How do you differentiate the following parametric equation: x(t)=tlnt, y(t)= cost-tsin^2t x(t)=tlnt,y(t)=costtsin2t?

1 Answer
Jul 13, 2016

(df(t))/dt = (ln(t) + 1 , -sin(t) - sin^2(t) - 2tsin(t)cos(t))df(t)dt=(ln(t)+1,sin(t)sin2(t)2tsin(t)cos(t))

Explanation:

Differentiating a parametric equation is as easy as differentiating each individual equation for its components.

If f(t) = (x(t), y(t))f(t)=(x(t),y(t)) then (df(t))/dt = ((dx(t))/dt , (dy(t))/dt)df(t)dt=(dx(t)dt,dy(t)dt)

So we first determine our component derivatives:
(dx(t))/dt = ln(t) + t/t = ln(t) + 1dx(t)dt=ln(t)+tt=ln(t)+1
(dy(t))/dt = -sin(t) - sin^2(t) - 2tsin(t)cos(t)dy(t)dt=sin(t)sin2(t)2tsin(t)cos(t)

Therefore the final parametric curve's derivatives is simply a vector of the derivatives:
(df(t))/dt = ((dx(t))/dt , (dy(t))/dt)df(t)dt=(dx(t)dt,dy(t)dt)
= (ln(t) + 1 , -sin(t) - sin^2(t) - 2tsin(t)cos(t))=(ln(t)+1,sin(t)sin2(t)2tsin(t)cos(t))