Calculus III Advanced (Course) (12.3) (Homework)
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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.
- Find the velocity, speed, and acceleration vectors for the object.
- 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.
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