Question

If $x=a(1-{\cos }^3\theta ),y=a{\sin }^3\theta$ and ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{\mathrm{\pi }/6}=\frac{A}{a}$ then $A=?$

• A. $\frac{27}{32}$
• B. $\frac{32}{27a}$
• C. $\frac{32}{27}$
• D. $\frac{27}{32a}$

Answer 1

The correct option is C

$\frac{32}{27}$

Explanation for the correct option.

Step 1: Differentiate $x\mathrm{and}y$ w.r.t $\theta$

$\begin{array}{rcl}x& =& a(1-{\cos }^3\theta )\\ & \Rightarrow & \frac{dx}{d\theta }=3a\left({\cos }^2\theta ·\sin \theta \right)\\ y& =& a{\sin }^3\theta \\ & \Rightarrow & \frac{dy}{d\theta }=3a\left({\sin }^2\theta ·\cos \theta \right)\end{array}$

Step 2: Find $\frac{dy}{dx}\mathrm{and}\frac{d^{2}y}{d{x}^2}$

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{3a\left({\sin }^2\theta ·\cos \theta \right)}{3a\left({\cos }^2\theta ·\sin \theta \right)}\\ & =& \tan \theta \end{array}$

Now,

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{d}{d\theta }\left(\frac{dy}{dx}\right)\times \frac{d\theta }{dx}=\frac{se{c}^2\theta }{3a\left({\cos }^2\theta ·\sin \theta \right)}\end{array}$

Step 3: Find $\frac{d^{2}y}{d{x}^2}$ At $\theta =\frac{\mathrm{\pi }}{6}$

$\begin{array}{rcl}{\left(\frac{d^{2}y}{d{x}^2}\right)}_{\theta =\mathrm{\pi }/6}& =& \frac{se{c}^2{\displaystyle \frac{\mathrm{\pi }}{6}}}{3a\left({\cos }^2{\displaystyle \frac{\mathrm{\pi }}{6}}·\sin {\displaystyle \frac{\mathrm{\pi }}{6}}\right)}\\ & =& \frac{{\left({\displaystyle \frac{2}{\sqrt{3}}}\right)}^2}{3a\times {\displaystyle {\left(\frac{\sqrt{3}}{2}\right)}^2}\times {\displaystyle \frac{1}{2}}}\\ & =& \frac{32}{27a}\end{array}$

Step 4: Find the value of $A$.

$\begin{array}{rcl}{\left(\frac{d^{2}y}{d{x}^2}\right)}_{\mathrm{\pi }/6}& =& \frac{A}{a}\\ \Rightarrow \frac{32}{27a}& =& \frac{A}{a}\\ \Rightarrow A& =& \frac{32}{27}\end{array}$

Hence, option C is correct.