mathbf { k }$ 
B)  $- \frac { \sqrt { 3 } } { 2 } \mathbf { i } - \frac { 1 } { 2 } \mathbf { k }$ 
C)  $\frac { \sqrt { 3 } } { 2 } \mathbf { i } + \frac { 1 } { 2 } \mathbf { k }$ 
D)  $\frac { \sqrt { 3 } } { 2 } \mathbf { i } - \frac { 1 } { 2 } \mathbf { k }$ 
E)  $- \frac { \sqrt { 3 } } { 2 } \mathbf { i } + \frac { 1 } { 2 } \mathbf { k }$ 
A)  $\frac { 1 } { 2 } \mathbf { i } + \frac { \sqrt { 3 } } { 2 } \mathbf { k }$ 
B)  $- \frac { \sqrt { 3 } } { 2 } \mathbf { i } - \frac { 1 } { 2 } \mathbf { k }$ 
C)  $\frac { \sqrt { 3 } } { 2 } \mathbf { i } + \frac { 1 } { 2 } \mathbf { k }$ 
D)  $\frac { \sqrt { 3 } } { 2 } \mathbf { i } - \frac { 1 } { 2 } \mathbf { k }$ 
E)  $- \frac { \sqrt { 3 } } { 2 } \mathbf { i } + \frac { 1 } { 2 } \mathbf { k }$

Question 11

Let $\mathbf { r } ( t ) = e ^ { t } \mathbf { i } - e ^ { - t } \mathbf { j } + 3 t \mathbf { k }$ Then the curvature K(0) is

Accepted Answer

A)  $\frac { \sqrt { 2 } } { 13 }$ 
B)  $\frac { \sqrt { 2 } } { 11 }$ 
C)  $\frac { 4 } { 5 }$ 
D)  $\frac { 4 } { 13 }$ 
E)  $\frac { 1 } { 2 }$ 
A)  $\frac { \sqrt { 2 } } { 13 }$ 
B)  $\frac { \sqrt { 2 } } { 11 }$ 
C)  $\frac { 4 } { 5 }$ 
D)  $\frac { 4 } { 13 }$ 
E)  $\frac { 1 } { 2 }$

Question 12

Kepler's Third Law 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 square of the period of revolution of a planet is proportional to the length of the semimajor axis of the planet's elliptical orbit. 
D) The square of the period of revolution of a planet is proportional to the square of the length of the semimajor axis of the planet's elliptical orbit. 
E) The square of the period of revolution of a planet is proportional to the cube of the length of the semimajor axis of the planet's elliptical orbit. 
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 square of the period of revolution of a planet is proportional to the length of the semimajor axis of the planet's elliptical orbit. 
D) The square of the period of revolution of a planet is proportional to the square of the length of the semimajor axis of the planet's elliptical orbit. 
E) The square of the period of revolution of a planet is proportional to the cube of the length of the semimajor axis of the planet's elliptical orbit.

Question 13

Let $\mathbf { r } ( t ) = \cos ( 2 t + 1 ) \mathbf { i } + \sqrt { t ^ { 2 } + 1 } \mathbf { j } + \frac { t ^ { 2 } } { 2 } \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

Accepted Answer

A)  $4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$ 
B)  $4 \cos ( 2 t + 1 ) \mathbf { i } - \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$ 
C)  $- 4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$ 
D)  $- 4 \cos ( 2 t + 1 ) \mathbf { i } - \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$ 
E)  $- 4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } - \mathbf { k }$ 
A)  $4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$ 
B)  $4 \cos ( 2 t + 1 ) \mathbf { i } - \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$ 
C)  $- 4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$ 
D)  $- 4 \cos ( 2 t + 1 ) \mathbf { i } - \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } + \mathbf { k }$ 
E)  $- 4 \cos ( 2 t + 1 ) \mathbf { i } + \frac { 1 } { \left( \sqrt { t ^ { 2 } + 1 } \right) ^ { 3 } } \mathbf { j } - \mathbf { k }$

Question 14

The magnitude of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = \sin ( 4 t ) \mathbf { i } - \cos ( 4 t ) \mathbf { j } + 7 \mathbf { k }$ is

Accepted Answer

A)  $2 \sqrt { 2 }$ 
B) 16 
C)  $4 \sqrt { 2 }$ 
D)  $\sqrt { 5 }$ 
E)  $9 \sqrt { 2 }$ 
A)  $2 \sqrt { 2 }$ 
B) 16 
C)  $4 \sqrt { 2 }$ 
D)  $\sqrt { 5 }$ 
E)  $9 \sqrt { 2 }$

