I am having some trouble trying to solve this vector functions problem. I can find the values and equations for the velocity and acceleration but I can't seem to understand how to sketch/graph them. I've looked through the solution that shows the sketch but I can't understand why the vectors are where they are. I've added an image to this post that shows the solution page that I referred to. It is the 5th exercise.
why are the vectors $v(\frac{\pi}{4}) , v(\frac{\pi}{2}), a(\frac{\pi}{4}), a(\frac{\pi}{2})$ in the positions that they are in. How do I understand why the vectors are in their specific positions.
I feel that I may be missing some prerequisites for sketching vectors for these kinds of problems, if anyone could recommend me some resources to understand these things better it would be greatly appreciated. Thanks for the help.
PROBLEM:
"give the position vectors of particles moving along various curves in the xy-plane. In each case, find the particle’s velocity and acceleration vectors at the stated times and sketch them as vectors on the curve."
5). Motion on the circle $x^2$ + $y^2$ = 1
$\vec{r(t)}$ = $\sin(t) \hat{i}$ + $\cos(t) \hat{j}$; $t = \frac{\pi}{4}$ and $\frac{\pi}{2}$
You've already found the position vector $\vec{r(t)}$ for you're problem so I'm assuming drawing that part is easy. For the velocity vectors, $v(t)$, you want to implement the following approach. First, look at the input, let's say you're finding $v(\frac{\pi}{4})$, so the input is $t=\frac{\pi}{4}$. First you want to evaluate $r(\frac{\pi}{4})$ to know what point the velocity vector's tail sits at. That should get you the point $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$. Next, calculate the velocity vector, that tells you what the direction and length of the vector should be. In you're case that's $v(\frac{\pi}{4})=(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$. So draw a vector that sits at $r(t)$, and moves $\frac{1}{\sqrt{2}}$ units in the $x$ direction and $-\frac{1}{\sqrt{2}}$ units in the $y$ direction. The approach is the same for drawing $a(t)$. $r(t)$ will tell you where the tail is (where the vector sits on the graph), $a(t)$ will tell you the magnitude and direction.