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

A
12
B
32
C
36
D
10

Solution:

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

$u = x^{2} + y^{2} and x = s + 3 t, y = 2 s - t, then \frac{d^{2} u}{d s^{2}}$ is

Solution:

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