Question

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

Short answer

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

Step-by-step solution

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 \( r_3(t) = 4t \).

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 \( r_3^\prime(t) = 4 \).

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.