Math Problem Statement

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. Give your answers in degrees, rounding to one decimal place. Enter your answers as a comma-separated list.) y = 7x2, y = 7x3

Solution

To find the acute angle between the curves y=7x2y = 7x^2 and y=7x3y = 7x^3 at their points of intersection, we follow these steps:

1. Find the points of intersection

To find the points where the two curves intersect, set the equations equal to each other:

7x2=7x37x^2 = 7x^3

Divide both sides by 7:

x2=x3x^2 = x^3

Factor the equation:

x2(x1)=0x^2(x - 1) = 0

This gives two solutions:

x=0orx=1x = 0 \quad \text{or} \quad x = 1

Thus, the points of intersection are (0,0)(0, 0) and (1,7)(1, 7).

2. Find the slopes of the tangent lines at each point

The slope of the tangent line to a curve at a given point is found by differentiating the equation of the curve.

For y=7x2y = 7x^2, the derivative is:

dydx=14x\frac{dy}{dx} = 14x

For y=7x3y = 7x^3, the derivative is:

dydx=21x2\frac{dy}{dx} = 21x^2

3. Calculate the angle between the tangent lines

The formula for the angle θ\theta between two curves at their point of intersection, based on their slopes m1m_1 and m2m_2, is given by:

tan(θ)=m1m21+m1m2\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|

For each point of intersection, we calculate the slopes and apply the formula.

At (0,0)(0, 0):

  • The slope of the tangent to y=7x2y = 7x^2 is m1=14(0)=0m_1 = 14(0) = 0.
  • The slope of the tangent to y=7x3y = 7x^3 is m2=21(0)2=0m_2 = 21(0)^2 = 0.

So, m1=m2=0m_1 = m_2 = 0, and the angle between them is:

tan(θ)=001+0×0=0\tan(\theta) = \left| \frac{0 - 0}{1 + 0 \times 0} \right| = 0

Thus, the angle between the curves at (0,0)(0, 0) is 00^\circ.

At (1,7)(1, 7):

  • The slope of the tangent to y=7x2y = 7x^2 is m1=14(1)=14m_1 = 14(1) = 14.
  • The slope of the tangent to y=7x3y = 7x^3 is m2=21(1)2=21m_2 = 21(1)^2 = 21.

Now, apply the formula for the angle:

tan(θ)=14211+14×21\tan(\theta) = \left| \frac{14 - 21}{1 + 14 \times 21} \right| tan(θ)=71+294=7295=0.0237\tan(\theta) = \left| \frac{-7}{1 + 294} \right| = \left| \frac{-7}{295} \right| = 0.0237

Now, find θ\theta:

θ=tan1(0.0237)1.36\theta = \tan^{-1}(0.0237) \approx 1.36^\circ

Final Answer:

The acute angles between the curves at their points of intersection are approximately:

0,1.40^\circ, 1.4^\circ

Would you like a detailed explanation of any of the steps, or have any additional questions?

Here are some related questions:

  1. How do we calculate the angle between two curves when their tangents are not perpendicular?
  2. How can we determine whether two curves are tangent at a point of intersection?
  3. How does the angle between the curves change if we have cubic vs quadratic functions?
  4. What is the significance of the angle between two curves in real-world applications?
  5. How do the slopes of tangents relate to the concavity of curves?

Tip: When finding the angle between curves, ensure you first find their points of intersection before calculating the slopes of their tangent lines.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Tangents
Intersection of Curves

Formulas

tan(θ) = |(m1 - m2) / (1 + m1 * m2)|

Theorems

Angle between two curves at a point of intersection

Suitable Grade Level

Grades 11-12