Question 15

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

Accepted Answer

A)  $\sqrt { 8 t ^ { 2 } - 1 }$ 
B)  $\sqrt { 13 }$ 
C)  $\sqrt { 8 t ^ { 2 } + 1 }$ 
D)  $\sqrt { 5 }$ 
E)  $\frac { 1 } { \sqrt { 7 } }$ 
A)  $\sqrt { 8 t ^ { 2 } - 1 }$ 
B)  $\sqrt { 13 }$ 
C)  $\sqrt { 8 t ^ { 2 } + 1 }$ 
D)  $\sqrt { 5 }$ 
E)  $\frac { 1 } { \sqrt { 7 } }$

Question 16

Let $\mathbf { r } ( t ) = - t \mathbf { i } + \sin ( t ) \mathbf { j } - \cos ( t ) \mathbf { k }$ Then the unit normal vector $\mathbf { n } ( 0 )$ is

Accepted Answer

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

Question 17

The solution $\mathbf { r } ( t )$ to the differential equation $\mathbf { r } ^ { \prime } ( t ) = 3 \cos t \mathbf { i } - 2 \sin t \mathbf { j } + \mathbf { k }$ with the condition $\mathbf { r } ( 0 ) = 2 \mathbf { i } + \mathbf { k }$ is

Accepted Answer

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

Question 18

Which of the following is Newton's Second Law of Motion?

Accepted Answer

A)  $F ( t ) = m \| \mathbf { v } ( t ) \|$ 
B)  $F ( t ) = m \| \mathbf { a } ( t ) \|$ 
C)  $\mathbf { F } ( t ) = m \| \mathbf { a } ( t ) \|$ 
D)  $\mathbf { F } ( t ) = m \| \mathbf { v } ( t ) \|$ 
E)  $F ( t ) = m \left\| \mathbf { a } ^ { \prime } ( t ) \right\|$ 
A)  $F ( t ) = m \| \mathbf { v } ( t ) \|$ 
B)  $F ( t ) = m \| \mathbf { a } ( t ) \|$ 
C)  $\mathbf { F } ( t ) = m \| \mathbf { a } ( t ) \|$ 
D)  $\mathbf { F } ( t ) = m \| \mathbf { v } ( t ) \|$ 
E)  $F ( t ) = m \left\| \mathbf { a } ^ { \prime } ( t ) \right\|$

Question 19

Let $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + 3 t \mathbf { k }$ Then the unit normal vector $\mathbf { n } \left( \frac { \pi } { 4 } \right)$ is

Accepted Answer

A) i 
B)  $- \frac { 1 } { \sqrt { 5 } } \mathbf { i } + \frac { 4 } { \sqrt { 5 } } \mathbf { k }$ 
C)  $\frac { 2 } { \sqrt { 29 } } \mathbf { j } + \frac { 5 } { \sqrt { 29 } } \mathbf { k }$ 
D)  $- \mathbf { j }$ 
E) k 
A) i 
B)  $- \frac { 1 } { \sqrt { 5 } } \mathbf { i } + \frac { 4 } { \sqrt { 5 } } \mathbf { k }$ 
C)  $\frac { 2 } { \sqrt { 29 } } \mathbf { j } + \frac { 5 } { \sqrt { 29 } } \mathbf { k }$ 
D)  $- \mathbf { j }$ 
E) k

Question 20

The radius of curvature of y = cos x at (?, -1) is

Accepted Answer

A) 1 
B)  $\frac { 1 } { 2 \sqrt { 2 } }$ 
C)  $\frac { 2 } { 5 \sqrt { 5 } }$ 
D)  $\frac { 2 } { 17 \sqrt { 17 } }$ 
E)  $\frac { 7 } { 125 }$ 
A) 1 
B)  $\frac { 1 } { 2 \sqrt { 2 } }$ 
C)  $\frac { 2 } { 5 \sqrt { 5 } }$ 
D)  $\frac { 2 } { 17 \sqrt { 17 } }$ 
E)  $\frac { 7 } { 125 }$

Let $\mathbf { r } ( t ) = - \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + 3 t \mathbf { k }$ Then the curvature K(t) is

The speed of a particle moving along the curve $\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } - t \mathbf { j } - t ^ { 2 } \mathbf { k }$ is

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

The normal component $a _ { \mathrm { N } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = e ^ { t } \mathbf { i } + e ^ { - t } \mathbf { j } + t \mathbf { k }$ is

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

Which of the following equation implies that the motion of a planet lies in a plane?

