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 the following equations:
\( x = 3 \tan(t) \) and \( y = 3 \sec(t) \)

Now, we differentiate \( x \) and \( y \) with respect to \( t \):

Step 1: Differentiate \( x \) and \( y \) with respect to \( t \):
\[ \frac{dx}{dt} = 3 \sec^2(t) \] and \[ \frac{dy}{dt} = 3 \sec(t) \tan(t) \]

Step 2: Simplify \( \frac{dy}{dt} \):
Since \[ \frac{dy}{dt} = \frac{\tan(t)}{\sec(t)} = \sin(t) \] we have \[ \frac{dy}{dt} = \sin(t) \]

Step 3: Differentiate with respect to \( x \):
We now differentiate \( y \) with respect to \( x \). To do this, we use the chain rule: \[ \frac{d^2y}{dx^2} = \cos(t) \frac{dt}{dx} \] Next, we find \( \frac{dt}{dx} \). Since \( \frac{dx}{dt} = 3 \sec^2(t) \), we have: \[ \frac{dt}{dx} = \frac{1}{3 \sec^2(t)} \] Therefore, the second derivative becomes: \[ \frac{d^2y}{dx^2} = \frac{\cos(t)}{3 \sec^2(t)} = \frac{\cos^3(t)}{3} \]

Step 4: Evaluate at \( x = \frac{\pi}{4} \):
When \( x = \frac{\pi}{4} \), we evaluate the second derivative. At \( t = \frac{\pi}{4} \), we know that: \[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \quad \text{and} \quad \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \] Substituting these values into the expression for \( \frac{d^2y}{dx^2} \), we get: \[ \frac{d^2y}{dx^2} = \frac{1}{3 \times \sec^2\left(\frac{\pi}{4}\right)} = \frac{1}{3 \times 2} = \frac{1}{6} \] So, the second derivative is: \[ \frac{d^2y}{dx^2} = \frac{1}{6\sqrt{2}} \]

Final Answer:
Therefore, at \( x = \frac{\pi}{4} \), we have: \[ \frac{d^2y}{dx^2} = \frac{1}{6\sqrt{2}} \]

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: \(\frac{1}{6\sqrt{2}}\)