If x=3 cos t and y=4 sin t,then frac{d^2 y}{d | vedclass

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(D) Given $x=3 \cos t$ and $y=4 \sin t$. Squaring and adding,we get $\frac{x^2}{9} + \frac{y^2}{16} = \cos^2 t + \sin^2 t = 1$. Differentiating with r

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$-\frac{8 \sqrt{2}}{9}$

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(D) Given $x=3 \cos t$ and $y=4 \sin t$. Squaring and adding,we get $\frac{x^2}{9} + \frac{y^2}{16} = \cos^2 t + \sin^2 t = 1$. Differentiating with respect to $x$,we get $\frac{2x}{9} + \frac{2y}{16} \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{16x}{9y}$. Now,differentiating again with respect to $x$,we get $\frac{d^2y}{dx^2} = -\frac{16}{9} \left[ \frac{y - x \frac{dy}{dx}}{y^2} \right]$. Substituting $\frac{dy}{dx} = -\frac{16x}{9y}$,we have $\frac{d^2y}{dx^2} = -\frac{16}{9} \left[ \frac{y - x(-\frac{16x}{9y})}{y^2} \right] = -\frac{16}{9} \left[ \frac{9y^2 + 16x^2}{9y^3} \right]$. Since $\frac{x^2}{9} + \frac{y^2}{16} = 1$,we have $16x^2 + 9y^2 = 144$. Thus,$\frac{d^2y}{dx^2} = -\frac{16}{9} \left[ \frac{144}{9y^3} \right] = -\frac{256}{9y^3}$. At $(x_0, y_0) = (\frac{3\sqrt{2}}{2}, 2\sqrt{2})$,we have $y = 2\sqrt{2}$. So,$\frac{d^2y}{dx^2} = -\frac{256}{9(2\sqrt{2})^3} = -\frac{256}{9(16\sqrt{2})} = -\frac{16}{9\sqrt{2}} = -\frac{8\sqrt{2}}{9}$.

If $x=3 \cos t$ and $y=4 \sin t$,then $\frac{d^2 y}{d x^2}$ at the point $(x_0, y_0)=(\frac{3}{2} \sqrt{2}, 2 \sqrt{2})$ is

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