Let [x(t)=2 √2 t 2 t] and [y(t)=2 √2 t 2 t, t (0, ()/(2))].

Question

Let $$x(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t}$$ and

$$y(t)=2 \sqrt{2} \sin t \sqrt{\sin 2 t}, t \in\left(0, \frac{\pi}{2}\right)$$.

Then $$\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}$$ at $$t=\frac{\pi}{4}$$ is equal to :

Accepted Answer

$$x = 2\sqrt 2 \cos t\sqrt {\sin 2t} ,\,y = 2\sqrt 2 \sin t\sqrt {\sin 2t} $$

$$\therefore$$ $${{dx} \over {dt}} = {{2\sqrt 2 \cos 3t} \over {\sqrt {\sin 2t} }},\,{{dy} \over {dt}} = {{2\sqrt 2 \sin 3t} \over {\sqrt {\sin 2t} }}$$

$$\therefore$$ $${{dy} \over {dx}} = \tan 3t,\,\left( {\mathrm{at}\,t = {\pi  \over 4},\,{{dy} \over {dx}} =  - 1} \right)$$

and $${{{d^2}y} \over {d{x^2}}} = 3{\sec ^2}3t\,.\,{{dt} \over {dx}} = {{3{{\sec }^2}3t\,.\,\sqrt {\sin 2t} } \over {2\sqrt 2 \cos 3t}}$$

$$\left( {\mathrm{At}\,t = {\pi  \over 4},\,{{{d^2}y} \over {d{x^2}}} =  - 3} \right)$$

$$\therefore$$ $${{1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \over {{{{d^2}y} \over {d{x^2}}}}} = {2 \over { - 3}} = {{ - 2} \over 3}$$

Solutions · 1