Question

find dy/dx in the following question when
(ii) x = sin t, y = cos 2t
(NCERT)
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Answer 1

x=sint

differentiating with respect to t

dx/dt=cost

y=cos2t

again differentiating with respect to t

dy/dt=-2sin2t

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

=-2*2sint*cost/cost

dy/dx =-4sint

Answer 2

Heya!!

Given perimetric Equations are

x = Sin t , y = Cos 2t

x = Sin t

Differentiate both sides with respect to t we have.

dx/dt = d(Sin t)/dt

dx/dt = Cos t ... ( i )

y = Cos 2t

Differentiate both sides with respect to t we have.

dy/dt = d(Cos2t)/dt

dy/dt = -Sin 2t × d(2t)/dt

By chain rule of Differentiation.

dy/dt = - Sin 2t × 2

dy/dt = -2 Sin 2t ...( ii )

Now, Divide both the Equations we have.

dy/dx = -( 2 Sin 2t )/ ( Cos t )

dy/dx = - 4 { Sin t × Cos t } / Cos t

Becoz Sin 2ß = 2 Sin ß × Cos ß

dy/dx = - 4 Sin t

Note:-

Differentiation of Sin x = Cos x

And

Differentiation of Cos x = - Sin x