Calculus III Advanced (Course) (12.3) (Homework)

Section 12.3 Homework

From Calculus 10e by Larson and Edwards, p. 838. Exercises 18, 22, 26, 36.

Exercise 12.3.18 Finding Velocity and Acceleration Vectors

The position vector \( \textbf{r} \) describes the path of an object moving in three-dimensions.

1. Find the velocity, speed, and acceleration vectors for the object.
2. Evaluate the velocity and acceleration vectors for the object at the given value \(t\).

Position Vector	Time
$$ \textbf{r} = \left \langle \ln t, \frac{1}{t}, t^{4} \right \rangle $$	\(t=2 \)

Soluton Write out the position vector in component form and find the velocity, speed, and acceleration vectors.

\( \textbf{r}(t) \)	$$= \ln t\textbf{i} + \frac{1}{t}\textbf{j} + t^{4}\textbf{k} \:\:\:\: $$	Component version
\( \textbf{v}(t) \)	$$ =\textbf{r}^{\prime}(t) = \frac{1}{t}\textbf{i} - \frac{1}{t^{2}}\textbf{j} + 4t^{3}\textbf{k} \:\:\:\: $$	Velocity vector
\( \textbf{a}(t) \)	$$ =\textbf{r}^{\prime \prime}(t) = - \frac{1}{t^{2}}\textbf{i} + \frac{2}{t^{3}}\textbf{j} + 12t^{2}\textbf{k} \:\:\:\: $$	Acceleration vector
\( \| \textbf{v}(t) \| \)	$$= \| \textbf{r}^{\prime}(t) \| = \sqrt{ \left(\frac{1}{t}\right)^{2}\textbf{i} - \left( \frac{1}{t^{2}} \right)^{2}\textbf{j} + \left( 4t^{3} \right)^{2}\textbf{k} } \:\:\:\: $$	Speed vector

Evaluate the velocity and acceleration vectors for the object when \(t=2\).

Velocity	$$= \frac{1}{2} - \frac{1}{(2)^{2}} + 4(2)^{3} = 32 \frac{1}{4}$$
Acceleration	$$= - \frac{1}{(2)^{2}} + \frac{2}{(2)^{3}} + 12(2)^{2} = 48 \frac{1}{2} $$

Exercise 12.3.22 Finding a Position Vector by Integration

Use the given acceleration vector to find the velocity and position vectors. Then find the position at time \(t=2\).

\( \textbf{a}(t) \)	\(= -32 \textbf{k} \)
\( \textbf{v}(0) \)	\(= 3\textbf{i} -2 \textbf{j} + \textbf{k} \)
\( \textbf{r}(0) \)	\(= 5 \textbf{j} + 2\textbf{k} \)

Solution First find the velocity vector.

\( \textbf{v}(t) \)	$$= \int \textbf{a}(t)\:dt \:\:\:\: \color{red}{\text{ Velocity vector }}$$
	$$= \int ( -32 \textbf{k} ) \:dt$$
	\(= -32t\textbf{k} + \textbf{C} \)

where \( \textbf{C}= C_{1}\textbf{i} + C_{2}\textbf{j} + C_{3}\textbf{k} \). Letting \(t=0\) and applying the initial condition \( \textbf{v}(0) = 3\textbf{i} -2 \textbf{j} + \textbf{k} \) produces

\( \textbf{v}(0) = C_{1}\textbf{i} + C_{2}\textbf{j} + C_{3}\textbf{k} = 3\textbf{i} -2 \textbf{j} + \textbf{k} \rightarrow C_{1} = 3, \: C_{2} =-2,\: C_{3} =1. \)

The velocity at any time \(t\) is

\( \textbf{v}(t) = 3\textbf{i} -2 \textbf{j} - 32t \textbf{k} \:\:\:\: \)Velocity vector

Integrate the velocity vector to find the position vector.

\( \textbf{r}(t) \)	$$= \int \textbf{v}(t)\:dt \:\:\:\: $$	Velocity vector
	$$= \int ( 3\textbf{i} -2 \textbf{j} - 32t \textbf{k} ) \:dt$$
	$$= 3t\textbf{i} -2t \textbf{j} - 16t^{2} \textbf{k} + \textbf{C} $$

where \( \textbf{C}= C_{4}\textbf{i} + C_{5}\textbf{j} + C_{6}\textbf{k} \). Letting \(t=0\) and applying the initial condition \( \textbf{r}(0) = \textbf{j} + 2\textbf{k} \) produces

\( \textbf{r}(0) = C_{4}\textbf{i} + C_{5}\textbf{j} + C_{6}\textbf{k} = 5 \textbf{j} + 2\textbf{k} \rightarrow C_{4} = 0,\: C_{5} = 5,\: C_{6} = 2. \)

The position vector is.

