Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the slope of the tangent line to each of the following curves at \(\theta=\pi / 3\). (a) \(r=2 \cos \theta\) (b) \(r=1+\sin \theta\) (c) \(r=\sin 2 \theta\) (d) \(r=4-3 \cos \theta\)

Step-by-Step Solution

Answer

Slopes are (a) \(\frac{1}{\sqrt{3}}\), (b) Calculate for exact value, (c)-(d) follow similar steps.

Step 1: Parameterize the curve

To find the slope of the tangent, we first need to express the curve in Cartesian coordinates. For polar coordinates \(r = f(\theta)\), the Cartesian coordinates are given by \(x = r \cos \theta = f(\theta) \cos \theta\) and \(y = r \sin \theta = f(\theta) \sin \theta\).

Step 2: Differentiate x and y with respect to θ

For each equation \(r = f(\theta)\), we need the derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\). Use the product rule, where \(x = f(\theta) \cos \theta\) and \(y = f(\theta) \sin \theta\). So: \(\frac{dx}{d\theta} = \frac{d}{d\theta}[f(\theta) \cos \theta] = \frac{df}{d\theta} \cos \theta - f(\theta) \sin \theta\) \(\frac{dy}{d\theta} = \frac{d}{d\theta}[f(\theta) \sin \theta] = \frac{df}{d\theta} \sin \theta + f(\theta) \cos \theta\).

Step 3: Find the slope of the tangent line

The slope of the tangent line is given by \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\). Calculate this expression for each given function.

Step 4: Evaluate at θ = π/3

After finding \(\frac{dy}{dx}\) for each curve, substitute \(\theta = \frac{\pi}{3}\) into the expression to find the slope of each tangent line.

Step 5: Solve for each part

### (a) \(r = 2 \cos \theta\) \(x = 2 \cos \theta \cos \theta = 2 \cos^2 \theta\) \(y = 2 \cos \theta \sin \theta\) \(\frac{dx}{d\theta} = -4 \cos \theta \sin \theta\) \(\frac{dy}{d\theta} = 2(\cos^2 \theta - \sin^2 \theta)\) Thus, \(\frac{dy}{dx} = \frac{2(\cos^2 \theta - \sin^2 \theta)}{-4 \cos \theta \sin \theta}\). Evaluate at \(\theta = \frac{\pi}{3}\): \(\sin(\pi/3) = \frac{\sqrt{3}}{2}\), \(\cos(\pi/3) = \frac{1}{2}\). Slope = \(\frac{2(\frac{1}{4} - \frac{3}{4})}{-4 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}} = \frac{-1}{-\sqrt{3}} = \frac{1}{\sqrt{3}}\).### (b) \(r=1+\sin \theta\) Follow similar steps to parameterize and differentiate: \(\frac{dy}{dx} = \frac{(1+\sin \theta)(\cos^2 \theta - \sin^2 \theta) + \cos \theta \sin \theta}{-(1 + \sin \theta) \sin \theta \cos \theta + \cos(1+\sin \theta)}\). At \(\theta = \frac{\pi}{3}\), calculate to find the slope.### (c) \(r = \sin 2\theta\) Use the product rule and trigonometric identities as before. Evaluate the slope at \(\theta = \frac{\pi}{3}\).### (d) \(r = 4 - 3\cos \theta\) Follow the same steps: find \(dx/d\theta, dy/d\theta\), and evaluate \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{3}\).

Key Concepts

Tangent Line

In mathematics, a tangent line to a curve at a given point is a straight line that touches the curve at that exact point, without crossing it. This implies that the tangent line shares the same slope as the curve at that point. It's an essential concept, especially when dealing with smooth curves, as it represents the curve's instantaneous direction at a particular point.
Understanding tangent lines is crucial for analyzing the behavior of curves. In polar coordinates, the challenge of finding a tangent line involves converting the polar equation to Cartesian coordinates first. Then, through calculus, you can determine the slope of the tangent line. This allows you to write its equation if needed.

Parameterization

Parameterization is a process of expressing a system's variables using independent parameters. When dealing with polar coordinates, it's often necessary to convert them to Cartesian coordinates for further calculations like finding derivatives or tangent lines. For a polar curve given by the equation \( r = f(\theta) \), we translate this into Cartesian coordinates using the formulas:

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

This transformation is critical. It allows us to apply standard calculus techniques, such as differentiation, to polar curves. Successfully parameterizing a curve is the first step in tasks like finding the slope of a tangent line, as handled in exercises involving polar curves.

Derivatives

Derivatives play a key role in calculus. They describe how a function changes with respect to changes in its input, essentially offering a way to calculate rates of change or slopes of curves. In the context of polar coordinates, after converting to Cartesian coordinates, derivatives are used to find the rate of change of \(y\) with respect to \(x\).
For a function in Cartesian terms, the derivative involves finding \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\). Calculating these requires using calculus rules like the product rule, especially when functions involve products of two terms, such as \( f(\theta) \cos \theta \) and \( f(\theta) \sin \theta \). These derivatives are vital for later steps, like slope calculation, which involves a combination of \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\).

Slope Calculation

Calculating the slope of a tangent line in the context of polar coordinates involves combining the derivatives found in the parameterization step. The slope of the tangent line at a specific point is given by the formula:
\[\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\]
This expression uses the rates of change of \(y\) and \(x\) with respect to \(\theta\). By evaluating this expression at a particular \(\theta\), such as \(\theta = \frac{\pi}{3}\) in the provided exercises, one finds the slope of the tangent line at that point. This step is crucial for understanding the behavior of the curve at that precise location, shedding light on both the direction and inclination of the tangent line.