Circular trajectory speed An object moves clockwise around a circle centered at the origin with radius 5m begnning at the point (0, 5).  Find a position function $\vec{r}$ that describes the motion if it occurs with speed $e^{-t}$
My attempt:
I know that $r(t) = (5\sin t, 5\cos t)$ for the general path of the function.  But I don't know exactly how I would change the speed of the function.
All I know is its going to result in changing the function sin and cos from $f(t)$ to $f(g(t))$.

Ok ill give it a go, but I might be misinterpreting. The particle in question can be described by the vector function $\mathbf{r}(\theta(t)) = (x,y) = [5\cos \theta(t), 5\sin \theta(t)]$ where $x,y$ are the coordinates in the plane, $\theta$ is the angle position of the particle at a particular time. 
\begin{align}
\frac{d\mathbf{r}}{dt} = \left( -5\frac{d\theta}{dt}\sin \theta(t), 5\frac{d\theta}{dt}\cos \theta(t) \right)
\end{align}
The speed is the magnitude of this vector which is easily simplified using basic trig to $|\frac{d\mathbf{r}}{dt}| = |5\frac{d\theta}{dt}|$ which equals $e^{-t}$ according to you. Hence $\theta(t)= C+\frac{e^{-t}}{5}$ in the equation above. I believe this is consistent with the fact that the particle is slowing down as it moves clockwise around the circle. Now we must find the value of $C$ using the initial condition that at $t=0$, $y = 5$. Hence we require that $\theta(t=0) = \pi/2$ which means that $C = \pi/2 -1/5$.
