Calculus III 12 Exam 1

Chapter 12 Exam

From Calculus 10e by Larson and Edwards, p. 863. Exercises 11, 25, 29, 35, 41.

Exercise 11 Sketching a Curve

Sketch the curve represented by the vector-valued function and give the curve's orientation.

\( \textbf{r}(t) = (t+1)\textbf{i} + (3t-1)\textbf{j} + 2t\textbf{k} \)

Solution Break it down into \(x=t+1\), \(y=3t-1\), and \(z=2t\), produces the sketch[1] shown in Figure 1.

The curve's orientation is to the right as \(t\) increases.

Exercise 25 Finding an Indefinite Integral

Find the indefinite integral.

$$ \int (\textbf{i} + 3\textbf{j} + 4t\textbf{k}) \:dt $$

Solution

$$ \int (\textbf{i} + 3\textbf{j} + 4t\textbf{k}) \:dt = t\textbf{i} + 3t\textbf{j} + 2t^{2}\textbf{k} + C$$

Exercise 29 Evaluating a Definite Integral

Find the indefinite integral.

$$ \int_{-2}^{2} (3t\textbf{i} + 2t^{2}\textbf{j} - t^{3}\textbf{k}) \:dt $$
	$$= \left[ \frac{3}{2}t^{2}\textbf{i} + \frac{2}{3}t^{3}\textbf{j} - \frac{1}{4}t^{4}\textbf{k} + C \right]_{-2}^{2}$$
	$$ = \left[\frac{3}{2}(2)^{2} + \frac{2}{3}(2)^{3} - \frac{1}{4}(2)^{4} \right] - \left[\frac{3}{2}(-2)^{2} + \frac{2}{3}(-2)^{3} - \frac{1}{4}(-2)^{4}\right] $$
	$$= \left[6 + \frac{16}{3} -4 \right] - \left[ 6 - \frac{16}{3}- 4 \right] $$
	$$= \left[2 + \frac{16}{3} \right] - \left[ 2 - \frac{16}{3} \right] $$
	$$= \left[ \frac{6}{3} + \frac{16}{3} \right] - \left[ \frac{6}{3} - \frac{16}{3} \right] $$
	$$= \frac{22}{3} + \frac{10}{3} = \frac{32}{3} $$

Exercise 35 Finding Velocity and Acceleration Vectors

The position vector describes the path for an object moving in three-dimensions.
(a) Find the velocity vector, speed, and acceleration vector of the object.
(b) Evaluate the velocity vector and acceleration vector for the object at the given \(t\) value.

\( \textbf{r}(t) = 4t\textbf{i} + t^{3}\textbf{j}-t\textbf{k}, \: t=1 \)

Solution By Definition 12.3.1, Velocity\(=\textbf{r}^{\prime}(t) \), and Acceleration\(=\textbf{r}^{\prime \prime}(t) \).

(a)	\(\textbf{r}^{\prime}(t) = 4\textbf{i} + 3t^{2}\textbf{j}-\textbf{k} \)	Velocity
	\(\textbf{r}^{\prime \prime}(t) = 6t\textbf{j} \)	Acceleration
(b)	\(\textbf{r}^{\prime}(1) = 4 + 3-1=6 \)
	\(\textbf{r}^{\prime \prime}(1) = 6 \)

Exercise 41 Projectile Motion

A projectile is fired from ground level at a 20° angle with the horizontal. The projectile has a 95 meter range. Find the minimum initial velocity.
Solution Using Theorem 12.3.1, where \(h=0\), \( \theta = 20^{\circ} \), and \(v_{0}\) is unknown.

\( v_{0} \)	$$= \sqrt{ \frac{Rg}{\sin 2 \theta} }$$	Formula for initial velocity.
	$$= \sqrt{ \frac{95(32)}{\sin 2( 20^{\circ})} }$$
	$$= \sqrt{ \frac{3040}{\sin \frac{2 \pi}{9}} }$$
	$$= \sqrt{ \frac{3040}{0.642787} }$$
	$$= \sqrt{ 4729.404919 } = 68.770669$$

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Parent Article: Calculus III Advanced (Course)