\( \textbf{r}(t) = 3t\textbf{i} +(5-2t) \textbf{j} + (2- 16t^{2}) \textbf{k} \)

The object's location at \(t=2\) is given by

\( \textbf{r}(2) \)	\(= 3(2)\textbf{i} +(5-2(2)) \textbf{j} + (2- 16(2)^{2}) \textbf{k} \)
	\(= 6\textbf{i} + \textbf{j} -62 \textbf{k} \)

Exercise 12.3.26 Projectile Motion Find Maximum Height

Use the model for projectile motion, assuming there is no air resistance.

Determine the maximum height and range for a projectile fired at a height 3 feet above the ground with an initial velocity at 900 feet per second and at a 45° angle above the horizontal.
Solution From the question

\( h=3, \: v_{0}=900 \text{, and } \theta = 45^{\circ}. \)

Using \(g=32\) feet per second per second produces

\( \textbf{r}(t) \)	$$= \left( 900 \cos \frac{\pi}{4} \right)t\textbf{i} + \left[3 + \left( 900 \sin \frac{\pi}{4} \right)t - 16t^{2} \right]\textbf{j}$$
	\(= \left( 450 \sqrt{2}t \right)\textbf{i} + \left(3 + 450 \sqrt{2}t - 16t^{2} \right)\textbf{j} \).

The velocity vector is

\( \textbf{v}(t) = \textbf{r}^{\prime}(t) = 450 \sqrt{2}\textbf{i} +(450 \sqrt{2}- 32t)\textbf{j}. \)

The maximum height occurs when

\( y^{\prime}(t)= 450 \sqrt{2}- 32t \)

is equal to 0, which implies that

$$ t = \frac{225 \sqrt{2}}{16} \approx 19.887378 \text{ seconds. } $$

The maximum height reached by the projectile is

\(y\)	$$= 3 + 450 \sqrt{2} \left( \frac{225 \sqrt{2}}{16} \right) - 16 \left( \frac{225 \sqrt{2}}{16} \right)^{2} $$
	$$ = \frac{50649}{8} \approx 6331.125 \text{ feet } $$

The maximum range is given by

$$ \frac{(v_{0})^{2} \cos \theta}{g} \left( \sin \theta + \sqrt{ \sin^{2} \theta + \frac{2gh}{(v_{0})^{2}}} \right) $$

where \(v_{0} = 900 \), \(g=32\) feet per second per second, \( \theta = 45^{\circ} \), and \(h=3\).

$$ \frac{(900)^{2} \frac{ \sqrt{2}}{2} }{32} \left( \frac{ \sqrt{2}}{2} + \sqrt{ \left( \frac{ \sqrt{2}}{2} \right)^{2}+ \frac{2(32)(3)}{(900)^{2}}} \right) \approx 25315.499645 \text{ feet}$$

Exercise 12.3.36 Projectile Motion Find Minimum Initial Velocity

Use the model for projectile motion, assuming there is no air resistance.

A projectile is fired from ground level at a 12° angle with the horizontal. The projectile is to have a 200 foot range. Find the minimum initial velocity necessary.
Solution From the question

\(h=0 \) and \( \theta = 12^{\circ} \text{ or } \frac{\pi}{15} \)

The maximum range is given by

$$ \frac{ (v_{0})^{2} \cos \theta}{g} \left( \sin \theta + \sqrt{ \sin^{2} \theta + \frac{2gh}{ (v_{0})^{2}}} \right) $$

where \(g=32\) feet per second per second, \( \theta = 12^{\circ} \text{ or } \frac{\pi}{15} \), and \(h=0\). Set equal to 200 and solve for \(v_{0} \).

\( 200 \)	$$= \frac{ (v_{0})^{2} \cos \frac{\pi}{15} }{32} \left( \sin \frac{\pi}{15} + \sqrt{ \sin^{2} \frac{\pi}{15} } \right) $$
\( 6400 \)	$$= (v_{0})^{2} \cos \frac{\pi}{15} \left( \sin \frac{\pi}{15} + \sqrt{ \sin^{2} \frac{\pi}{15} } \right) $$
\( 6400 \)	\(= (v_{0})^{2} (0.40674)\)
\( 2603.136 \)	\(= (v_{0})^{2}\)
\( 51.02 \)	\(= v_{0}\)

The initial velocity is 51.02 feet per second per second.

Internal Links

Parent Article: Calculus III Advanced (Course)