suppose r(t) = <cos(πt),sin(πt),4t> is the position vector function for an insect crawling up a helix. units…

Explanation:

Step1: Find the time when height is 12

The height of the insect is given by the $z -$ component of the position vector $\mathbf{r}(t)=\langle\cos(\pi t),\sin(\pi t),4t\rangle$. Set the $z -$ component equal to 12. $4t = 12$ Solve for $t$: $t=\frac{12}{4}=3$ seconds.

Step2: Find the velocity vector

The velocity vector $\mathbf{v}(t)$ is the derivative of the position vector $\mathbf{r}(t)$. If $\mathbf{r}(t)=\langle\cos(\pi t),\sin(\pi t),4t\rangle$, then using the derivative rules: The derivative of $\cos(\pi t)$ with respect to $t$ is $-\pi\sin(\pi t)$ (by the chain - rule, where if $u = \pi t$, $\frac{d}{dt}\cos(u)=-\sin(u)\cdot\frac{du}{dt}=-\pi\sin(\pi t)$). The derivative of $\sin(\pi t)$ with respect to $t$ is $\pi\cos(\pi t)$ (by the chain - rule, $\frac{d}{dt}\sin(u)=\cos(u)\cdot\frac{du}{dt}=\pi\cos(\pi t)$). The derivative of $4t$ with respect to $t$ is 4. So, $\mathbf{v}(t)=\langle-\pi\sin(\pi t),\pi\cos(\pi t),4\rangle$. When $t = 3$, we substitute $t = 3$ into $\mathbf{v}(t)$: $\sin(3\pi)=0$ and $\cos(3\pi)=- 1$. $\mathbf{v}(3)=\langle-\pi\times0,\pi\times(-1),4\rangle=\langle0,-\pi,4\rangle$

Answer:

(a) $t = 3$ seconds (b) $\mathbf{v}(3)=\langle0,-\pi,4\rangle$