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Question

If x=acos4θ,y=asin4θ, then dydx at θ=π4 is

A
1
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B
1
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C
a2
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a2
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Solution

The correct option is A 1
Given, x=acos4θ and y=asin4θ
On differentiating with respect to θ respectively, we get
dxdθ=4acos3θ(sinθ)
=4asinθcos3θ
and dydθ=4asin3θcosθ
dydx=dydθdxdθ=4asin3θcosθ4asinθcos3θ
dydx=sin2θcos2θ=tan2θ
Now, (dydx)θ=3π4=tan2(3π4)=1

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