Question

The acute angle between the curves \( y = 3x^2 - 2x - 1 \) and \( y = x^3 - 1 \) at their point of intersection which lies in the first quadrant is

• A. \( \tan^{-1}\left(\frac{2}{121}\right) \)
• B. \( \tan^{-1}(2) \)
• C. \( \tan^{-1}\left(\frac{1}{13}\right) \)
• D. \( \frac{\pi}{2} \)

Hint:

Ensure you pick the correct intersection point. "First quadrant" usually implies \( x \textgreater 0, y \textgreater 0 \). Points on axes like \( (1,0) \) might be candidates in some contexts, but \( (2,7) \) is clearly the intended one here.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

To find the angle between two curves, we first need to find their point(s) of intersection. Then, we calculate the slope of the tangent to each curve (derivative) at the intersection point. Finally, we use the formula for the angle between two lines with slopes \( m_1 \) and \( m_2 \).
Step 2: Key Formula or Approach:

1. Angle formula: \( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \). 2. Slope \( m = \frac{dy}{dx} \).
Step 3: Detailed Explanation:

Find the intersection points by equating the two expressions for \( y \): \[ 3x^2 - 2x - 1 = x^3 - 1 \] \[ x^3 - 3x^2 + 2x = 0 \] Factor the equation: \[ x(x^2 - 3x + 2) = 0 \] \[ x(x-1)(x-2) = 0 \] The solutions are \( x = 0, 1, 2 \). We need the point in the first quadrant. - If \( x=0 \), \( y = -1 \) (Not Q1). - If \( x=1 \), \( y = 1^3 - 1 = 0 \) (On boundary, strictly speaking not inside Q1, but let's check). - If \( x=2 \), \( y = 2^3 - 1 = 7 \) (Inside Q1). The point of interest is \( P(2, 7) \). Calculate the slopes at \( x=2 \): Curve 1: \( y = 3x^2 - 2x - 1 \) \[ \frac{dy}{dx} = 6x - 2 \] At \( x=2 \), \( m_1 = 6(2) - 2 = 10 \). Curve 2: \( y = x^3 - 1 \) \[ \frac{dy}{dx} = 3x^2 \] At \( x=2 \), \( m_2 = 3(2)^2 = 12 \). Calculate the angle \( \theta \): \[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \] \[ \tan \theta = \left| \frac{10 - 12}{1 + (10)(12)} \right| \] \[ \tan \theta = \left| \frac{-2}{1 + 120} \right| = \frac{2}{121} \] \[ \theta = \tan^{-1}\left(\frac{2}{121}\right) \]
Step 4: Final Answer:

The acute angle is \( \tan^{-1}\left(\frac{2}{121}\right) \).