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

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Answer 1 · 2018-04-08T22:22:21

The derivative #dy/dx# of the parametric equation is #(3t^2-2t+3)/(2t-1)#.

To find #dy/dx# of the parametric equation #(t^2-t,t^3-t^2+3t)#, find #dx# from #x(t)# and #dy# from #y(t)# in terms of #dt#:

#color(white)=>x(t)=t^2-t#

#=>dx=(2t-1)dt#

and

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

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

Now, #dy/dx# 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!