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

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(B) Given $x=3 	an t$ and $y=3 \sec t$. First,find the derivatives with respect to $t$: $\frac{dx}{dt} = 3 \sec^2 t$ $\frac{dy}{dt} = 3 \sec t 	an t

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

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(B) Given $x=3 	an t$ and $y=3 \sec t$. First,find the derivatives with respect to $t$: $\frac{dx}{dt} = 3 \sec^2 t$ $\frac{dy}{dt} = 3 \sec t 	an t$ Now,find the first derivative $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3 \sec t 	an t}{3 \sec^2 t} = \frac{	an t}{\sec t} = \sin t$ Now,find the second derivative $\frac{d^2y}{dx^2}$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(\sin t) = \frac{d}{dt}(\sin t) \cdot \frac{dt}{dx} = \cos t \cdot \frac{1}{3 \sec^2 t} = \frac{\cos t}{3 \sec^2 t} = \frac{\cos^3 t}{3}$ Finally,evaluate at $t = \frac{\pi}{4}$: $\left(\frac{d^2y}{dx^2}ight)_{t=\frac{\pi}{4}} = \frac{(\cos(\pi/4))^3}{3} = \frac{(1/\sqrt{2})^3}{3} = \frac{1/(2\sqrt{2})}{3} = \frac{1}{6\sqrt{2}}$

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

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