Problem
Find the normal vector $\bf{N}$ to $\bf{r}(t)=\langle t,\cos t \rangle$ at $t=\frac{9\pi}{4}.$
(Use symbolic notation and fractions where needed.)
Feedback: Recall that the normal vector of $\bf{r}(t)$ is $\bf{T'}(t),$ where $\bf{T}(t)=\frac{\bf{r'}(t)}{||\bf{r'}(t)||}$ is a unit tangent vector.
How do I find this normal vector? So basically I did what the feedback said. I found the derivative of each function in the vector and got.
$$(1, -\sin(t))$$
Then I got the magnitude:
$$\sqrt{1+\sin^2\left(\frac{9\pi \:}{4}\right)}$$
Then I divided everything in my vector by that magnitude while putting in $$\frac{9\pi }{4}$$ like so:
$$\frac{-\sin\left(\frac{9\pi }{4}\right)}{\sqrt{1+\sin^2\left(\frac{9\pi \:}{4}\right)}}$$
What my problem? Thank you. Here's my answer:
$$\left\langle\sqrt{\frac{2}{3}},-\frac{\sqrt{3}}{3}\right\rangle$$
Best Answer
The unit tangent vector is $$T(t)=(\frac{1}{\sqrt{1+\sin^2(t)}},-\frac{\sin(t)}{\sqrt{\sin^2(t)+1}})$$
So the normal vector of $r(t)$ is $$T'(t)=(-\frac{\sqrt{2}\sin(2t)}{({3-\cos(2t)})^{\frac{3}{2}}},-\frac{\cos(t)}{(1+\sin^{2}(t))^{\frac{3}{2}}})$$
Now let $t=\frac{9\pi}{4}$ and you should get $$T'(\frac{9\pi}{4})=(-\frac{\sqrt{6}}{9},-\frac{2\sqrt{3}}{9}).$$