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Chapter 10: Problem 42

Use the definition of the derivative to find \(\mathbf{r}^{\prime}(t)\). $$ \mathbf{r}(t)=\langle 0, \sin t, 4 t\rangle $$

Short Answer

Expert verified
The derivative \( \mathbf{r}^\prime(t) \) of the vector is \( \langle 0, cos(t), 4 \rangle \).

Step by step solution

01

Break Down the Vector

Firstly, separate the vector into its components. The vector \( \mathbf{r}(t) \) is given by \( \langle 0, \sin(t), 4t \rangle \). Thus, it has three components: \( r_1(t) = 0, r_2(t) = \sin(t) \), and \( r_3(t) = 4t \).
02

Derivate Each Component

Considering each of these components as separate functions, we now find their derivatives. The derivative of \( r_1(t) = 0 \) is simply \( r_1^\prime(t) = 0 \). For \( r_2(t) = sin(t) \), using the standard derivative of \( sin(t) \), we find \( r_2^\prime(t) = cos(t) \). Finally, the derivative of \( r_3(t) = 4t \) is \( r_3^\prime(t) = 4 \).
03

Recombine the Components

With each of the derivative components calculated, we simply form the derivative vector \( \mathbf{r}^\prime(t) \) by recombining these, yielding \( \mathbf{r}^\prime(t) = \langle 0, cos(t), 4 \rangle \) as the answer.

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Most popular questions from this chapter

Find (a) \(\quad D_{t}[\mathbf{r}(t) \cdot \mathbf{u}(t)] \quad\) and (b) \(D_{t}[\mathbf{r}(t) \times \mathbf{u}(t)]\) by differentiating the product, then applying the properties of Theorem 10.2. $$ \mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+t \mathbf{k}, \quad \mathbf{u}(t)=\mathbf{j}+t \mathbf{k} $$

Consider a particle moving on a circular path of radius \(b\) described by $$ \begin{aligned} &\mathbf{r}(t)=b \cos \omega t \mathbf{i}+b \sin \omega t \mathbf{j}\\\ &\text { where } \omega=d \theta / d t \text { is the constant angular velocity. } \end{aligned} $$ Find the acceleration vector and show that its direction is always toward the center of the circle.

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