Question

How do you differentiate the following parametric equation: `x(t)=lnt-2t, y(t)= cos^2t`?

Answer 1

`dy/dx=(t*sin 2t)/(2t-1)`

Explanation:

Start from the given `x(t)=ln t -2t` and `y(t)=cos^2 t`

to find `dy/dx` the equivalent of `dy/dt` and `dx/dt` must be obtained then proceed to determine `dy/dx=((dy/dt))/((dx/dt))`

`dy/dt=d/dt(cos^2 t)=2*cos t*(-sin t)*dt/dt=-2*sin t*cos t`

`dy/dt=-2*sint * cos t`

Next, find `dx/dt`

`dx/dt=d/dt(ln t - 2t)=1/t-2=(1-2t)/t`

`dx/dt=(1-2t)/t`

Finally, divide `dy/dt` by `dx/dt`

`dy/dx=(-2*sint *cos t)/((1-2t)/t)`

`dy/dx=(-2t*sin t* cos t)/(1-2t)`

from Plane Trigonometry, `2*sin t* cos t = sin 2t`

`dy/dx=(-t*sin 2t)/(1-2t)`

placing the negative sign (-) at the bottom so that the final answer becomes:

`dy/dx=(t*sin 2t)/(2t-1)`

I hope the explanation helps ...