Question 1

Solve the problem.
-Show that a planet in a circular orbit moves with constant speed.

Accepted Answer

Kepler's first law says that the eccentr

Question 2

Calculate the arc length of the indicated portion of the curve r(t).
-Following the curve $r ( t ) = 5 t i + ( 12 \cos t ) j + ( 12 \sin t ) k$ in the direction of increasing of arc length, find the point that lies $26 \pi$ units away from the point where $t = 0$.

Accepted Answer

A)  $( 10 \pi , 0,0 )$ 
B)  $( 10,12,0 )$ 
C)  $( 10 \pi , 12,0 )$ 
D)  $( 10 \pi , 0,12 )$ 
A)  $( 10 \pi , 0,0 )$ 
B)  $( 10,12,0 )$ 
C)  $( 10 \pi , 12,0 )$ 
D)  $( 10 \pi , 0,12 )$

Question 3

Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface
-A projectile is fired at a speed of $720 \mathrm {~m} / \mathrm { sec }$ at an angle of $45 ^ { \circ }$. How long will it take to get $18 \mathrm {~km}$ downrange? Round your answer to the nearest whole number.

Accepted Answer

A)  33 sec 
B)  35 sec 
C)  37 sec 
D)  It will never get that far downrange. 
A)  33 sec 
B)  35 sec 
C)  37 sec 
D)  It will never get that far downrange.

Question 4

Provide an appropriate response.
-Derive the equations
$$\begin{array} { l } 
x = x _ { 0 } + \frac { v _ { 0 } } { k } \left( 1 - e ^ { - k t } \right) \cos \alpha \
y = y _ { 0 } + \frac { v _ { 0 } } { k } \left( 1 - e ^ { - k t } \right) \sin \alpha + \frac { g } { k ^ { 2 } } \left( 1 - k t - e ^ { - k t } \right)
\end{array}$$
by solving the following initial value problem for a vector $\mathbf { r }$ in the plane.
$$\begin{aligned}
\text { Differential equation } \frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm { dt } ^ { 2 } } & = - g \mathbf { j } - \mathrm { k } \mathbf { v } = - \mathrm { g } \mathbf { j } - \mathrm { k } \frac { \mathrm { d } \mathbf { r } } { \mathrm { dt } } \
\text { Initial conditions: } \quad \mathrm { r } ( 0 ) & = \mathrm { x } _ { 0 } \mathbf { i } + \mathrm { y } _ { 0 } \mathbf { j } \
& \frac { \mathrm { d } \mathbf { r } } { \mathrm { dt } } ( 0 ) = \mathbf { v } _ { 0 } = \left( \mathrm { v } _ { 0 } \cos \alpha \right) \mathbf { i } + \left( \mathrm { v } _ { 0 } \sin \alpha \right) \mathbf { j }
\end{aligned}$$
The drag coefficient $\mathrm { k }$ is a positive constant representing resistance due to air density, vo and $\alpha$ are the projectile's initial speed and launch angle, and $g$ is the acceleration of gravity.

Accepted Answer

The answer of Provide an appropriate response.
-Derive the equations
\[\begin{array} {...

Question 5

Provide an appropriate response.
-A human cannonball is to be fired with an initial speed of $80 \mathrm { ft } / \mathrm { sec }$. The circus performer hopes to land on a cushion located 190 feet downrange at the same height as the muzzle of the cannon. The circus is being held in a large room with a flat ceiling 33 feet higher than the muzzle. Can the performer be fired to the cushion without striking the ceiling? If so, what is the proper firing angle? ( $\mathrm { g } = 32 \mathrm { ft } / \sec ^ { 2 }$ )

Accepted Answer

The answer of Provide an appropriate response.
-A human cannonball is...

Question 6

Find the curvature of the space curve.
-$$\mathbf { r } ( \mathrm { t } ) = 5 \mathrm { ti } + \left( 4 + 8 \cos \frac { 3 } { 2 } \mathrm { t } \right) \mathbf { j } + \left( 4 + 8 \sin \frac { 3 } { 2 } \right) \mathbf { k }$$

Accepted Answer

A)  $\kappa = \frac { 18 } { 13 }$ 
B)  $\kappa = \frac { 12 } { 169 }$ 
C)  $\kappa = \frac { 18 } { 169 }$ 
D)  $\kappa = \frac { 12 } { 13 }$ 
A)  $\kappa = \frac { 18 } { 13 }$ 
B)  $\kappa = \frac { 12 } { 169 }$ 
C)  $\kappa = \frac { 18 } { 169 }$ 
D)  $\kappa = \frac { 12 } { 13 }$

