Question

If \(x = a(\cos t + t\sin t)\) and \(y = a(\sin t - t\cos t)\), then the value of \(\frac{d^2y}{dx^2}\) is given by :

• A. \(\frac{at}{\cos^3 t}\)
• B. \(\frac{1}{at\cos^3 t}\)
• C. \(\frac{at}{\sin^3 t}\)
• D. \(\frac{1}{at\sin^3 t}\)

Hint:

For parametric differentiation, never forget that:
\[ \frac{d^2y}{dx^2} \ne \frac{d^2y/dt^2}{d^2x/dt^2} \]
You must always divide the derivative of the first slope by \(\frac{dx}{dt}\):
\[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To find the second derivative of a parametric function, we must first find the first derivative \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]
Then, we find the second derivative \(\frac{d^2y}{dx^2}\) using the relation:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} \]

Step 2: Detailed Explanation:
Let us compute the derivatives step-by-step:
-

Step 1: Find \(\frac{dx}{dt}\).
Given \(x = a(\cos t + t\sin t)\):
Using the product rule for \(t\sin t\):
\[ \frac{dx}{dt} = a\left(-\sin t + \left(1 \cdot \sin t + t\cos t\right)\right) = a\left(-\sin t + \sin t + t\cos t\right) = at\cos t \]
- Find \(\frac{dy}{dt}\).
Given \(y = a(\sin t - t\cos t)\):
Using the product rule for \(t\cos t\):
\[ \frac{dy}{dt} = a\left(\cos t - \left(1 \cdot \cos t + t(-\sin t)\right)\right) = a\left(\cos t - \cos t + t\sin t\right) = at\sin t \]
-

Step 2: Find the first derivative \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at\sin t}{at\cos t} = \tan t \]
-

Step 3: Find the second derivative \(\frac{d^2y}{dx^2}\):
Differentiate \(\frac{dy}{dx} = \tan t\) with respect to \(x\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dt}(\tan t) \cdot \frac{dt}{dx} \]
We know that \(\frac{d}{dt}(\tan t) = \sec^2 t\) and \(\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{at\cos t}\):
\[ \frac{d^2y}{dx^2} = \sec^2 t \cdot \frac{1}{at\cos t} \]
Since \(\sec^2 t = \frac{1}{\cos^2 t}\), we can rewrite this as:
\[ \frac{d^2y}{dx^2} = \frac{1}{\cos^2 t} \cdot \frac{1}{at\cos t} = \frac{1}{at\cos^3 t} \]
This matches Option (B).

Step 3: Final Answer:
The correct option is (B).