u=x2+y2 and x=s+3t,y=2s−t, then d2uds2 is

# is

1. A

12

2. B

32

3. C

36

4. D

10

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### Solution:

Given $u={x}^{2}+{y}^{2},x=s+3t,y=2s-t$
$\begin{array}{l}\frac{\mathrm{dx}}{\mathrm{ds}}=1,\frac{\mathrm{dy}}{\mathrm{ds}}=2;\frac{{\mathrm{d}}^{2}\mathrm{x}}{{\mathrm{ds}}^{2}}=0,\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{ds}}^{2}}=0\\ \mathrm{u}={\mathrm{x}}^{2}+{\mathrm{y}}^{2}⇒\frac{\mathrm{du}}{\mathrm{ds}}=2\mathrm{x}\frac{\mathrm{dx}}{\mathrm{ds}}+2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{ds}}\end{array}$
$\begin{array}{r}=\frac{{d}^{2}u}{d{s}^{2}}=2{\left(\frac{dx}{ds}\right)}^{2}+2x\frac{{d}^{2}x}{d{s}^{2}}+2{\left(\frac{dy}{ds}\right)}^{2}+2y\left(\frac{{d}^{2}y}{d{s}^{2}}\right)\\ ⇒\frac{{d}^{2}u}{d{s}^{2}}=2\left(1{\right)}^{2}+2x\left(0\right)+2\left(2{\right)}^{2}+2y\left(0\right)=2+8=10\end{array}$  +91

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