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  • Find dy/dx of each of the functions expressed in parametric form x = 3 cos theta - 2 cos cube theta

Find dy/dx of each of the functions expressed in parametric form in

x=3 \cos \theta-2 \cos ^{3} \theta, y=3 \sin \theta-2 \sin ^{3} \theta

Answers (1)

We have given,  x=3 \cos \theta-2 \cos ^{3} \theta, y=3 \sin \theta-2 \sin ^{3} \theta

Now, differentiate both the equation w.r.t. x then we get

 x=3 \cos \theta-2 \cos ^{3} \theta

$$ \frac{d x}{d \theta}=\left(-3 \sin \theta-2 \times 3 \cos ^{2} \theta \cdot \frac{d}{d \theta}(\cos \theta)\right)

\\\frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\left(-3 \sin \theta+6 \cos ^{2} \theta \cdot(\sin \theta)\right)$ \\And, for $y=3 \sin \theta-2 \sin ^{3} \theta

\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{d} \theta}=\left(3 \cos \theta-2 \times 3 \sin ^{2} \theta \cdot \frac{\mathrm{d}}{\mathrm{d} \theta}(\sin \theta)\right)\\ &\frac{\mathrm{dy}}{\mathrm{d} \theta}=\left(3 \cos \theta-6 \sin ^{2} \theta \cdot \cos \theta\right)\\ &\text { Therefore, }\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d} \theta} \times \frac{\mathrm{d} \theta}{\mathrm{d} \mathrm{x}}\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\left(3 \cos \theta-6 \sin ^{2} \theta \cdot \cos \theta\right) \times \frac{1}{\left(-3 \sin \theta+6 \cos ^{2} \theta \cdot(\sin \theta)\right.} \end{aligned}

\\ \frac{d y}{d x}=\frac{3 \cos \theta\left(1-2 \sin ^{2} \theta\right)}{3 \sin \theta\left(-1+2 \cos ^{2} \theta\right)} \\ \frac{d y}{d x}=\frac{\cos 2 \theta}{\cos 2 \theta} \times \cot \theta \\ \frac{d y}{d x}=\cot \theta \\ \text { Hence, } d y / d x=\cot \theta

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