Question

How do you differentiate the following parametric equation: `x(t)=(t-1)^2-e^t, y(t)= (t+2)^2+t^2`?

Answer 1

To find derivatives with respect to t, differentiate normally. To find `(dy)/dx`, use `(dy)/dx = ((dy)/dt)/((dx)/dt)`. Solutions below.

Explanation:

Provided you mean finding `(dx)/dt` and `(dy)/dt` , we do it the same way that we would if t were x and x (t) and y (t) were both functions of x. The usual rules apply, including the chain rule (`f (x)=g (h (x))-> f'(x)=h'(x)g'(h)`. Thus...

`x(t)=(t-1)^2-e^t -> (dx)/dt = 2 (t-1)-e^t`

And

`y(t)=(t+2)^2 +t^2 -> (dy)/dt = 2 (t+2)+2t = 4t+2`

If instead you want `(dy)/dx`, we can solve that as follows:

`(dy)/dt = (dy)/dx (dx)/dt` (chain rule).

Rearranging, we can get...

`(dy)/dx = ((dy)/dt)/((dx)/dt) -> (dy)/dx =(4t+2)/(2(t-1)-e^t`