If x = 3 tan t and y = 3 sec t, then the valu | vedclass

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(D) Given $x = 3 	an t$ and $y = 3 \sec t.$ Differentiating with respect to $t$: $\frac{dx}{dt} = 3 \sec^2 t$ $\frac{dy}{dt} = 3 \sec t 	an t$ Using

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$\frac{1}{6\sqrt{2}}$

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(D) Given $x = 3 	an t$ and $y = 3 \sec t.$ Differentiating with respect to $t$: $\frac{dx}{dt} = 3 \sec^2 t$ $\frac{dy}{dt} = 3 \sec t 	an t$ Using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3 \sec t 	an t}{3 \sec^2 t} = \frac{	an t}{\sec t} = \sin t$ Now,differentiating $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(\sin t) = \cos t \cdot \frac{dt}{dx}$ Since $\frac{dx}{dt} = 3 \sec^2 t,$ then $\frac{dt}{dx} = \frac{1}{3 \sec^2 t}.$ Substituting this: $\frac{d^2y}{dx^2} = \cos t \cdot \frac{1}{3 \sec^2 t} = \frac{\cos^3 t}{3}$ At $t = \frac{\pi}{4},$ $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}.$ Therefore,$\frac{d^2y}{dx^2} = \frac{(1/\sqrt{2})^3}{3} = \frac{1}{3 \cdot 2\sqrt{2}} = \frac{1}{6\sqrt{2}}.$

If $x = 3 \tan t$ and $y = 3 \sec t,$ then the value of $\frac{d^2y}{dx^2}$ at $t = \frac{\pi}{4}$ is

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