Question

If \( x = at^4 \) and \( y = 2at^2 \), then \( \frac{d^2y}{dx^2} \) is equal to :

• A. \( -\frac{1}{4at^4} \)
• B. \( -\frac{2}{t^3} \)
• C. \( -\frac{1}{t} \)
• D. \( -\frac{1}{2at^6} \)

Hint:

Remember: \( \frac{d^2y}{dx^2} \neq \frac{y''(t)}{x''(t)} \).
A helpful mnemonic for the parametric second derivative is:
\( \text{Derivative of Slope} \div \text{Derivative of x} \).
Always express your final answer in the simplest power form to avoid confusion with the options.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem deals with the differentiation of parametric equations. In many physical or geometric contexts, coordinates \( x \) and \( y \) are not expressed directly as functions of each other but are instead defined in terms of a third variable, often called a parameter (denoted here as \( t \)). This is common in kinematics where \( x \) and \( y \) positions depend on time.
To find the second derivative of \( y \) with respect to \( x \), denoted as \( \frac{d^2y}{dx^2} \), we cannot simply differentiate \( y \) twice and divide by the second derivative of \( x \). Instead, we must apply the chain rule carefully. The second derivative represents the rate of change of the first derivative (\( \frac{dy}{dx} \)) with respect to \( x \). Since our expression for \( \frac{dy}{dx} \) will be in terms of \( t \), we must differentiate it with respect to \( t \) and then multiply by \( \frac{dt}{dx} \).

Step 2: Key Formula or Approach:
The step-by-step approach involves:
1. Finding the first derivatives of \( x \) and \( y \) with respect to the parameter: \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
2. Computing the first derivative of \( y \) with respect to \( x \) using the formula:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]
3. Computing the second derivative using the formula:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{\frac{dx}{dt}} \]
This final step is where most errors occur, as it is easy to forget to divide by \( \frac{dx}{dt} \) a second time.

Step 3: Detailed Explanation:
Step 3.1: Differentiating the parametric equations with respect to \( t \).
Given \( x = at^4 \):
\[ \frac{dx}{dt} = \frac{d}{dt}(at^4) = 4at^3 \]
Given \( y = 2at^2 \):
\[ \frac{dy}{dt} = \frac{d}{dt}(2at^2) = 4at \]
Step 3.2: Finding the first derivative \( \frac{dy}{dx} \).
\[ \frac{dy}{dx} = \frac{4at}{4at^3} \]
Canceling common factors (\( 4a \) and one \( t \)):
\[ \frac{dy}{dx} = \frac{1}{t^2} = t^{-2} \]
Step 3.3: Finding the second derivative \( \frac{d^2y}{dx^2} \).
First, differentiate \( \frac{dy}{dx} \) with respect to \( t \):
\[ \frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d}{dt}(t^{-2}) = -2t^{-3} = -\frac{2}{t^3} \]
Now, apply the parametric second derivative formula:
\[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{-\frac{2}{t^3}}{4at^3} \]
Simplifying the fraction:
\[ \frac{d^2y}{dx^2} = -\frac{2}{4at^3 \cdot t^3} = -\frac{2}{4at^6} \]
Reducing the fraction \( \frac{2}{4} \) to \( \frac{1}{2} \):
\[ \frac{d^2y}{dx^2} = -\frac{1}{2at^6} \]
This matches option (4) provided in the question.

Step 4: Final Answer:
By systematically differentiating the parametric equations and correctly applying the chain rule for the second derivative, we find that \( \frac{d^2y}{dx^2} = -\frac{1}{2at^6} \). This is a classical result for such power-form parametric equations. Therefore, Option (4) is the correct answer.