Question

What is the derivative of `f(t) = (tcos^2t , t^2-t sect )`?

Answer 1

`(dy)/(dx)=(2t-sect-tsect tant)/(cos^2t-tsin2t)`

Explanation:

When a functiion is given in parametric form such as `f(t)=(x(t),y(t))`, its dervative is given by `(dy)/(dx)=((dy)/(dt))/((dx)/(dt))`

Here we have `y(t)=t^2-tsect` hence `(dy)/(dt)=2t-sect-tsect tant`

and `x(t)=tcos^2t` hence `(dx)/(dt)=cos^2t+txx2costxx(-sint)`

= `cos^2t-tsin2t`

Hence `(dy)/(dx)=(2t-sect-tsect tant)/(cos^2t-tsin2t)`