If $x=4 \cos ^3 \theta$ and $y=3 \sin ^2 \theta$, then $\frac{d^2 y}{d x^2}$ at $\theta=\frac{\pi}{4}$, is

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Aicharya · Accepted Answer

Given, $x=4 \cos ^3 \theta$...(i) and $y=3 \sin ^2 \theta$...(ii) Differentiating Eq. (i) and Eq. (ii) w. r. t. $t \theta$ $\begin{aligned} \frac{d x}{d \theta} & =4 \times 3 \cos ^2 \theta(-\sin \theta) \ & =-12 \cos ^2 \theta \sin \theta \quad \ldots (iii) \end{aligned}$ $\begin{aligned} \text{and } \frac{d y}{d \theta} & =3 \times 2 \sin \theta \cos \theta \ & =6 \sin \theta \cos \theta \quad \ldots (iv) \end{aligned}$ $\begin{aligned} Now,\frac{d y}{d x} & =\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \ & =-\frac{1}{2} \sec \theta ~[\text {by Eqs. (iii) and (iv)] } \end{aligned}$ So, $\frac{d^2 y}{d x^2}=-\frac{1}{2} \sec \theta \tan \theta \cdot \frac{d \theta}{d x}$ $\begin{aligned} & =-\frac{1}{2} \times \sec \theta \tan \theta \times \frac{-1}{12 \cos ^2 \theta \sin \theta} \quad \text { [by Eq. (iii)] } \ & =\frac{1}{24} \times \frac{1}{\cos \theta} \times \frac{\sin \theta}{\cos \theta} \times \frac{1}{\cos ^2 \theta \sin \theta} \ & =\frac{1}{24(\cos \theta)^4}=\frac{1}{24\left(\cos \frac{\pi}{4}\right)^4} \quad\left(\because \theta=\frac{\pi}{4}\right) \ & =\frac{1}{24} \times \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^4}=\frac{1}{24} \times 4=\frac{1}{6} \end{aligned}$

If \(x=4 \cos ^3 \theta\) and \(y=3 \sin ^2 \theta\), then \(\frac{d^2 y}{d x^2}\) at…

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