Question

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

• A. $\frac{3}{a^2}$
• B. $\frac{-12}{a^2}$
• C. $\frac{-3}{a^2}$
• D. $\frac{12}{a^2}$

Answer 1

The correct option is B $\frac{-12}{a^2}$

Clearly $x^2+y^2=a^2$ and $y\left(\pi /4\right)=a/\sqrt{2}$, $x\left(\pi /4\right)=a/\sqrt{2}$.
Differentiating we get
$2x+2y{y}_1=0\Rightarrow y_1=-\frac{x}{y}$, so $y_1\left(\pi /4\right)=-1$
Now $x+y{y}_1=0\Rightarrow 1+y_1^2+y{y}_2=0$
$\Rightarrow$ $y_2=\left(\frac{\pi }{4}\right)=-\frac{1+{\left(y_1\left(\frac{\pi }{4}\right)\right)}^2}{y\left(\frac{\pi }{4}\right)}=\frac{-2\sqrt{2}}{a}$
Differentiating again, we have
$2y_1{y}_2+y_1{y}_2+y{y}_3=0\Rightarrow y_3=-\frac{3y_1{y}_2}{y}$
$\Rightarrow$ $y_3\left(\frac{\pi }{4}\right)=-\frac{3(-1)(-2)\sqrt{2}}{a.\frac{a}{\sqrt{2}}}=-\frac{12}{a^2}$