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } + \left( 2 t ^ { 2 } - 1 \right) \mathbf { j } - 3 t ^ { 2 } \mathbf { k }$ is

The radius of curvature of $\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k }$ at the point corresponding to $t = \frac { \pi } { 2 }$ is

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

Let $\mathbf { r } ( t ) = - \sin ( 2 t ) \mathbf { i } + 2 t \mathbf { j } - \cos ( 2 t ) \mathbf { k }$ Then the unit normal vector $\mathbf { n } \left( \frac { \pi } { 6 } \right)$ is

Let $\mathbf { r } ( t ) = e ^ { t } \mathbf { i } - e ^ { - t } \mathbf { j } + 3 t \mathbf { k }$ Then the curvature K(0) is

Kepler's Third Law of Planetary Motion states that

Let $\mathbf { r } ( t ) = \cos ( 2 t + 1 ) \mathbf { i } + \sqrt { t ^ { 2 } + 1 } \mathbf { j } + \frac { t ^ { 2 } } { 2 } \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

The magnitude of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = \sin ( 4 t ) \mathbf { i } - \cos ( 4 t ) \mathbf { j } + 7 \mathbf { k }$ is

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

Let $\mathbf { r } ( t ) = - t \mathbf { i } + \sin ( t ) \mathbf { j } - \cos ( t ) \mathbf { k }$ Then the unit normal vector $\mathbf { n } ( 0 )$ is

The solution $\mathbf { r } ( t )$ to the differential equation $\mathbf { r } ^ { \prime } ( t ) = 3 \cos t \mathbf { i } - 2 \sin t \mathbf { j } + \mathbf { k }$ with the condition $\mathbf { r } ( 0 ) = 2 \mathbf { i } + \mathbf { k }$ is

Which of the following is Newton's Second Law of Motion?

Let $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + 3 t \mathbf { k }$ Then the unit normal vector $\mathbf { n } \left( \frac { \pi } { 4 } \right)$ is

The radius of curvature of y = cos x at (?, -1) is

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Exam 12: Vector Functions

Let r(t)=−cos⁡(2t)i+sin⁡(2t)j+3tk\mathbf { r } ( t ) = - \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + 3 t \mathbf { k }r(t)=−cos(2t)i+sin(2t)j+3tk Then the curvature K(t) is

The speed of a particle moving along the curve r(t)=t2i−tj−t2k\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } - t \mathbf { j } - t ^ { 2 } \mathbf { k }r(t)=t2i−tj−t2k is

The normal component aN(t)a _ { \mathrm { N } } ( t )aN​(t) of the acceleration of a particle moving along the curve r(t)=eti+e−tj+tk\mathbf { r } ( t ) = e ^ { t } \mathbf { i } + e ^ { - t } \mathbf { j } + t \mathbf { k }r(t)=eti+e−tj+tk is

The tangential component aT(t)a _ { \mathrm { T } } ( t )aT​(t) of the acceleration of a particle moving along the curve r(t)=t2i−tj−t2k\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } - t \mathbf { j } - t ^ { 2 } \mathbf { k }r(t)=t2i−tj−t2k is

Which of the following equation implies that the motion of a planet lies in a plane?

The tangential component aT(t)a _ { \mathrm { T } } ( t )aT​(t) of the acceleration of a particle moving along the curve r(t)=t2i+(2t2−1)j−3t2k\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } + \left( 2 t ^ { 2 } - 1 \right) \mathbf { j } - 3 t ^ { 2 } \mathbf { k }r(t)=t2i+(2t2−1)j−3t2k is

The radius of curvature of r(t)=cos⁡ti+sin⁡tj+tk\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k }r(t)=costi+sintj+tk at the point corresponding to t=π2t = \frac { \pi } { 2 }t=2π​ is

Let r(t)=−sin⁡(2t)i+2tj−cos⁡(2t)k\mathbf { r } ( t ) = - \sin ( 2 t ) \mathbf { i } + 2 t \mathbf { j } - \cos ( 2 t ) \mathbf { k }r(t)=−sin(2t)i+2tj−cos(2t)k Then the unit normal vector n(π6)\mathbf { n } \left( \frac { \pi } { 6 } \right)n(6π​) is