Question 7

Find the curvature of the curve r(t).
-r(t) = (8 + cos 9t - sin 9t)i + (5 + sin 9t + cos 9t)j + 7k

Accepted Answer

A)  $\kappa = \frac { \sqrt { 2 } } { 2 }$ 
B)  $\kappa = \sqrt { 2 }$ 
C)  $\kappa = \frac { 9 } { 2 }$ 
D)  $\kappa = \frac { 9 } { 2 } \sqrt { 2 }$ 
A)  $\kappa = \frac { \sqrt { 2 } } { 2 }$ 
B)  $\kappa = \sqrt { 2 }$ 
C)  $\kappa = \frac { 9 } { 2 }$ 
D)  $\kappa = \frac { 9 } { 2 } \sqrt { 2 }$

Question 8

The position vector of a particle is r(t). Find the requested vector.
-The velocity at $t = 0$ for $\mathbf { r } ( \mathrm { t } ) = \ln \left( \mathrm { t } ^ { 3 } - 3 \mathrm { t } ^ { 2 } + 4 \right) \mathbf { i } - \sqrt { 2 \mathrm { t } + 36 } \mathbf { j } - 5 \cos ( \mathrm { t } ) \mathbf { k }$

Accepted Answer

A)  $\mathbf { v } ( 0 ) = \frac { 1 } { 4 } \mathbf { i } - \frac { 1 } { 6 } \mathrm { j }$ 
B)  $\mathbf { v } ( 0 ) = \frac { 1 } { 6 } \mathbf { j }$ 
C)  $\mathbf { v } ( 0 ) = - \frac { 1 } { 6 } \mathrm { j }$ 
D)  $\mathbf { v } ( 0 ) = \frac { 1 } { 4 } \mathbf { i } - \frac { 1 } { 12 } \mathbf { j } + 5 \mathbf { k }$ 
A)  $\mathbf { v } ( 0 ) = \frac { 1 } { 4 } \mathbf { i } - \frac { 1 } { 6 } \mathrm { j }$ 
B)  $\mathbf { v } ( 0 ) = \frac { 1 } { 6 } \mathbf { j }$ 
C)  $\mathbf { v } ( 0 ) = - \frac { 1 } { 6 } \mathrm { j }$ 
D)  $\mathbf { v } ( 0 ) = \frac { 1 } { 4 } \mathbf { i } - \frac { 1 } { 12 } \mathbf { j } + 5 \mathbf { k }$

Question 9

Calculate the arc length of the indicated portion of the curve r(t).
-$$r ( t ) = \left( 2 \cos ^ { 3 } 5 t \right) \mathbf { j } + \left( 2 \sin ^ { 3 } 5 t \right) \mathbf { k } ; \frac { 3 } { 5 } \pi \leq t \leq \frac { 7 } { 10 } \pi$$

Accepted Answer

A)  1 
B)  3 
C)  0 
D)  6 
A)  1 
B)  3 
C)  0 
D)  6

Question 10

$\text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. }$
-r(t) = (t + 10)i + (ln(cos t) - 6)j + 2k

Accepted Answer

A)  a = (sec t tan t)T + (sec t)N 
B)  a = (sec t tan t)T - (sec t)N 
C)  a = (-sec t tan t)T + (sec t)N 
D)  a = (-sec t tan t)T - (sec t)N 
A)  a = (sec t tan t)T + (sec t)N 
B)  a = (sec t tan t)T - (sec t)N 
C)  a = (-sec t tan t)T + (sec t)N 
D)  a = (-sec t tan t)T - (sec t)N

Question 11

Solve the problem.
-The orbit of a satellite had a semimajor axis of $a = 6955 \mathrm {~km}$. Calculate the period of the satellite. (Earth's mass $= 5.975 \times 10 ^ { 24 } \mathrm {~kg}$ and $\mathrm { G } = 6.6720 \times 10 ^ { - 11 } \mathrm { Nm } ^ { 2 } \mathrm {~kg} - 2$ ).

Accepted Answer

A)  0.18 sec 
B)  96.2 min 
C)  124 min 
D)  9250 hr 
A)  0.18 sec 
B)  96.2 min 
C)  124 min 
D)  9250 hr

Question 12

Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from the horizontal, and that the projectile is launched from the origin over a horizontal surface
-An athlete puts a 16-lb shot at an angle of $39 ^ { \circ }$ to the horizontal from $6.1 \mathrm { ft }$ above the ground at an initial speed of $47 \mathrm { ft } / \mathrm { sec }$. How far forward does the shot travel before it hits the ground? Round your answer to the nearest tenth.

