Question

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

Never forget to multiply by \(\frac{dt}{dx}\) (or divide by \(\frac{dx}{dt}\)) when finding the second derivative of parametric functions:
\[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \] A common mistake is to just write \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\), which would incorrectly yield \(\sec^2 t\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given parametric equations of a curve: \(x(t)\) and \(y(t)\).
We need to find the second derivative \(\frac{d^2y}{dx^2}\) with respect to \(x\).
Key Formula or Approach:
1. Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
2. Calculate \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] 3. Calculate the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \]

Step 2: Detailed Explanation:

• Differentiate \(x = a(\cos t + t\sin t)\) with respect to \(t\):
Using the product rule for the second term:
\[ \frac{dx}{dt} = a\left(-\sin t + \sin t + t\cos t\right) = at\cos t \]

• Differentiate \(y = a(\sin t - t\cos t)\) with respect to \(t\):
Using the product rule for the second term:
\[ \frac{dy}{dt} = a\left(\cos t - (\cos t - t\sin t)\right) = at\sin t \]

• Find the first derivative \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{at\sin t}{at\cos t} = \tan t \]

• Find the second derivative \(\frac{d^2y}{dx^2}\):
\[ \frac{d^2y}{dx^2} = \frac{d}{dt}(\tan t) \cdot \frac{dt}{dx} = \sec^2 t \cdot \frac{1}{\frac{dx}{dt}} \] Substitute \(\frac{dx}{dt} = at\cos t\):
\[ \frac{d^2y}{dx^2} = \sec^2 t \cdot \frac{1}{at\cos t} \] Since \(\sec^2 t = \frac{1}{\cos^2 t}\), we have:
\[ \frac{d^2y}{dx^2} = \frac{1}{\cos^2 t \cdot at\cos t} = \frac{1}{at\cos^3 t} \]

Step 3: Final Answer:
The second derivative is \(\frac{d^2 y}{dx^2} = \frac{1}{at\cos^3 t}\).