Question

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

• A. $\frac{a}{2\sqrt{2}}$
• B. $-\frac{a}{2\sqrt{2}}$
• C. $\frac{2\sqrt{2}}{a}$
• D. $-\frac{2\sqrt{2}}{a}$

Answer 1

The correct option is D $-\frac{2\sqrt{2}}{a}$

Clearly $x^2+y^2=a^{2}andy\left(\frac{\pi }{4}\right)=,x\left(\frac{\pi }{4}\right)=\frac{a}{\sqrt{2}}.$
Differentiating we get,$2x+2y{y}_1=0$
$\Rightarrow y_1=-\frac{x}{y}$, so $y_1\left(\frac{\pi }{4}\right)=-1.$
Now $x+y{y}_1=0$
$\Rightarrow 1+y_1^2+y{y}_2=0$
$\Rightarrow y_2\left(\frac{\pi }{4}\right)=-\frac{1+\left(y_1\right(\frac{\pi }{4}){)}^2}{y\left(\frac{\pi }{4}\right)}=\frac{-2\sqrt{2}}{a}$