Let r(t)=eti−e−tj+3tk\mathbf { r } ( t ) = e ^ { t } \mathbf { i } - e ^ { - t } \mathbf { j } + 3 t \mathbf { k }r(t)=eti−e−tj+3tk Then the curvature K(0) is

Kepler's Third Law of Planetary Motion states that ​

Let r(t)=cos⁡(2t+1)i+t2+1j+t22k\mathbf { r } ( t ) = \cos ( 2 t + 1 ) \mathbf { i } + \sqrt { t ^ { 2 } + 1 } \mathbf { j } + \frac { t ^ { 2 } } { 2 } \mathbf { k }r(t)=cos(2t+1)i+t2+1​j+2t2​k Then r′′(t)\mathbf { r } ^ { \prime \prime } ( t )r′′(t) is

The magnitude of the acceleration of a particle moving along the curve r(t)=sin⁡(4t)i−cos⁡(4t)j+7k\mathbf { r } ( t ) = \sin ( 4 t ) \mathbf { i } - \cos ( 4 t ) \mathbf { j } + 7 \mathbf { k }r(t)=sin(4t)i−cos(4t)j+7k is

The speed of a particle moving along the curve r(t)=(t+2)i+2tj+3k\mathbf { r } ( t ) = ( t + 2 ) \mathbf { i } + 2 t \mathbf { j } + 3 \mathbf { k }r(t)=(t+2)i+2tj+3k is

Let r(t)=−ti+sin⁡(t)j−cos⁡(t)k\mathbf { r } ( t ) = - t \mathbf { i } + \sin ( t ) \mathbf { j } - \cos ( t ) \mathbf { k }r(t)=−ti+sin(t)j−cos(t)k Then the unit normal vector n(0)\mathbf { n } ( 0 )n(0) is

Which of the following is Newton's Second Law of Motion?

Let r(t)=cos⁡(2t)i+sin⁡(2t)j+3tk\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + 3 t \mathbf { k }r(t)=cos(2t)i+sin(2t)j+3tk Then the unit normal vector n(π4)\mathbf { n } \left( \frac { \pi } { 4 } \right)n(4π​) is

The radius of curvature of y = cos x at (?, -1) is

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Let $\mathbf { r } ( t ) = - \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + 3 t \mathbf { k }$ Then the curvature K(t) is

The speed of a particle moving along the curve $\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } - t \mathbf { j } - t ^ { 2 } \mathbf { k }$ is

The normal component $a _ { \mathrm { N } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = e ^ { t } \mathbf { i } + e ^ { - t } \mathbf { j } + t \mathbf { k }$ is

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

The tangential component $a _ { \mathrm { T } } ( t )$ of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = t ^ { 2 } \mathbf { i } + \left( 2 t ^ { 2 } - 1 \right) \mathbf { j } - 3 t ^ { 2 } \mathbf { k }$ is

The radius of curvature of $\mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k }$ at the point corresponding to $t = \frac { \pi } { 2 }$ is

Let $\mathbf { r } ( t ) = - \sin ( 2 t ) \mathbf { i } + 2 t \mathbf { j } - \cos ( 2 t ) \mathbf { k }$ Then the unit normal vector $\mathbf { n } \left( \frac { \pi } { 6 } \right)$ is

Let $\mathbf { r } ( t ) = e ^ { t } \mathbf { i } - e ^ { - t } \mathbf { j } + 3 t \mathbf { k }$ Then the curvature K(0) is

Kepler's Third Law of Planetary Motion states that

Let $\mathbf { r } ( t ) = \cos ( 2 t + 1 ) \mathbf { i } + \sqrt { t ^ { 2 } + 1 } \mathbf { j } + \frac { t ^ { 2 } } { 2 } \mathbf { k }$ Then $\mathbf { r } ^ { \prime \prime } ( t )$ is

The magnitude of the acceleration of a particle moving along the curve $\mathbf { r } ( t ) = \sin ( 4 t ) \mathbf { i } - \cos ( 4 t ) \mathbf { j } + 7 \mathbf { k }$ is

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

Let $\mathbf { r } ( t ) = - t \mathbf { i } + \sin ( t ) \mathbf { j } - \cos ( t ) \mathbf { k }$ Then the unit normal vector $\mathbf { n } ( 0 )$ is

Let $\mathbf { r } ( t ) = \cos ( 2 t ) \mathbf { i } + \sin ( 2 t ) \mathbf { j } + 3 t \mathbf { k }$ Then the unit normal vector $\mathbf { n } \left( \frac { \pi } { 4 } \right)$ is