x=cos3θ,y=sin3θthen1+(dydx)2

dydx=dydθdxdθ=ddθ(asin3θ)ddθ(acos3θ)=3sin2θcosθ−3cos2θsinθ=−tanθ(1+(dydx)2)=1+(−tanθ)2=1+tan2θ=|secθ|

x=cos3θ,y=sin3θthen1+(dydx)2

dydx=dydθdxdθ=ddθ(asin3θ)ddθ(acos3θ)=3sin2θcosθ−3cos2θsinθ=−tanθ(1+(dydx)2)=1+(−tanθ)2=1+tan2θ=|secθ|

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$x = \cos^{3} θ, y = \sin^{3} θ t h e n \sqrt{1 + {(\frac{d y}{d x})}^{2}}$

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$\tan^{2} θ$

$S e c^{2} θ$

$\sec θ$

$| \sec θ |$

answer is D.

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$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{\frac{d}{d θ} (a \sin^{3} θ)}{\frac{d}{d θ} (a \cos^{3} θ)} = \frac{3 \sin^{2} θ \cos θ}{- 3 \cos^{2} θ \sin θ} = - \tan θ$

$\sqrt{(1 + {(\frac{d y}{d x})}^{2})} = \sqrt{1 + {(- \tan θ)}^{2}} = \sqrt{1 + \tan^{2} θ} = | \sec θ |$

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$x = \cos^{3} θ, y = \sin^{3} θ t h e n \sqrt{1 + {(\frac{d y}{d x})}^{2}}$