Question

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

• A. $\frac{3}{2 \sqrt{2}}$
• B. $\frac{1}{3 \sqrt{2}}$
• C. $\frac{1}{6}$
• D. $\frac{1}{6 \sqrt{2} }$

Answer 1

The Correct Option is D

To find the value of \(\frac{d^2 y}{dx^2}\) at \(t = \frac{\pi}{4}\), given \(x = 3 \tan t\) and \(y = 3 \sec t\), we will first determine expressions for \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), and then use them to find \(\frac{d^2 y}{dx^2}\).

1. First, differentiate \(x\) with respect to \(t\):
   \[ x = 3 \tan t \quad \Rightarrow \quad \frac{dx}{dt} = 3 \sec^2 t \]
2. Next, differentiate \(y\) with respect to \(t\):
   \[ y = 3 \sec t \quad \Rightarrow \quad \frac{dy}{dt} = 3 \sec t \tan t \]
3. To find \(\frac{dy}{dx}\), use the chain rule:
   \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3 \sec t \tan t}{3 \sec^2 t} = \frac{\tan t}{\sec t} = \sin t \]
4. Differentiate \(\frac{dy}{dx} = \sin t\) with respect to \(t\) again to find \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\):
   \[ \frac{d}{dt}(\sin t) = \cos t \]
5. Now, find \(\frac{d^2 y}{dx^2}\) using the expression:
   \[ \frac{d^2 y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\cos t}{3 \sec^2 t} = \frac{\cos t}{3 \cdot \frac{1}{\cos^2 t}} = \frac{\cos^3 t}{3} \]
6. Substitute \(t = \frac{\pi}{4}\) into the expression:
   \[ \cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad \cos^3 \left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^3 = \frac{1}{2\sqrt{2}} \]
   Therefore,
   \[ \frac{d^2 y}{dx^2} = \frac{\frac{1}{2\sqrt{2}}}{3} = \frac{1}{6\sqrt{2}} \]

Thus, the value of \(\frac{d^2 y}{dx^2}\) at \(t = \frac{\pi}{4}\) is \(\frac{1}{6\sqrt{2}}\).