Question

If x = 3 tan t and y = 3 sec t, then find \(\frac{d^2y}{dx^2} \, at\,t=\frac{\pi}{4}\), is:

Answer 1

We have x=3tan t and y=3sec t

∴\(\frac{dx}{dt}=3sec^2t\)
 
and \(\frac{dy}{dt}=3\,sec\,t\,tan\,t\)

since, \(\frac{dy}{dt}=\frac{tan\,t}{sec\,t}=sin\,t\)

Again differentiating w.r.t.  x

\(\frac{d^2y}{dx^2}=cost\,\frac{dt}{dx}\)

\(\frac{d^2y}{dx^2}=\frac{cos\,t}{3sec^2\,t}=\frac{cos^3t}{3}\)

since,  at \(x=\frac{\pi}{4}\)

\(=\frac{1}{3\times2\sqrt2}=\frac{1}{6\sqrt2}\)

Answer 2

Given:
\(x = 3 \tan t\)
\(y = 3 \sec t\)

Find \(\frac{dx}{dt}\):
\(\frac{dx}{dt} = 3 \sec^2 t\)

Find \(\frac{dy}{dt}\):
\(\frac{dy}{dt} = 3 \sec t \tan t\)

Calculate \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3 \sec t \tan t}{3 \sec^2 t} = \frac{\tan t}{\sec t} = \sin t\)

Now, differentiate\(\frac{dy}{dx}\) with respect to x:
\(\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(\sin t)\)

To find \(\frac{d}{dx}(\sin t)\), we need \(\frac{dt}{dx}\). From \(\frac{dx}{dt} = 3 \sec^2 t\), we get:
\(\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{3 \sec^2 t} = \frac{\cos^2 t}{3}\)

Therefore,
\(\frac{d}{dx}(\sin t) = \cos t \cdot \frac{dt}{dx} = \cos t \cdot \frac{\cos^2 t}{3} = \frac{\cos^3 t}{3}\)

\(\frac{d^2y}{dx^2} = \frac{\cos^3 t}{3}\)

Evaluate\(\frac{d^2y}{dx^2}\) at \(t = \frac{\pi}{4}\):
\(\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}\)

Therefore,
\(\frac{d^2y}{dx^2} \bigg|_{t=\frac{\pi}{4}} = \frac{\left(\frac{1}{\sqrt{2}}\right)^3}{3} = \frac{\frac{1}{2\sqrt{2}}}{3} = \frac{1}{6\sqrt{2}}\)

So, the answer is: \(\frac{1}{6\sqrt{2}}\)