If x = 3 tant and y = 3 sec t, then the value of `(d^2y)/(dx^2)" at" t=pi/4` is

Options

A. `3/(2sqrt2)`
B. `1/(3sqrt2)`
C. `1/6`
D. `1/(6sqrt2)`

Answer

To solve the problem, we need to find the second derivative of \( y \) with respect to \( x \) given the equations \( x = 3 \tan t \) and \( y = 3 \sec t \). We will evaluate this at \( t = \frac{\pi}{4} \).

### Step-by-Step Solution:

1. **Differentiate \( x \) with respect to \( t \)**:
\[ x = 3 \tan t \implies \frac{dx}{dt} = 3 \sec^2 t \]

2. **Differentiate \( y \) with respect to \( t \)**:
\[ y = 3 \sec t \implies \frac{dy}{dt} = 3 \sec t \tan t \]

3. **Find \( \frac{dy}{dx} \)** using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3 \sec t \tan t}{3 \sec^2 t} = \frac{\tan t}{\sec t} = \sin t \]

4. **Differentiate \( \frac{dy}{dx} \) with respect to \( t \)** to find \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dt} \left( \frac{dy}{dx} \right) \cdot \frac{1}{\frac{dx}{dt}} = \frac{d}{dt}(\sin t) \cdot \frac{1}{\frac{dx}{dt}} \]
\[ \frac{d}{dt}(\sin t) = \cos t \]
\[ \frac{d^2y}{dx^2} = \cos t \cdot \frac{1}{3 \sec^2 t} = \frac{\cos t}{3 \sec^2 t} = \frac{\cos^3 t}{3} \]

5. **Evaluate \( \frac{d^2y}{dx^2} \) at \( t = \frac{\pi}{4} \)**:
\[ \frac{d^2y}{dx^2} \bigg|_{t = \frac{\pi}{4}} = \frac{\cos^3\left(\frac{\pi}{4}\right)}{3} \]
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \implies \cos^3\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^3 = \frac{1}{2\sqrt{2}} \]
\[ \frac{d^2y}{dx^2} \bigg|_{t = \frac{\pi}{4}} = \frac{1}{3} \cdot \frac{1}{2\sqrt{2}} = \frac{1}{6\sqrt{2}} \]

### Final Answer:
\[ \frac{d^2y}{dx^2} \bigg|_{t = \frac{\pi}{4}} = \frac{1}{6\sqrt{2}} \]