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

Question

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

Christopher Agresta · Accepted Answer

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

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#

Cameron Alexander · Answer

undefined To differentiate the given parametric equations $x(t) = (t - 1)^2 - e^t$ and $y(t) = (t + 2)^2 + t^2$, follow these steps:

1. Differentiate each parametric equation separately with respect to $t$.
2. Apply the chain rule where necessary.
3. Express $dx/dt$ and $dy/dt$ as functions of $t$.

Here are the derivatives:

For $x(t) = (t - 1)^2 - e^t$:

$$ \frac{dx}{dt} = 2(t - 1) - e^t $$

For $y(t) = (t + 2)^2 + t^2$:

$$ \frac{dy}{dt} = 2(t + 2) + 2t $$

These are the derivatives of the given parametric equations with respect to $t$.