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Question

If x=asin2t(1+cos2t) and y=bcos2t(1cos2t), then find dydx at t=π4.

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Solution

x=asin2t(1+cos2t)

Differentiating w.r.t t, we get

dxdt=2acos2t(1+cos2t)+asin2t(2sin2t)

=2acos2t(1+cos2t)2asin22t

[dxdt]t=π4=2acos(2×π4){1+cos(2×π4)}2asin2(2×π4)

=02a=2a

y=bcos2t(1cos2t)

Differentiating w.r.t t, we get,

dydt=2bsin2t(1cos2t)+bcos2t(2sin2t)

=2bsin2t(1cos2t)+2bcos2tsin2t

[dydt]t=π4=2bsinπ2(1cosπ2)+2bcosπ2sinπ2

=2b

Now, dydx=dy/dtdx/dt=2b2a=ba

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