MCQ
If $u = {x^2} + {y^2}$ and $x = s + 3t,$$y = 2s - t,$ then ${{{d^2}u} \over {d{s^2}}} = $
- A$12$
- B$32$
- C$36$
- ✓$10$
✓
Answer
Correct option: D.
$10$
d
(d) $u = {x^2} + {y^2},$ $x = s + 3t,$ $y = 2s - t$
(d) $u = {x^2} + {y^2},$ $x = s + 3t,$ $y = 2s - t$
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 $(ii),$
$\frac{{{d^2}u}}{{d{s^2}}} = 2 \times 1 + 0 + 2 \times 4 + 0 = 10$.
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