Question

1. let $vec{r}(t):=langlecos(3t),sin(3t)

angle$ and $f(t):=e^{2t}$. compute $\frac{d}{dt}(vec{r}(f(t)))$ first by composing the functions and then taking the derivative, and second by using the multivariable chain rule.

Explanation:

Step1: Recall the multivariable chain - rule

The multivariable chain - rule for $\frac{d}{dt}(\vec{r}(f(t)))$ is $\vec{r}'(f(t))\cdot f'(t)$. First, find $\vec{r}'(t)$ and $f'(t)$.
Given $\vec{r}(t)=\langle\cos(3t),\sin(3t)
angle$. Using the derivative rules for trigonometric functions, if $x(t)=\cos(3t)$ and $y(t)=\sin(3t)$, then $x'(t)=- 3\sin(3t)$ and $y'(t)=3\cos(3t)$. So, $\vec{r}'(t)=\langle - 3\sin(3t),3\cos(3t)
angle$.
Given $f(t)=e^{2t}$, by the chain - rule for single - variable functions, $f'(t) = 2e^{2t}$.

Step2: Apply the multivariable chain - rule

Substitute $f(t)$ into $\vec{r}'(t)$ and multiply by $f'(t)$.
$\vec{r}'(f(t))=\langle - 3\sin(3e^{2t}),3\cos(3e^{2t})
angle$ and $f'(t)=2e^{2t}$.
Then $\frac{d}{dt}(\vec{r}(f(t)))=\vec{r}'(f(t))\cdot f'(t)=2e^{2t}\langle - 3\sin(3e^{2t}),3\cos(3e^{2t})
angle=\langle - 6e^{2t}\sin(3e^{2t}),6e^{2t}\cos(3e^{2t})
angle$.

Answer:

$\langle - 6e^{2t}\sin(3e^{2t}),6e^{2t}\cos(3e^{2t})
angle$