Question

If x = cos t (3 – 2 cos² t) and y = sin t (3 – 2 sin² t), find the value of dx/dy at t =4/π.

Answer 1

x = cos t (3 – 2 cos² t) and y = sin t (3 – 2 sin² t)

We need to find dy/dx :

`dy/dx=(dy/dt)/(dx/dt)`

Let us find dx/dt:

x = cos t (3 – 2 cos² t)

`dx/dt=cost(4costsint)+(3-2cos^2t)(-sint)`

`=>dx/dt=-3sint+4cos^2tsint+2cos^2tsint`

Let us find dy/dx:

y = sin t (3 – 2 sin² t)

`dy/dt=sint(-4sintcost)+(3-2sin^2t)(cost)`

`=>dy/dt=3cost-4sin^2tcost-2sin^2tcost`

thus,

`dy/dx=(3cost-4sin^2tcost-2sin^2tcost)/(-3sint+4cos^2tsint+2cos^2tsint)`

`=>dy/dx=(3cost-6sin^2tcost)/(-3sint+6cos^2tsint)`

`=>dy/dx=(3cost(1-2sin^2t))/(-3sint(1-2cos^2t))`

`=>dy/dx=(3cost(1-2sin^2t))/(3sint(2cos^2t-1))`

`=>dy/dx=cost/sint [because 2cos^2t-1=1-2sin^2t]`

`=>dy/dx=cott`

`=>(dy/dx)_(t=pi/4)=cot(pi/4)=1`