#### Question

If *x* = a sin 2t (1 + cos 2t) and *y* = b cos 2t (1 – cos 2t), find the values of `dy/dx `

#### Solution

*x* = a sin 2t (1 + cos 2t)*y* = b cos 2t (1 – cos 2t)

`dy/dx=(dy/dx)/(dx/dt)=(b[-2sin2t+2cos 2t sin2t xx2 ))/(a[2cos2t+2cos4t])`

`dx/dy=b/a[(-2sin2+2sin 4t)/(2cos2t+2cos4t)]`

`dy/dx|_(t=pi/4)=b/a[(-2+0)/(0-2)]=b/a`

and `dy/dx|_(t=pi/3)=b/a[(-2sqrt3)/-2]=(sqrt3b)/a`

Is there an error in this question or solution?

Solution If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), find the values of dy/dx Concept: Derivatives of Functions in Parametric Forms.