If x=32cos3θ and y=4tan2θ then (dydx)θ=π/4=

x=32cos3θdxdθ=-323cos2θsinθy=4tan2θdydθ=42tanθsec2θdydx=8tanθsec2θ-92cos2θsinθdydx=812-921212dydx=-329

If x=32cos3θ and y=4tan2θ then (dydx)θ=π/4=

x=32cos3θdxdθ=-323cos2θsinθy=4tan2θdydθ=42tanθsec2θdydx=8tanθsec2θ-92cos2θsinθdydx=812-921212dydx=-329

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If $x = 3 \sqrt{2} \cos^{3} θ$ and $y = 4 \tan^{2} θ$ then ${(\frac{dy}{dx})}_{θ = π / 4} =$

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$\frac{32 \sqrt{2}}{9}$

$- \frac{16}{9}$

$\frac{16}{9}$

$- \frac{32}{9}$

answer is D.

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Detailed Solution

$x = 3 \sqrt{2} \cos^{3} θ$

$\frac{dx}{dθ} = - 3 \sqrt{2} (3 \cos^{2} θ \sin θ)$

$y = 4 \tan^{2} θ$

$\frac{dy}{dθ} = 4 (2 {tanθsec}^{2} θ)$

$\frac{d y}{d x} = \frac{8 \tan θ s e c^{2} θ}{- 9 \sqrt{2} \cos^{2} θ \sin θ}$

$\frac{d y}{d x} = \frac{8 (1) (2)}{- 9 \sqrt{2} (\frac{1}{2}) (\frac{1}{\sqrt{2}})}$

$\frac{d y}{d x} = - \frac{32}{9}$

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If $x = 3 \sqrt{2} \cos^{3} θ$ and $y = 4 \tan^{2} θ$ then ${(\frac{dy}{dx})}_{θ = π / 4} =$