Calculus III Advanced (Course) (12.2) (Homework)

Section 12.2 Homework

From Calculus 10e by Larson and Edwards, p. 830. Exercises 4, 26, 42, 46, 52, 58.

Exercise 12.2.4 Differentiate Vector-Valued Functions

Find \( \textbf{r}^{\prime}(t), \: \textbf{r}(t_{0}), \: \textbf{r}^{\prime}(t_{0}) \) for

$$ \textbf{r}(t) = 3 \sin t \textbf{i} + 4 \cos t \textbf{j}, \: t_{0}=\frac{\pi}{2} $$

Solution First, find \( \textbf{r}^{\prime}(t) \)

\( \textbf{r}^{\prime}(t) = 3 \cos t \textbf{i} - 4 \sin t \textbf{j}\)

Second, plug in \( t_{0} \) into both \( \textbf{r}(t) \) and \( \textbf{r}^{\prime}(t) \).

$$ \textbf{r} \left( \frac{\pi}{2} \right) $$	$$= 3 \sin \frac{\pi}{2} \textbf{i} + 4 \cos \frac{\pi}{2} \textbf{j} $$
	\(=3(1)\textbf{i} + 4 (0)\textbf{j} \)
	\(= 3\textbf{i}\)

$$ \textbf{r}^{\prime} \left( \frac{\pi}{2} \right) $$	$$= 3 \cos \frac{\pi}{2} \textbf{i} - 4 \sin \frac{\pi}{2} \textbf{j} $$
	\(= 3(0) \textbf{i} - 4(1) \textbf{j} \)
	\(= -4 \textbf{j}\)

Exercise 12.2.26 Higher-Order Differentiation

Find

a. \( \textbf{r}^{\prime}(t) \)

b. \( \textbf{r}^{\prime \prime}(t) \)

c. \( \textbf{r}^{\prime}(t) \cdot \textbf{r}^{\prime \prime}(t) \)

d. \( \textbf{r}^{\prime}(t) \times \textbf{r}^{\prime \prime}(t) \)

where

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

Solution First find the first and second derivatives.

a. \( \textbf{r}^{\prime}(t) = 3t^{2}\textbf{i} + 4t\textbf{j} + 3 \textbf{k} \)

b. \( \textbf{r}^{\prime \prime}(t) = 6t\textbf{i} + 4\textbf{j}\)

c. \( \textbf{r}^{\prime}(t) \cdot \textbf{r}^{\prime \prime}(t) = (3t^{2})( 6t)\textbf{i} + (4t)(4) \textbf{j} + (3)(0) \textbf{k} = 18t^{3} \textbf{i} + 16t \textbf{j} + 0 \textbf{k}\)

d. \( \textbf{r}^{\prime}(t) \times \textbf{r}^{\prime \prime}(t) \)	$$= \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 3t^{2} & 4t & 3 \\ 6t & 4 & 0 \end{vmatrix} $$
	$$= \begin{vmatrix} 4t & 3 \\ 4 & 0 \end{vmatrix}\textbf{i}-\begin{vmatrix} 3t^{2} & 3 \\ 6t & 0 \end{vmatrix}\textbf{j} $$	$$+\begin{vmatrix} 3t^{2} & 4t \\ 6t & 4 \end{vmatrix}\textbf{k}$$
	\(= -12\textbf{i} + 18t\textbf{j} - 12t^{2}\textbf{k} \)

Exercise 12.2.42 Using Two Methods

Find

$$\textbf{a} \:\:\:\: \frac{d}{dt} \left[ \textbf{r}(t) \cdot \textbf{u}(t) \right] $$

$$\textbf{b} \:\:\:\: \frac{d}{dt} \left[ \textbf{r}(t) \times \textbf{u}(t) \right] $$

(i) Find the product first, then differentiate.
(ii) Apply Theorem 12.2.2.
where

\( \textbf{r}(t) = \cos t\textbf{i} + \sin t\textbf{j} + t\textbf{k}\)

\( \textbf{u}(t) = \textbf{j} + t \textbf{k} \)

Solution
a i

$$ \frac{d}{dt} \left[ \textbf{r}(t) \cdot \textbf{u}(t) \right] $$	$$= \frac{d}{dt} \left[\cos t\textbf{i} + \sin t\textbf{j} + t\textbf{k} \cdot \textbf{j} + t \textbf{k} \right] $$
	$$= \frac{d}{dt} \left[ t^{2}+\sin t \right]= 2t + \cos t $$

a ii Applying Theorem 12.2.2 Property 4 produces

\( \textbf{r}^{\prime}(t) = -\sin t\textbf{i} + \cos t\textbf{j} + \textbf{k}\)

