Exam 14: Vector-Valued Functions and Motion in Space

Find the torsion of the space curve. - $\mathbf { r } ( \mathrm { t } ) = \left( 2 + 10 \sin \frac { 2 } { 5 } \mathrm { t } \right) \mathbf { i } + 3 \mathrm { tj } + \left( 2 + 10 \cos \frac { 2 } { 5 } \mathrm { t } \right) \mathbf { k }$

The position vector of a particle is r(t). Find the requested vector. -The acceleration at $t = 0$ for $r ( t ) = t ^ { 2 } i + \left( 8 t ^ { 3 } - 10 \right) j + \sqrt { 4 - 3 t } k$

Solve the initial value problem. -Differential Equation: $\frac { \mathrm { d } \mathbf { r } } { \mathrm { dt } } = ( 3 \sec 3 \mathrm { t } \tan 3 \mathrm { t } ) \mathrm { i } + \frac { \mathrm { t } } { \left( \mathrm { t } ^ { 2 } + 7 \right) ^ { 2 } } \mathrm { j } + \mathrm { t } ^ { 2 } \mathbf { k }$ Initial Condition: $\mathbf { r } ( 0 ) = 9 \mathrm { i } + \frac { 55 } { 14 } \mathrm { j } - 6 \mathbf { k }$

Calculate the arc length of the indicated portion of the curve r(t). - $\mathbf { r } ( \mathrm { t } ) = 4 \mathrm { ti } + \left( 6 \cos \frac { 1 } { 2 } \mathrm { t } \right) \mathbf { j } + \left( 6 \sin \frac { 1 } { 2 } \mathrm { t } \right) \mathbf { k } ; - 8 \leq \mathrm { t } \leq 1$

Provide an appropriate response. -A baseball is hit when it is $3.1$ feet above the ground. It leaves the bat with an initial speed of $150 \mathrm { ft } / \mathrm { sec }$ , making an angle of $24 ^ { \circ }$ with the horizontal. Assuming a drag coefficient $k = 0.12$ , find the range and the flight time of the ball. For projectiles with linear drag: x=+ 1- \alpha, y=+ 1- \alpha+ 1-kt- where $\mathrm { k }$ is the drag coefficient, $\mathrm { v } _ { 0 }$ and $\alpha$ are the projectile's initial speed and launch angle, and $g$ is the acceleratic of gravity $\left( 32 \mathrm { ft } / \mathrm { sec } ^ { 2 } \right)$ .

The position vector of a particle is r(t). Find the requested vector. -The acceleration at $t = \frac { \pi } { 8 }$ for $\mathbf { r } ( \mathrm { t } ) = \left( 10 \mathrm { t } - 3 \mathrm { t } ^ { 3 } \right) \mathbf { i } + 2 \tan ( 2 \mathrm { t } ) \mathbf { j } + \mathrm { e } ^ { 5 } \mathrm { t } \mathbf { k }$

The position vector of a particle is r(t). Find the requested vector. -The velocity at $t = 4$ for $\mathbf { r } ( \mathrm { t } ) = \left( 8 - 4 \mathrm { t } ^ { 2 } \right) \mathbf { i } + ( 6 \mathrm { t } + 2 ) \mathbf { j } - \mathrm { e } ^ { - 3 \mathrm { t } } \mathbf { k }$

Find the length of the indicated portion of the curve. - $\mathbf { r } ( \mathrm { t } ) = ( 5 + 2 \mathrm { t } ) \mathbf { i } + ( 6 + 3 \mathrm { t } ) \mathbf { j } + ( 4 - 6 \mathrm { t } ) \mathbf { k } , - 1 \leq \mathrm { t } \leq 0$

The position vector of a particle is r(t). Find the requested vector. -The acceleration at $t = 1$ for $\mathbf { r } ( \mathrm { t } ) = \mathrm { t } ^ { 5 } \mathbf { i } + 7 \ln \left( \frac { 1 } { 4 + \mathrm { t } } \right) \mathbf { j } + \frac { 3 } { \mathrm { t } } \mathbf { k }$

$\text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. }$ - $\mathbf { r } ( \mathrm { t } ) = \frac { 4 } { 3 } ( 1 + \mathrm { t } ) ^ { 3 / 2 } \mathbf { i } + \frac { 4 } { 3 } ( 1 - t ) ^ { 3 / 2 } \mathbf { j } + 1 \mathrm { t } \mathbf { k }$

Find the torsion of the space curve. -r(t) = (5t sin t + 5 cos t)i + (5t cos t) - 5 sin t)j - 6k

Find the torsion of the space curve. - $\mathbf { r } ( \mathrm { t } ) = ( \mathrm { t } - 9 ) \mathbf { i } + ( \ln ( \sec t ) + 8 ) \mathbf { j } - 10 \mathbf { k } , - \pi / 2 < \mathrm { t } < \pi / 2$

Find the unit tangent vector of the given curve. - $\mathbf { r } ( \mathrm { t } ) = \left( 5 + 2 \mathrm { t } ^ { 8 } \right) \mathbf { i } + \left( 3 + 10 \mathrm { t } ^ { 8 } \right) \mathbf { j } + \left( 5 + 11 t ^ { 8 } \right) \mathbf { k }$

$\text { Find the acceleration vector in terms of } u _ { r } \text { and } u _ { \theta } \text {. }$ - $r = a ( 5 - \cos \theta ) \text { and } \frac { d \theta } { d t } = 9$

The position vector of a particle is r(t). Find the requested vector. -The velocity at $t = 0$ for $\mathbf { r } ( t ) = \cos ( 5 t ) \mathbf { i } + 7 \ln ( t - 2 ) \mathbf { j } - \frac { t ^ { 3 } } { 10 } \mathbf { k }$

$\text { For the curve } r ( t ) \text {, write the acceleration in the form aTT } + \text { anN. }$ - $\mathbf { r } ( \mathrm { t } ) = \left( 8 \sin \frac { 5 } { 8 } \mathrm { t } + 8 \right) \mathrm { i } + \left( 8 \cos \frac { 5 } { 8 } \mathrm { t } - 3 \right) \mathrm { j } + 12 \mathrm { tk }$

Calculate the arc length of the indicated portion of the curve r(t). - $r ( t ) = 2 t ^ { 4 } \mathbf { i } + 10 t ^ { 4 } \mathbf { j } + 11 t ^ { 4 } k _ { , } - 1 \leq t \leq 3$

Calculate the arc length of the indicated portion of the curve r(t). - $\mathbf { r } ( t ) = ( 4 - 6 t ) \mathbf { i } + ( 9 + 9 t ) \mathbf { j } + ( 2 t - 2 ) \mathbf { k } , - 10 \leq t \leq 1$

$\text { Find the arc length parameter along the curve from the point where } t = 0 \text { by evaluating } s = \int _ { 0 } ^ { t } | v ( \tau ) | d \tau \text {. }$ -r(t) = (2 + 2t)i + (5 + 3t)j + (5 - 6t)k

For the curve r(t), find an equation for the indicated plane at the given value of t. -r(t) = (t2 - 8)i + (2t - 5)j + 8k; osculating plane at t = 6.