Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all points on the limaçon \(r=1-2 \sin \theta\) where the tangent line is horizontal.

Step-by-Step Solution

Answer

Points: \((-1, \frac{\pi}{2})\), \((3, \frac{3\pi}{2})\), \(\left(\frac{1}{2}, \arcsin \frac{1}{4}\right)\), and \(\left(\frac{1}{2}, \pi - \arcsin \frac{1}{4}\right)\)."

Step 1: Understand the Tangent Line Conditions

A tangent line to a polar curve is horizontal when the slope of the tangent is zero. For polar coordinates, the slope \( \frac{dy}{dx} \) is given by \( \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta} \). For a horizontal tangent line, the numerator must be zero.

Step 2: Differentiate r with respect to θ

Differentiate the given polar equation \( r = 1 - 2 \sin \theta \) with respect to \( \theta \): \[\frac{dr}{d\theta} = -2 \cos \theta\]

Step 3: Set the Numerator of the Slope to Zero

Plug \( r \) and \( \frac{dr}{d\theta} \) into the numerator \( r' \sin \theta + r \cos \theta \): \[-2 \cos \theta \sin \theta + (1 - 2 \sin \theta) \cos \theta = 0\]

Step 4: Simplify the Numerator Expression

Simplify the expression: \[\cos \theta - 2 \cos \theta \sin \theta - 2 \sin \theta \cos \theta = 0 \] This can be rearranged to: \[ \cos \theta (1 - 4\sin \theta) = 0 \]

Step 5: Solve for θ

The equation \( \cos \theta (1 - 4 \sin \theta) = 0 \) gives two cases: 1. \( \cos \theta = 0 \), giving \( \theta = \frac{\pi}{2}, \frac{3\pi}{2} \)2. \( 1 - 4 \sin \theta = 0 \), giving \( \sin \theta = \frac{1}{4} \). Solving for \( \theta \) gives \( \theta = \arcsin \frac{1}{4} \) and its supplement.

Step 6: Find Corresponding Points on the Limaçon

For \( \theta = \frac{\pi}{2} \) and \( \theta = \frac{3\pi}{2} \), calculate \( r_{1} = -1 \) and \( r_{2} = 3 \), respectively.For \( \theta = \arcsin \frac{1}{4} \), find \( r = 1 - 2 \left( \frac{1}{4} \right) = \frac{1}{2} \).For the supplementary angle \( \theta = \pi - \arcsin \frac{1}{4} \), similarly, find \( r = \frac{1}{2} \).

Step 7: Conclude the Solutions

The points where the tangent line is horizontal are given by the pairs \((r, \theta)\): - \((-1, \frac{\pi}{2})\)- \((3, \frac{3\pi}{2})\)- \(\left(\frac{1}{2}, \arcsin \frac{1}{4}\right)\)- \(\left(\frac{1}{2}, \pi - \arcsin \frac{1}{4}\right)\)

Key Concepts

Horizontal Tangent

A horizontal tangent is a critical concept in calculus used to identify points on a curve where the slope of the tangent line is zero. In simpler terms, it is where the curve is perfectly flat at a point. To find such points for a polar curve, like the limaçon, we focus on the formula for the slope of the tangent in polar coordinates.

For polar curves, the tangent line is horizontal when the numerator of the derivative \[\frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}\]is zero. This zero condition means the vertical change (\(dy\)) relative to the horizontal change (\(dx\)) is zero, resulting in a flat slope. Identifying horizontal tangents involves finding when \(r' \sin \theta + r \cos \theta = 0\).

In the example of the limaçon \(r = 1 - 2 \sin \theta\), after differentiation, setting the numerator to zero helps us determine the necessary angles where these tangents occur.

Differentiation in Polar Coordinates

Differentiation in polar coordinates is essential to analyze curves not easily handled in Cartesian coordinates. While Cartesian coordinates represent points as \((x, y)\), polar coordinates express them through a radius \(r\) and angle \(\theta\). To find derivatives, one typically moves between polar and Cartesian representations.

For a polar equation like \(r = 1 - 2 \sin \theta\), differentiation focuses on \(\theta\). We calculate the derivative of the radius \(r\) with respect to the angle. In our example, it simplifies to the derivative \[\frac{dr}{d\theta} = -2 \cos \theta.\]This derivation gives us the rate of change of \(r\) as \(\theta\) changes, crucial for determining the slope of the curve.

By plugging this derivative back into the polar equations, we find the correct conditions for specific features of the curve, such as horizontal tangents.

Limaçon Curve

Limaçons are a fascinating family of polar curves defined generally as \(r = a + b \sin \theta\) or \(r = a + b \cos \theta\). They exhibit different forms depending on the values of \(a\) and \(b\). Traits of limaçons include dimpled shapes, loops, or even a convex form. Variations depend on the relationship between \(a\) and \(b\).

In our exercise, the limaçon equation \(r = 1 - 2 \sin \theta\) implies a unique shape. Because \(b\) (the coefficient of the sine function) is two times greater than \(a\), the curve boasts an inner loop. This shape would differ if \(a\) exceeded \(b\).

Understanding the differentiations and transformations of such curves in polar form is key to analyzing their graphical and mathematical properties, which help in determining characteristics like tangent directions and areas enclosed by these intriguing shapes.