Question

Acute angle between the curves \(y=x^{2}-2\) and \(y=x^{3}-3 x\) at their point of intersection \((2,2)\) is (1) \(\tan ^{-1}\left(\frac{1}{21}\right)\) (2) \(\frac{\pi}{4}\) (3) \(\tan ^{-1}\left(\frac{5}{37}\right)\) (4) \(\tan ^{-1}\left(\frac{4}{41}\right)\)

Short answer

(3) \( \tan^{-1} \left( \frac{5}{37} \right) \).

Step-by-step solution

- Find the derivatives of the curves

First, compute the derivative of each curve at the point of intersection. For the curve given by the equation \( y = x^2 - 2 \), the derivative is \( \frac{dy_1}{dx} = 2x \). For the curve given by \( y = x^3 - 3x \), the derivative is \( \frac{dy_2}{dx} = 3x^2 - 3 \).

- Evaluate the derivatives at the point (2,2)

Next, substitute \( x = 2 \) into the derivatives to find their values at the point of intersection. For the first curve, \( \frac{dy_1}{dx} = 2(2) = 4 \). For the second curve, \( \frac{dy_2}{dx} = 3(2)^2 - 3 = 12 - 3 = 9 \). Thus, the slopes of the tangents at the intersection are 4 and 9 respectively.

- Calculate the tangent of the angle between the tangents

The formula to calculate the tangent of the angle \(\theta\) between two lines with slopes \( m_1 \) and \( m_2 \) is \( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \). Using \( m_1 = 4 \) and \( m_2 = 9 \), we get \( \tan \theta = \left| \frac{4 - 9}{1 + (4)(9)} \right| = \left| \frac{-5}{37} \right| = \frac{5}{37} \).

- Find the acute angle

Since the calculated tangent value \( \frac{5}{37} \) gives us the tangent of the angle between the curves, the acute angle is \( \theta = \tan^{-1} \left( \frac{5}{37} \right) \).

Key concepts

Understanding Derivatives

Derivatives help us understand how a function changes as its input changes. They are crucial for finding slopes of curves.
In our exercise, we have two curves:

• The first curve is expressed as: \( y = x^2 - 2 \)
• The second curve is expressed as: \( y = x^3 - 3x \)

To find the angle between these curves at their intersection point, we first need their slopes. The slope of a curve at a point is given by its derivative at that point.

For the first curve, \( y = x^2 - 2 \), its derivative will be \( \frac{dy_1}{dx} = 2x \).
For the second curve, \( y = x^3 - 3x \), its derivative will be \( \frac{dy_2}{dx} = 3x^2 - 3 \).

By deriving these functions, we now have formulas to calculate the slopes of the curves at any point.

Finding Intersection Points

Intersection points are where two curves meet. To find these points, we solve the equations of both curves simultaneously.

In our case, we are given the intersection point (2,2) for the curves \( y = x^2 - 2 \) and \( y = x^3 - 3x \). This point means that both curves pass through (2,2).
Next, we substitute \( x = 2 \) into the derivatives we calculated earlier to get the slopes at the intersection point.
For the first curve, the slope at \( x = 2 \) is \( \frac{dy_1}{dx} = 2(2) = 4 \).
For the second curve, the slope at \( x = 2 \) is \( \frac{dy_2}{dx} = 3(2)^2 - 3 = 12 - 3 = 9 \).
Thus, at the intersection point (2,2), the slopes of the tangents to the curves are 4 and 9 respectively.

Calculating Tangent Angle Between Curves

To find the angle between two curves at a point of intersection, we need the angles of their tangents. The formula for the tangent of the angle \( \theta \) between two tangents with slopes \( m_1 \) and \( m_2 \) is:
\( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \).
In our case, we have \( m_1 = 4 \) and \( m_2 = 9 \).
Using the formula, we get: \( \tan \theta = \left| \frac{4 - 9}{1 + 4*9} \right| = \left| \frac{-5}{37} \right| = \frac{5}{37} \).
The calculated value \( \frac{5}{37} \) is the tangent of the angle between the tangents of the curves.
Therefore, the acute angle between the curves is \( \theta = \tan^{-1} \left( \frac{5}{37} \right) \). This helps us understand the spatial relationship between the curves at their intersection point and solve the given exercise successfully.