How do you differentiate the following parametric equation: x(t)=t^2-t, y(t)=t^3-t^2+3t x(t)=t2t,y(t)=t3t2+3t?

1 Answer
Apr 8, 2018

The derivative dy/dxdydx of the parametric equation is (3t^2-2t+3)/(2t-1)3t22t+32t1.

Explanation:

To find dy/dxdydx of the parametric equation (t^2-t,t^3-t^2+3t)(t2t,t3t2+3t), find dxdx from x(t)x(t) and dydy from y(t)y(t) in terms of dtdt:

color(white)=>x(t)=t^2-tx(t)=t2t

=>dx=(2t-1)dtdx=(2t1)dt

and

color(white)=>y(t)=t^3-t^2+3ty(t)=t3t2+3t

=>dy=(3t^2-2t+3)dtdy=(3t22t+3)dt

Now, dy/dxdydx will be the quotient of these two:

dy/dx=((3t^2-2t+3)color(red)cancelcolor(black)(dt))/((2t-1)color(red)cancelcolor(black)(dt))=(3t^2-2t+3)/(2t-1)

That's the derivative. Hope this helped!