Problem 70 - Calculus Early Transcendentals

Question

Show that the curves \(r=a \sin \theta\) and \(r=a \cos \theta\) intersect at right angles.

Step-by-Step Solution

Answer

The curves intersect at right angles when both slopes multiply to -1.

Step 1: Express the Curves in Cartesian Coordinates

To begin, convert the polar equations of the curves into Cartesian coordinates. For the curve \(r = a \sin \theta\), in Cartesian coordinates, this is \(x = r \cos \theta = a \sin \theta \cos \theta\) and \(y = r \sin \theta = a \sin^2 \theta\). Similarly, for \(r = a \cos \theta\), \(x = a \cos^2 \theta\) and \(y = a \cos \theta \sin \theta\). However, we will simplify these using polar conversions \(x = r \cos \theta\) and \(y = r \sin \theta\).

Step 2: Parametrize the Curves

For more straightforward differentiation, express \(x\) and \(y\) using parametric forms. For \(r = a \sin \theta\), we get \(x = a \sin \theta \cos \theta\) and \(y = a \sin^2 \theta\). For \(r = a \cos \theta\), we get \(x = a \cos^2 \theta\) and \(y = a \sin \theta \cos \theta\).

Step 3: Calculate the Slopes using Derivatives

Find the derivatives \(\frac{dy}{dx}\) for both parametric forms to find the slopes of the tangents at any point. For \(r = a \sin \theta\), using \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\), \(\frac{dy}{dx} = \frac{2a \sin \theta \cos \theta}{a(\cos^2 \theta - \sin^2 \theta)} = \tan \theta\). For \(r = a \cos \theta\), \(\frac{dy}{dx} = \frac{a(\cos^2 \theta - \sin^2 \theta)}{2a \sin \theta \cos \theta} = \cot \theta\).

Step 4: Evaluate Slopes at Intersection Points

The curves intersect when \(\theta\) satisfies both \(x = 0\) and \(y = 0\) for both curves. Solving for \(\theta\), these equations intersect at \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{3\pi}{4}\). At these angles, \(\tan \theta = \cot \theta = 1\) or \(-1\).

Step 5: Confirm Orthogonality of Tangents

If two curves intersect at right angles, the product of their slopes at the intersection must equal \(-1\). At \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{3\pi}{4}\), the slopes are \(\tan \theta = 1\) and \(\cot \theta = 1\). Thus, their product is \(1 \cdot 1 = 1\) and \(1 \cdot (-1) = -1\), indicating right angles at only some points.

Key Concepts

Polar Coordinates

Polar coordinates provide an alternative to the more commonly used Cartesian coordinate system. In polar coordinates, each point on a plane is determined by a distance from a reference point and an angle from a reference direction. The reference point is typically the origin, called the pole, and the angle is measured from the positive x-axis, known as the polar axis.

The primary advantage of using polar coordinates is their suitability for dealing with problems and equations that are naturally circular or involve radial symmetry.

• The radius (r) measures the distance from the pole to the point.
• The angle (θ) provides the direction.

For example, the equations given in the exercise, such as \(r = a \sin \theta\) and \(r = a \cos \theta\), are easily expressed and interpreted using polar coordinates, helping to illustrate the geometric properties of curves.

Parametric Forms

Parametric forms express the coordinates \(x\) and \(y\) as functions of a third variable, often serving as a time parameter in applications such as motion analysis. This form allows us to write the coordinates of points on a curve in terms of one or more parameters, which can simplify the differentiation and integration processes.

In the context of the given curves, converting to parametric form involves expressing \(x\) and \(y\) for the polar equations in terms of \(\theta\). For example, using the curve \(r = a \sin \theta\):

• \(x = a \sin \theta \cos \theta\)
• \(y = a \sin^2 \theta\)

These parametric equations provide a convenient way to examine how the curve behaves as \(\theta\) varies. They also help prepare the functions for taking derivatives, which are crucial in finding slopes and analyzing curve intersections.

Derivative Calculation

Calculating derivatives is a fundamental technique to determine how functions change and to find slopes of curves at given points. In problems involving parametric equations, we often use the chain rule to compute derivatives like \(\frac{dy}{dx}\).

For the exercise curves, this involves calculating \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) and dividing:

• For \(r = a \sin \theta\), \(\frac{dy}{dx} = \frac{2a \sin \theta \cos \theta}{a(\cos^2 \theta - \sin^2 \theta)} = \tan \theta\).
• For \(r = a \cos \theta\), \(\frac{dy}{dx} = \frac{a(\cos^2 \theta - \sin^2 \theta)}{2a \sin \theta \cos \theta} = \cot \theta\).

This calculation shows how the slope of the tangent line to the curve varies with \(\theta\), essential for determining where and how curves intersect.

Orthogonal Curves

Orthogonal curves are curves that intersect at right angles. To check for orthogonality, we look at the tangents at their intersection points. If the tangent lines of two curves at their intersection are perpendicular, their slopes' product must be \(-1\).

In this exercise, the curves \(r = a \sin \theta\) and \(r = a \cos \theta\) are shown to have intersection points at \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{3\pi}{4}\). At these points:

• The slopes for both curves are \(\tan \theta = 1\) and \(\cot \theta = 1\) when considered at their natural intersections.
• Their products: \(1 \cdot 1 = 1\) and \(1 \cdot (-1) = -1\) satisfy the orthogonality condition at one of the pairs of angles.

Understanding orthogonal curves aids in visualizing the geometric configuration and confirming the nature of the intersection.