Question

If the rate of change of the slope of the tangent drawn to the curve \( y = x^3 - 2x^2 + 3x - 2 \) at the point (2,4) is k times the rate of change of its abscissa, then k =

• A. 2
• B. 4
• C. 6
• D. 8

Hint:

When a problem states "rate of change of A is k times rate of change of B", it mathematically translates to \( \frac{dA}{dt} = k \frac{dB}{dt} \), which simplifies via chain rule to \( \frac{dA}{dB} = k \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

The problem relates the rate of change of the slope (\( m \)) to the rate of change of the abscissa (\( x \)). Both are changing with respect to time \( t \). We need to find the constant \( k \) in the relation \( \frac{dm}{dt} = k \frac{dx}{dt} \).
Step 2: Key Formula or Approach:

1. Slope \( m = \frac{dy}{dx} \). 2. Chain rule: \( \frac{dm}{dt} = \frac{dm}{dx} \cdot \frac{dx}{dt} \).
Step 3: Detailed Explanation:

Given curve: \( y = x^3 - 2x^2 + 3x - 2 \). Find the slope \( m \) as a function of \( x \): \[ m = \frac{dy}{dx} = 3x^2 - 4x + 3 \] We are given that the rate of change of slope is \( k \) times the rate of change of abscissa: \[ \frac{dm}{dt} = k \frac{dx}{dt} \] Using the chain rule on \( \frac{dm}{dt} \): \[ \frac{dm}{dx} \cdot \frac{dx}{dt} = k \frac{dx}{dt} \] Assuming \( \frac{dx}{dt} \neq 0 \), we can divide by it: \[ \frac{dm}{dx} = k \] Now, find \( \frac{dm}{dx} \): \[ \frac{dm}{dx} = \frac{d}{dx}(3x^2 - 4x + 3) \] \[ \frac{dm}{dx} = 6x - 4 \] We need to evaluate this at the point \( (2, 4) \), so substitute \( x = 2 \): \[ k = 6(2) - 4 \] \[ k = 12 - 4 = 8 \]
Step 4: Final Answer:

The value of \( k \) is 8.