Question

suppose r(t) = <cos(πt),sin(πt),4t> is the position vector function for an insect crawling up a helix. units are inches for distance and seconds for the time - parameter t. at t = 0, the insect is on the floor (i.e., the xy - plane). (a) at what time t does the insect reach a height of 12 inches above the floor? t = (b) what is the insects velocity vector when its height is 12? v(t) =

Explanation:

Step1: Find the time when height is 12

The height of the insect is given by the $z -$ component of the position vector $\mathbf{r}(t)=\langle\cos(\pi t),\sin(\pi t),4t
angle$. 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 $\mathbf{r}(t)=\langle\cos(\pi t),\sin(\pi t),4t
angle$, 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, $\mathbf{v}(t)=\langle-\pi\sin(\pi t),\pi\cos(\pi t),4
angle$.
When $t = 3$, we substitute $t = 3$ into $\mathbf{v}(t)$:
$\sin(3\pi)=0$ and $\cos(3\pi)=- 1$.
$\mathbf{v}(3)=\langle-\pi\times0,\pi\times(-1),4
angle=\langle0,-\pi,4
angle$

Answer:

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