\( \textbf{u}^{\prime}(t) = \textbf{k} \)

$$ \frac{d}{dt} \left[ \textbf{r}(t) \cdot \textbf{u}(t) \right] $$	\(= \textbf{r}(t) \cdot \textbf{u}^{\prime}(t) + \textbf{r}^{\prime}(t) \cdot \textbf{u}(t)\)
	\(= (\cos t\textbf{i} + \sin t\textbf{j} + t\textbf{k}) \cdot ( \textbf{k} ) + ( -\sin t\textbf{i} + \cos t\textbf{j} + \textbf{k}) \cdot (\textbf{j} + t \textbf{k})\)
	\(= t + (t + \cos t) = 2t+ \cos t\)

b i

$$ \frac{d}{dt} \left[ \textbf{r}(t) \times \textbf{u}(t) \right] $$	$$= \frac{d}{dt} \left[\cos t\textbf{i} + \sin t\textbf{j} + t\textbf{k} \times \textbf{j} + t \textbf{k} \right] $$
	$$= \frac{d}{dt} \left[ t \sin t -t \right] = \sin t + t \cos t - 1 $$

b ii Applying Theorem 12.2.2 Property 5 produces

$$ \frac{d}{dt} \left[ \textbf{r}(t) \times \textbf{u}(t) \right] $$	\(= \textbf{r}(t) \times \textbf{u}^{\prime}(t) + \textbf{r}^{\prime}(t) \times \textbf{u}(t)\)
	\(= (\cos t\textbf{i} + \sin t\textbf{j} + t\textbf{k}) \times ( \textbf{k} ) + ( -\sin t\textbf{i} + \cos t\textbf{j} + \textbf{k}) \times (\textbf{j} + t \textbf{k})\)
	\(= t + (t + \cos t) = 2t+ \cos t\)

Exercise 12.2.46 Find an Indefinite Integral

Find the indefinite integral for

$$ \int \left( \ln t \textbf{i} + \frac{1}{t}\textbf{j} + \textbf{k} \right) \: dt$$

Solution Integrating on a component-by-component basis produces

$$ \int \left( \ln t \textbf{i} + \frac{1}{t}\textbf{j} + \textbf{k} \right) \: dt = (t \ln t - t + C )\textbf{i} + (\ln t + C)\textbf{j}+ (t+C)\textbf{k}$$

Exercise 12.2.52 Evaluate a Definite Integral

Evaluate the definite integral.

$$ \int_{-1}^{1} \left( t \textbf{i} + t^{3} \textbf{j} + \sqrt[3]{t} \textbf{k} \right) \: dt $$

Solution

$$ \int_{-1}^{1} \left( t \textbf{i} + t^{3} \textbf{j} + \sqrt[3]{t} \textbf{k} \right) \: dt $$	$$= \left( \int_{-1}^{1} t \: dt \right) \textbf{i} + \left( \int_{-1}^{1} t^{3} \: dt \right) \textbf{j} + \left( \int_{-1}^{1} \sqrt[3]{t} \: dt \right) \textbf{k} $$
	$$= \left[ \frac{1}{2} t^{2} \right]_{-1}^{1} \textbf{i} + \left[ \frac{1}{4} t^{4} \right]_{-1}^{1} \textbf{j} + \left[ \frac{3}{4} t^{3/4} \right]_{-1}^{1} \textbf{k} $$
	$$= \left( \frac{1}{2} (1)^{2} - \frac{1}{2} (-1)^{2} \right) \textbf{i} + \left( \frac{1}{4} (1)^{4} - \frac{1}{4} (-1)^{4} \right) \textbf{j} + \left( \frac{3}{4} (1)^{3/4} - \frac{3}{4} (-1)^{3/4} \right) \textbf{k} $$
	$$= 0 \textbf{i} + 0 \textbf{j} + \frac{3}{4} - \left( -\frac{3 \sqrt[3]{-1}}{4} \right) \textbf{k} $$

Exercise 12.2.58 Find the Antiderivative

Find the antiderivative where

$$ \textbf{r}^{\prime}(t) = 3 t^{2} \textbf{j} + 6 \sqrt{t} \textbf{k}, \: \textbf{r}(0) = \textbf{i} + 2 \textbf{j}$$

Solution First step, find \( \textbf{r}(t) \).

\( \textbf{r}(t) \)	$$= \int \textbf{r}^{\prime}(t) \: dt $$
	$$= \left( \int 3 t^{2} \: dt \right) \textbf{j} + \left( \int 6 \sqrt{t} \: dt \right) \textbf{k}$$
	$$= \left( \vphantom{t^{3/2} } t^{3} + C_{1} \right) \textbf{j} + \left( 4 t^{3/2} + C_{2} \right) \textbf{k}$$
\( \textbf{r}(0) \)	\(= \textbf{i} + \left( 0 + C_{1} \right) \textbf{j} + \left( 0 + C_{2} \right) \textbf{k}\)
\(\textbf{i} + 2 \textbf{j} \)	\( = \textbf{i} + \left( 0 + C_{1} \right) \textbf{j} + \left( 0 + C_{2} \right) \textbf{k} \)

Therefore, \(C_{1} = 2 \) and \(C_{2} = 0 \).
The antiderivative is
\( \textbf{r}(t) = (t + 1) \textbf{i} + \left( \vphantom{t^{3/2} } t^{3} + 2 \right) \textbf{j} + \left( 4 t^{3/2} \right) \textbf{k} \)

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Parent Article: Calculus III Advanced (Course)