Find the acute angles between the curves at their points of intersection. (The angle between two curves is the angle between their tangent lines at the point of intersection.) y=x^2, y=x^3

We are tasked with finding the acute angles between the curves \( y = x^2 \) and \( y = x^3 \) at their points of intersection. The angle between two curves at a point of intersection is the angle between their tangent lines at that point.

### Step-by-step Solution:

#### 1. Find the points of intersection:
The points of intersection occur when \( y = x^2 \) and \( y = x^3 \) have the same value. Set the two equations equal to each other:

\[
x^2 = x^3
\]

Simplify the equation:

\[
x^2 (1 - x) = 0
\]

This gives two solutions:

\[
x = 0 \quad \text{or} \quad x = 1
\]

Thus, the curves intersect at \( x = 0 \) and \( x = 1 \).

#### 2. Find the derivatives (slopes of the tangent lines):
The slopes of the tangent lines at a given point are given by the derivatives of the functions \( y = x^2 \) and \( y = x^3 \).

- For \( y = x^2 \), the derivative is:

\[
\frac{dy}{dx} = 2x
\]

- For \( y = x^3 \), the derivative is:

\[
\frac{dy}{dx} = 3x^2
\]

#### 3. Calculate the slopes at the points of intersection:
Now, we evaluate the derivatives at the points of intersection \( x = 0 \) and \( x = 1 \).

- At \( x = 0 \):
    - Slope of \( y = x^2 \): \( 2(0) = 0 \)
    - Slope of \( y = x^3 \): \( 3(0)^2 = 0 \)

Both slopes are 0, which means the tangent lines are horizontal at \( x = 0 \), so the angle between the curves at this point is \( 0^\circ \).

- At \( x = 1 \):
    - Slope of \( y = x^2 \): \( 2(1) = 2 \)
    - Slope of \( y = x^3 \): \( 3(1)^2 = 3 \)

#### 4. Find the angle between the tangent lines at \( x = 1 \):
The formula for the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:

\[
\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|
\]

Substitute \( m_1 = 2 \) and \( m_2 = 3 \):

\[
\tan(\theta) = \left| \frac{2 - 3}{1 + 2 \times 3} \right| = \left| \frac{-1}{1 + 6} \right| = \left| \frac{-1}{7} \right| = \frac{1}{7}
\]

Now, find the angle \( \theta \):

\[
\theta = \tan^{-1} \left( \frac{1}{7} \right)
\]

Using a calculator:

\[
\theta \approx 8.13^\circ
\]

Thus, the acute angle between the curves at \( x = 1 \) is approximately \( 8.13^\circ \).

### Final Answer:
- The angle at \( x = 0 \) is \( 0^\circ \).
- The acute angle at \( x = 1 \) is approximately \( 8.13^\circ \).

Would you like more details, or do you have any other questions?

### Related Questions:
1. How do you find the angle between two vectors in 3D space?
2. How can you calculate the angle between two planes in space?
3. What is the geometric interpretation of a derivative in calculus?
4. How can we find the intersection points of two parametric curves?
5. How does the method change when curves are in polar coordinates?

### Tip:
Remember, the derivative of a function gives the slope of the tangent line, which is key when analyzing angles between curves at points of intersection!

This detailed solution explains how to find the acute angle between the curves y = x^2 and y = x^3 at their points of intersection. Key calculus concepts such as differentiation and the angle between two lines formula are covered. Ideal for students in Grades 11-12 or early college calculus courses.

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