What is the derivative of #f(t) = (t-sint , cost/t^2 ) #?

Question

1556 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-18T07:10:21

#f'(t)=(1-cost,-(tsint+2cost)/t^3)#.

Let #f(t)=(t-sint, cost/t^2)=(x(t),y(t)), say#. Then,

#f'(t)=(x'(t),y'(t))#

Now, #x(t)=t-sint rArr x'(t)=1-cost#.

#y(t)=cost/t^2=t^-2*cost#

#rArr y'(t)=t^-2*(-sint)+(-2t^-3*cost)#

#=-sint/t^2-(2cost)/t^3=-(tsint+2cost)/t^3#.

Hence, #f'(t)=(1-cost,-(tsint+2cost)/t^3)#.