5. continuity and differentiation — Maths STD 12 Science — Question

Now $\frac{{dx}}{{ds}} = 1,$ $\frac{{dy}}{{ds}} = 2$ .....$(i)$

$\frac{{{d^2}x}}{{d{s^2}}} = 0,$ $\frac{{{d^2}y}}{{d{s^2}}} = 0$ ......$(ii)$

Now $u = {x^2} + {y^2}$, $\frac{{du}}{{ds}} = 2x\,.\,\frac{{dx}}{{ds}} + 2y\,.\,\frac{{dy}}{{ds}}$

$\frac{{{d^2}u}}{{d{s^2}}} = 2{\left( {\frac{{dx}}{{ds}}} \right)^2} + 2x\frac{{{d^2}x}}{{d{s^2}}} + 2{\left( {\frac{{dy}}{{ds}}} \right)^2} + 2y\left( {\frac{{{d^2}y}}{{d{s^2}}}} \right)$

From $(i)$ and $(ii),$ 

$\frac{{{d^2}u}}{{d{s^2}}} = 2 \times 1 + 0 + 2 \times 4 + 0 = 10$.