Accepted Answer

A)  2 ft 
B)  74.4 ft 
C)  227.8 ft 
D)  6.8 ft 
A)  2 ft 
B)  74.4 ft 
C)  227.8 ft 
D)  6.8 ft

Question 13

Find the principal unit normal vector N for the curve r(t).
-r(t) = (10 sin t)i + (10 cos t)j + 9k

Accepted Answer

A)  N = (sin t)i + (cos t)j 
B)  N = (-sin t)i + (cos t)j 
C)  N = (-sin t)i - (cos t)j 
D)  N = (-cos t)i - (sin t)j 
A)  N = (sin t)i + (cos t)j 
B)  N = (-sin t)i + (cos t)j 
C)  N = (-sin t)i - (cos t)j 
D)  N = (-cos t)i - (sin t)j

Question 14

For the curve r(t), find an equation for the indicated plane at the given value of t.
-$\left. \mathbf { r } ( \mathrm { t } ) = \left( 3 \sin \frac { 5 } { 3 } \mathrm { t } + 7 \right) \mathbf { i } + ( 3 \cos 20 \mathrm { t } ) - 6 \right) \mathbf { j } + 12 \mathrm { k } \mathbf { k }$; osculating plane at $t = 2.5 \pi$.

Accepted Answer

A)  $\frac { 12 } { 169 } ( y - 6 ) + \frac { 60 } { 169 } ( z - 30 \pi ) = 0$ 
B)  $\frac { 12 } { 13 } ( y + 6 ) + \frac { 60 } { 13 } ( z + 30 ) = 0$ 
C)  $\frac { 12 } { 13 } ( y + 6 ) + \frac { 60 } { 13 } ( z - 30 \pi ) = 0$ 
D)  $\frac { 12 } { 13 } ( x - 7 ) + \frac { 60 } { 13 } ( z - 30 \pi ) = 0$ 
A)  $\frac { 12 } { 169 } ( y - 6 ) + \frac { 60 } { 169 } ( z - 30 \pi ) = 0$ 
B)  $\frac { 12 } { 13 } ( y + 6 ) + \frac { 60 } { 13 } ( z + 30 ) = 0$ 
C)  $\frac { 12 } { 13 } ( y + 6 ) + \frac { 60 } { 13 } ( z - 30 \pi ) = 0$ 
D)  $\frac { 12 } { 13 } ( x - 7 ) + \frac { 60 } { 13 } ( z - 30 \pi ) = 0$

Question 15

$\text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. }$
-r(t) = (6t sin t + 6 cos t)i + (6t cos t - 6 sin t)j + 3k

Accepted Answer

A)  $a = 6 T + 6 t N$ 
B)  $a = \frac { 1 } { 6 t } N$ 
C)  $a = 6 T + \frac { 1 } { 6 t } N$ 
D)  $a = 6 \mathrm { t }$ 
A)  $a = 6 T + 6 t N$ 
B)  $a = \frac { 1 } { 6 t } N$ 
C)  $a = 6 T + \frac { 1 } { 6 t } N$ 
D)  $a = 6 \mathrm { t }$

Question 16

For the curve r(t), find an equation for the indicated plane at the given value of t.
-$r ( t ) = \frac { 40 } { 9 } ( 1 + t ) ^ { 3 / 2 } \mathbf { i } + \frac { 40 } { 9 } ( 1 - t ) ^ { 3 / 2 } \mathbf { j } + \frac { 10 } { 3 } t \mathbf { k }$; osculating plane $a t \mathrm { t } = 0$

Accepted Answer

A)  $x - y = 0$ 
B)  $x - y = - \frac { 40 } { 9 }$ 
C)  $x - y = \frac { 40 } { 9 }$ 
D)  $x - y = \frac { 80 } { 9 }$ 
A)  $x - y = 0$ 
B)  $x - y = - \frac { 40 } { 9 }$ 
C)  $x - y = \frac { 40 } { 9 }$ 
D)  $x - y = \frac { 80 } { 9 }$

Question 17

For the curve r(t), find an equation for the indicated plane at the given value of t.
-$\mathbf { r } ( \mathrm { t } ) = ( \mathrm { t } + 6 ) \mathbf { i } + ( \ln ( \cos \mathrm { t } ) - 8 ) \mathbf { j } + 7 \mathbf { k } , - \pi / 2 < \mathrm { t } < \pi \cdot 2 ;$ rectifying plane $\mathrm { at } \mathrm {} \mathrm { t } = \frac { \pi } { 4 }$.

