Problem 13 - The Calculus with Analytic Geometry

Question

Find a measurement of the angle between the tangent lines of the given pair of curves at all points of intersection.\(\left\\{\begin{array}{l}r=1-\sin \theta \\\ r=1+\sin \theta\end{array}\right.\)

Step-by-Step Solution

Answer

The angle between the tangent lines is 0 degrees (parallel).

Step 1: Find the points of intersection

First, set the equations equal to find their points of intersection.Solve for \( \theta \) in the system of equations:\[ 1 - \sin\theta = 1 + \sin\theta \]This simplifies to:\[ -\sin\theta = \sin\theta \]Which results in:\[ \sin\theta = 0 \]Thus, \( \theta = 0 \) or \( \theta = \pi \).

Step 2: Parametrize the curves

Using the polar form, convert the curves to their Cartesian coordinates for deeper analysis.For \( r = 1 - \sin\theta \):\[ x = r\cos\theta = (1 - \sin\theta) \cos\theta \]\[ y = r\sin\theta = (1 - \sin\theta) \sin\theta \]For \( r = 1 + \sin\theta \):\[ x = r\cos\theta = (1 + \sin\theta) \cos\theta \]\[ y = r\sin\theta = (1 + \sin\theta) \sin\theta \]

Step 3: Find derivatives at points of intersection

The slopes (tangent) at points of intersection are given by derivatives \( \frac{dy}{dx} \).For \( r = 1 - \sin\theta \):\[ \frac{dy}{d\theta} = (1 - \sin\theta)\cos\theta + (-\cos\theta)(\sin\theta) \]\[ \frac{dx}{d\theta} = -(1 - \sin\theta)\sin\theta + (-\sin\theta)(\cos\theta) \]For \( r = 1 + \sin\theta \):\[ \frac{dy}{d\theta} = (1 + \sin\theta)\cos\theta + (\cos\theta)(\sin\theta) \]\[ \frac{dx}{d\theta} = -(1 + \sin\theta)\sin\theta + (\sin\theta)(\cos\theta) \]

Step 4: Calculate the slopes at θ = 0 and θ = π

Evaluate derivatives:For \( r = 1 - \sin\theta \) and \( r = 1 + \sin\theta \) when \( \theta = 0 \):\[ \frac{dy}{dx} (\theta = 0) = -\sin(0) / \cos(0) = 0/1 = 0 \]For \( r = 1 - \sin\theta \) and \( r = 1 + \sin\theta \) when \( \theta = \pi \):\[ \frac{dy}{dx} (\theta = \pi) = -\sin(\pi) / \cos(\pi) = 0/\cos(\pi) = 0 \]Thus, the slopes at both points are equal.

Step 5: Find the angle between the tangent lines

The angle \( \phi \) between intersecting tangents can be found using the tangent slopes.\[ \tan(\phi) = \left| \frac{m1 - m2}{1 + m1*m2} \right| \]Since both slopes are equal (0), substitution gives:\[ \tan(\phi) = \left| \frac{0 - 0}{1 + 0*0} \right| = 0 \]Thus, \( \phi = 0 \).

Key Concepts

calculus problems

This problem deals with finding the angle between the tangent lines of two curves at their points of intersection. It is a classic example of calculus applications in geometry.
To begin with, you'll need a strong grasp of differentiation and how to handle equations in their various forms.

Identifying points of intersection is the first crucial step. In these kinds of problems, you set the equations of the curves equal to find common values, typically resulting in angles or coordinates where the curves meet.
Once you find the points of intersection, the next step involves finding the derivative. The derivative tells you the slope of the tangent line at any given point on the curve.

Finally, using the slopes, you can determine the angle between the tangent lines. The relationship is given by a specific formula involving arctangents, which simplifies when the slopes are equal.

polar coordinates

Polar coordinates are particularly useful for curves defined in terms of distance from a fixed point and an angle from a fixed direction. For this exercise, the curves are given as:

• \(r = 1 - \sin\theta\)
• \(r = 1 + \sin\theta\)

In polar coordinates, \(r\) represents the radius, and \(\theta\) represents the angle.
Transforming these to Cartesian coordinates simplifies finding derivatives and slopes.

For example, you convert the curve \(r = 1 - \sin\theta\) to Cartesian using:

• \(x = r\cos\theta = (1 - \sin\theta)\cos\theta\)
• \(y = r\sin\theta = (1 - \sin\theta)\sin\theta\)

This step is essential for clearer differentiation and slope calculation. Similarly, for \(r = 1 + \sin\theta\), the conversion steps are:

• \(x = (1 + \sin\theta)\cos\theta\)
• \(y = (1 + \sin\theta)\sin\theta\)

Polar coordinates provide an elegant way to represent and analyze curves, especially in circular or spiral forms.

derivative calculation

Finding the derivative is key to understanding how the function behaves. For this problem, you need to determine the slopes of the tangent lines at the intersection points.
Start by differentiating the x and y components of the curves.For \(r = 1 - \sin\theta\):

• \(\frac{dy}{d\theta} = (1 - \sin\theta)\cos\theta + (-\cos\theta)(\sin\theta)\)
• \(\frac{dx}{d\theta} = -(1 - \sin\theta)\sin\theta + (-\sin\theta)(\cos\theta)\)

By calculating these derivatives, you get the slopes of the tangent lines.
Similarly, for \(r = 1 + \sin\theta\):

• \(\frac{dy}{d\theta} = (1 + \sin\theta)\cos\theta + (\cos\theta)(\sin\theta)\)
• \(\frac{dx}{d\theta} = -(1 + \sin\theta)\sin\theta + (\sin\theta)(\cos\theta)\)

At the points of intersection, \(\theta = 0\) and \(\theta = \pi\), substitute these values back into the derivatives to find the slopes.

points of intersection

Finding the points of intersection is the foundation of solving this problem. Set the equations of the curves equal to each other:

• \(1 - \sin\theta = 1 + \sin\theta\)

Solving for \(\theta\), we get:

• \(-\sin\theta = \sin\theta\)
• \(\sin\theta = 0\)

This simplifies to \(\theta = 0\) or \(\theta = \pi\).
Once you know these intersection points, you use them in derivative calculations to find the slope of tangent lines.
In this exercise, both points yield slopes of 0.
The angle between the tangent lines at these points is found using the slopes and the formula:

• \(\tan(\phi) = \left| \frac{m1 - m2}{1 + m1*m2} \right|\).

Since the slopes are equal (both are 0), \(\phi\) is 0 degrees. Thus, the tangent lines are parallel at these intersection points.