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

Question

2321 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-11T10:00:16

#dy/dx=(9t^2+4t-3)/{2-(t+1)e^t}#.

Let #x=x(t), and, y=y(t)# be the given parametric eqns.

Then, #dy/dx=(dy/dt)/(dx/dt), (dx)/(dt)!=0......(star)#.

Now, #y(t)=-3t^3-2t^2+3t rArr dy/dt=-9t^2-4t+3.....(1)#.

#x(t)=te^t-2t=t(e^t-2)#.

#rArr x'(t)=t(e^t-0)+(e^t-2)1=(t+1)e^t-2........(2)#.

Using #(1) & (2)# in #(star)#, we get,

#dy/dx=(-9t^2-4t+3)/((t+1)e^t-2)=(9t^2+4t-3)/{2-(t+1)e^t}#.