Accepted Answer

A)  $\frac { \sqrt { 2 } } { 2 } \left( x - \left( \frac { \pi } { 4 } + 6 \right) \right) - \frac { \sqrt { 2 } } { 2 } \left( y - \left( \ln \left( \frac { \sqrt { 2 } } { 2 } \right) - 8 \right) \right) = 0$ 
B)  $- \frac { \sqrt { 2 } } { 2 } \left( x - \left( \frac { \pi } { 4 } + 6 \right) \right) + \frac { \sqrt { 2 } } { 2 } \left( y - \left( \ln \left( \frac { \sqrt { 2 } } { 2 } \right) - 8 \right) \right) = 0$ 
C)  $\frac { \sqrt { 2 } } { 2 } ( x - ( 1 + 6 ) ) = 0$ 
D)  $\left. \frac { \sqrt { 2 } } { 2 } \left( x - \left( \frac { \pi } { 4 } + 6 \right) \right) \right) ^ { \prime } + \frac { \sqrt { 2 } } { 2 } \left( y - \left( \ln \left( \frac { \sqrt { 2 } } { 2 } \right) - 8 \right) \right) = 0$ 
A)  $\frac { \sqrt { 2 } } { 2 } \left( x - \left( \frac { \pi } { 4 } + 6 \right) \right) - \frac { \sqrt { 2 } } { 2 } \left( y - \left( \ln \left( \frac { \sqrt { 2 } } { 2 } \right) - 8 \right) \right) = 0$ 
B)  $- \frac { \sqrt { 2 } } { 2 } \left( x - \left( \frac { \pi } { 4 } + 6 \right) \right) + \frac { \sqrt { 2 } } { 2 } \left( y - \left( \ln \left( \frac { \sqrt { 2 } } { 2 } \right) - 8 \right) \right) = 0$ 
C)  $\frac { \sqrt { 2 } } { 2 } ( x - ( 1 + 6 ) ) = 0$ 
D)  $\left. \frac { \sqrt { 2 } } { 2 } \left( x - \left( \frac { \pi } { 4 } + 6 \right) \right) \right) ^ { \prime } + \frac { \sqrt { 2 } } { 2 } \left( y - \left( \ln \left( \frac { \sqrt { 2 } } { 2 } \right) - 8 \right) \right) = 0$

Question 18

Find the unit tangent vector of the given curve.
-$$r ( t ) = \left( 5 \sin ^ { 3 } 9 t \right) i + \left( 5 \cos ^ { 3 } 9 t \right) \mathbf { j }$$

Accepted Answer

A)  T = (5 sin 9t)i - (5 cos 9t)j 
B)  T = (sin 9t)i - (cos 9t)j 
C)  T = (135 sin 9t)i -(135 cos 9t)j 
D)  T = (5 cos 9t)i - (5 sin 9t)j 
A)  T = (5 sin 9t)i - (5 cos 9t)j 
B)  T = (sin 9t)i - (cos 9t)j 
C)  T = (135 sin 9t)i -(135 cos 9t)j 
D)  T = (5 cos 9t)i - (5 sin 9t)j

Question 19

Provide an appropriate response.
-The following equations each describe the motion of a particle. For which path is the particle's speed constant?
(1) $r ( t ) = t ^ { 3 } i + t ^ { 2 } j$
(2) $r ( t ) = \cos ( 8 t ) i + \sin ( 6 t ) j$
(3) $r ( t ) = t i + t j$
(4) $r ( t ) = \cos \left( - 5 t ^ { 2 } \right) \mathbf { i } + \sin \left( - 5 t ^ { 2 } \right) \mathbf { j }$

Accepted Answer

A)  Path (4) and Path (2) 
B)  Path (1) 
C)  Path (2) and Path (3) 
D)  Path (3) 
A)  Path (4) and Path (2) 
B)  Path (1) 
C)  Path (2) and Path (3) 
D)  Path (3)

Question 20

Solve the initial value problem.
-Differential Equation: $\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm { dt } ^ { 2 } } = ( 8 \mathrm { t } - 3 ) \mathbf { i }$
Initial Conditions: $\left. \frac { \mathrm { d } \mathbf { r } } { \mathrm { dt } } \right| _ { \mathrm { r } = 0 } = - \mathbf { k } , \mathrm { r } ( 0 ) = 8 \mathbf { i } + 7 \mathbf { j } + 6 \mathbf { k }$

