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Chapter 10: Problem 14

Find \(\mathbf{r}^{\prime}(t)\). $$ \mathbf{r}(t)=\left\langle\sin t-t \cos t, \cos t+t \sin t, t^{2}\right\rangle $$

Short Answer

Expert verified
\(\mathbf{r}^{\prime}(t) = \left\langle t \sin t, t \cos t, 2t \right\rangle\)

Step by step solution

01

Identify the functions to be differentiated

The vector function \(\mathbf{r}(t)\) comprises three components, each depending on the variable \(t\): \(\sin t-t \cos t\), \(\cos t+t \sin t\), \(t^{2}\). Each of these will be differentiated separately.
02

Differentiate the first component

The first component \(\sin t - t \cos t\) can be separated into two parts: \(y = \sin t\) and \(z = -t \cos t\). The derivative of \(\sin t\) is \(\cos t\) and the derivative of \(-t \cos t\) which uses the product rule is \(-\cos t + t \sin t\). Adding those two parts gives the derivative of the first component: \(\cos t - \cos t + t \sin t = t \sin t\).
03

Differentiate the second component

The second component \(\cos t + t \sin t\) can be separated into two parts: \(y = \cos t\) and \(z = t \sin t\). The derivative of \(\cos t\) is \(-\sin t\) and the derivative of \(t \sin t\) using the product rule is \(\sin t + t \cos t\). Adding those two parts gives the derivative of the second component: \(-\sin t + \sin t + t \cos t = t \cos t\).
04

Differentiate the third component

The third component is the simple function \(t^2\). Its derivative is \(2t\).
05

Combine the derivatives into a vector

combine the derivatives found in the earlier steps into a new vector. This vector represents the derivative of the vector function \(\mathbf{r}(t)\), and is written as \(\mathbf{r}^{\prime}(t) = \left\langle t \sin t, t \cos t, 2t \right\rangle\)

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Most popular questions from this chapter

Find \(\mathbf{r}(t)\) for the given conditions. $$ \mathbf{r}^{\prime}(t)=t e^{-t^{2}} \mathbf{i}-e^{-t} \mathbf{j}+\mathbf{k}, \quad \mathbf{r}(0)=\frac{1}{2} \mathbf{i}-\mathbf{j}+\mathbf{k} $$

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