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. - Mathematics and Statistics

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.

Solution

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

Concept: Derivatives of Functions in Parametric Forms
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