Accepted Answer

A)  $r ( t ) = \left( \frac { 8 } { 3 } t ^ { 3 } - 3 t ^ { 2 } + 8 \right) i + 7 j + ( 6 - t ) \mathbf { k }$ 
B)  $r ( t ) = \left( \frac { 8 } { 3 } t ^ { 3 } + \frac { 3 } { 2 } t ^ { 2 } + 8 \right) i - 7 j + ( t - 6 ) \mathbf { k }$ 
C)  $ r ( t ) = \left( \frac { 4 } { 3 } t ^ { 3 } - \frac { 3 } { 2 } t ^ { 2 } + 8 \right) i + 7 j + ( 6 - t ) k$ 
D)  $r ( t ) = \left( \frac { 4 } { 3 } t ^ { 3 } + \frac { 3 } { 2 } t ^ { 2 } + 8 \right) i - 7 j + ( 6 - t ) k$ 
A)  $r ( t ) = \left( \frac { 8 } { 3 } t ^ { 3 } - 3 t ^ { 2 } + 8 \right) i + 7 j + ( 6 - t ) \mathbf { k }$ 
B)  $r ( t ) = \left( \frac { 8 } { 3 } t ^ { 3 } + \frac { 3 } { 2 } t ^ { 2 } + 8 \right) i - 7 j + ( t - 6 ) \mathbf { k }$ 
C)  $ r ( t ) = \left( \frac { 4 } { 3 } t ^ { 3 } - \frac { 3 } { 2 } t ^ { 2 } + 8 \right) i + 7 j + ( 6 - t ) k$ 
D)  $r ( t ) = \left( \frac { 4 } { 3 } t ^ { 3 } + \frac { 3 } { 2 } t ^ { 2 } + 8 \right) i - 7 j + ( 6 - t ) k$

Solve the problem. -Show that a planet in a circular orbit moves with constant speed.

Calculate the arc length of the indicated portion of the curve r(t). -Following the curve $r ( t ) = 5 t i + ( 12 \cos t ) j + ( 12 \sin t ) k$ in the direction of increasing of arc length, find the point that lies $26 \pi$ units away from the point where $t = 0$ .

Find the curvature of the space curve. - $\mathbf { r } ( \mathrm { t } ) = 5 \mathrm { ti } + \left( 4 + 8 \cos \frac { 3 } { 2 } \mathrm { t } \right) \mathbf { j } + \left( 4 + 8 \sin \frac { 3 } { 2 } \right) \mathbf { k }$

Find the curvature of the curve r(t). -r(t) = (8 + cos 9t - sin 9t)i + (5 + sin 9t + cos 9t)j + 7k

Calculate the arc length of the indicated portion of the curve r(t). - $r ( t ) = \left( 2 \cos ^ { 3 } 5 t \right) \mathbf { j } + \left( 2 \sin ^ { 3 } 5 t \right) \mathbf { k } ; \frac { 3 } { 5 } \pi \leq t \leq \frac { 7 } { 10 } \pi$

$\text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. }$ -r(t) = (t + 10)i + (ln(cos t) - 6)j + 2k

Solve the problem. -The orbit of a satellite had a semimajor axis of $a = 6955 \mathrm {~km}$ . Calculate the period of the satellite. (Earth's mass $= 5.975 \times 10 ^ { 24 } \mathrm {~kg}$ and $\mathrm { G } = 6.6720 \times 10 ^ { - 11 } \mathrm { Nm } ^ { 2 } \mathrm {~kg} - 2$ ).

Find the principal unit normal vector N for the curve r(t). -r(t) = (10 sin t)i + (10 cos t)j + 9k

$\text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. }$ -r(t) = (6t sin t + 6 cos t)i + (6t cos t - 6 sin t)j + 3k

For the curve r(t), find an equation for the indicated plane at the given value of t. - $r ( t ) = \frac { 40 } { 9 } ( 1 + t ) ^ { 3 / 2 } \mathbf { i } + \frac { 40 } { 9 } ( 1 - t ) ^ { 3 / 2 } \mathbf { j } + \frac { 10 } { 3 } t \mathbf { k }$ ; osculating plane $a t \mathrm { t } = 0$

Find the unit tangent vector of the given curve. - $r ( t ) = \left( 5 \sin ^ { 3 } 9 t \right) i + \left( 5 \cos ^ { 3 } 9 t \right) \mathbf { j }$

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Exam 14: Vector-Valued Functions and Motion in Space