Question

What is the instantaneous velocity of an object moving in accordance to `f(t)= (e^(t^2),2t-e^t)` at `t=-1`?

Answer 1

Find the velocity vector by taking the derivatives of the `x` and `y` portions separately:

`f'(t)=(2te^(t^2),2-e^t)`

So at `t=-1` the velocity vector is:

`f'(-1)=(-2e^((-1)^2),2-e^-1)=(-2e,2-1/e)`

We could find the speed at this point by finding the magnitude of the vector:

`"speed"=sqrt((-2e)^2+(2-1/e)^2)`