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

Question

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

Christopher Agresta · Accepted Answer

# dy/dx = (-t^3sint + 3t^2cost - tsin2t -sin^2t) /(1 + lnt) #

We have two parametric equations: # x = tlnt # # y = t^3cost - tsin^2t # We differentiate wrt #t# (and apply the product rule): # dx/dt = (t)(1/t) + (1)(lnt) # # \ \ \ \ \ \ = 1 + lnt # And: # dy/dt = t^3cost - tsin^2t # # \ \ \ \ \ \ = (t^3)(-sint) + (3t^2)(cost) - {(t)(2sintcost) + (1)(sin^2t)} # # \ \ \ \ \ \ = -t^3sint + 3t^2cost - tsin2t -sin^2t # And from the chain rule we have: # dy/dx = (dy/dt)/(dy/dx) # # \ \ \ \ \ \ = (-t^3sint + 3t^2cost - tsin2t -sin^2t) /(1 + lnt) #

Samantha Adkison · Answer

undefined To differentiate the parametric equations $ x(t) = t \ln(t) $ and $ y(t) = t^3 \cos(t) - t \sin^2(t) $ with respect to $ t $, you can use the chain rule and product rule where necessary:

$ \frac{dx}{dt} = \ln(t) + 1 $

$ \frac{dy}{dt} = 3t^2 \cos(t) - 3t^2 \sin(t) \cos(t) - \sin^2(t) - 2t \sin(t) \cos(t) $

This simplifies to:

$ \frac{dy}{dt} = 3t^2 \cos(t) - 3t^2 \sin(t) \cos(t) - \sin^2(t) - 2t \sin(t) \cos(t) $