If x^3+y^3=3axy,then at left(frac{3a}{2}, fra | vedclass

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(C) Given the equation $x^3+y^3=3axy$. Differentiating with respect to $x$: $3x^2+3y^2 \frac{dy}{dx} = 3ay + 3ax \frac{dy}{dx}$ $(y^2-ax) \frac{dy}{dx

Vedclass · Accepted Answer

$8$

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(C) Given the equation $x^3+y^3=3axy$. Differentiating with respect to $x$: $3x^2+3y^2 \frac{dy}{dx} = 3ay + 3ax \frac{dy}{dx}$ $(y^2-ax) \frac{dy}{dx} = ay-x^2$ $\frac{dy}{dx} = \frac{ay-x^2}{y^2-ax}$ At point $\left(\frac{3a}{2}, \frac{3a}{2}\right)$,$\frac{dy}{dx} = \frac{a(\frac{3a}{2}) - (\frac{3a}{2})^2}{(\frac{3a}{2})^2 - a(\frac{3a}{2})} = \frac{\frac{3a^2}{2} - \frac{9a^2}{4}}{\frac{9a^2}{4} - \frac{3a^2}{2}} = \frac{-\frac{3a^2}{4}}{\frac{3a^2}{4}} = -1$. Differentiating $\frac{dy}{dx}(y^2-ax) = ay-x^2$ again with respect to $x$: $(2y \frac{dy}{dx} - a) \frac{dy}{dx} + (y^2-ax) \frac{d^2y}{dx^2} = a \frac{dy}{dx} - 2x$ At $\left(\frac{3a}{2}, \frac{3a}{2}\right)$ with $\frac{dy}{dx} = -1$: $(2(\frac{3a}{2})(-1) - a)(-1) + ((\frac{3a}{2})^2 - a(\frac{3a}{2})) y^{\prime \prime} = a(-1) - 2(\frac{3a}{2})$ $(-3a-a)(-1) + (\frac{9a^2}{4} - \frac{6a^2}{4}) y^{\prime \prime} = -a - 3a$ $4a + \frac{3a^2}{4} y^{\prime \prime} = -4a$ $\frac{3a^2}{4} y^{\prime \prime} = -8a$ $3ay^{\prime \prime} = -32$ Therefore,$3ay^{\prime \prime} + 40 = -32 + 40 = 8$.

If $x^3+y^3=3axy$,then at $\left(\frac{3a}{2}, \frac{3a}{2}\right)$ the value of $3ay^{\prime \prime}+40$ is

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