If u = x^2 + y^2 and x = s + 3t, y = 2s - t, | vedclass

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(D) Given: $u = x^2 + y^2$,$x = s + 3t$,$y = 2s - t$.First,find the first derivatives with respect to $s$:$\frac{dx}{ds} = 1$ and $\frac{dy}{ds} = 2$.

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Given: $u = x^2 + y^2$,$x = s + 3t$,$y = 2s - t$.First,find the first derivatives with respect to $s$:$\frac{dx}{ds} = 1$ and $\frac{dy}{ds} = 2$.Next,find the second derivatives with respect to $s$:$\frac{d^2x}{ds^2} = 0$ and $\frac{d^2y}{ds^2} = 0$.Now,differentiate $u$ with respect to $s$ using the chain rule:$\frac{du}{ds} = 2x \frac{dx}{ds} + 2y \frac{dy}{ds}$.Differentiate again with respect to $s$:$\frac{d^2u}{ds^2} = 2 \left( \frac{dx}{ds} \right)^2 + 2x \frac{d^2x}{ds^2} + 2 \left( \frac{dy}{ds} \right)^2 + 2y \frac{d^2y}{ds^2}$.Substitute the values:$\frac{d^2u}{ds^2} = 2(1)^2 + 2x(0) + 2(2)^2 + 2y(0) = 2(1) + 8 = 10$.

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

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