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 find the value of \(\frac{d^2y}{dx^2}\) at \(t = \frac{\pi}{4}\) given \(x = 3 \tan t\) and \(y = 3 \sec t\), we will follow these steps:

### Step 1: Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)

Given:
- \(y = 3 \sec t\)

To find \(\frac{dy}{dt}\):
\[ \frac{dy}{dt} = 3 \sec t \tan t \]

Now, for \(x\):
- \(x = 3 \tan t\)

To find \(\frac{dx}{dt}\):
\[ \frac{dx}{dt} = 3 \sec^2 t \]

### Step 2: Find \(\frac{dy}{dx}\)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \]

We know:
\[ \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{3 \sec^2 t} \]

Thus, substituting:
\[ \frac{dy}{dx} = \left(3 \sec t \tan t\right) \cdot \left(\frac{1}{3 \sec^2 t}\right) = \frac{\tan t}{\sec t} = \sin t \]

### Step 3: Find \(\frac{d^2y}{dx^2}\)

To find \(\frac{d^2y}{dx^2}\), we differentiate \(\frac{dy}{dx}\) with respect to \(x\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\sin t\right) = \frac{d}{dt}\left(\sin t\right) \cdot \frac{dt}{dx} \]

Calculating \(\frac{d}{dt}\left(\sin t\right)\):
\[ \frac{d}{dt}\left(\sin t\right) = \cos t \]

Now substituting \(\frac{dt}{dx}\):
\[ \frac{d^2y}{dx^2} = \cos t \cdot \frac{1}{3 \sec^2 t} = \frac{\cos t}{3 \sec^2 t} = \frac{\cos^3 t}{3} \]

### Step 4: Evaluate at \(t = \frac{\pi}{4}\)

Now, substituting \(t = \frac{\pi}{4}\):
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus:
\[ \frac{d^2y}{dx^2} = \frac{\left(\frac{1}{\sqrt{2}}\right)^3}{3} = \frac{1}{3 \cdot 2\sqrt{2}} = \frac{1}{6\sqrt{2}} \]

### Final Answer:
The value of \(\frac{d^2y}{dx^2}\) at \(t = \frac{\pi}{4}\) is \(\frac{1}{6\sqrt{2}}\).

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