Question

For the following exercises, find the slope of a tangent line to a polar curve \(r=f(\theta) .\) Let \(x=r \cos \theta=f(\theta) \cos \theta\) and \(y=r \sin \theta=f(\theta) \sin \theta\), so the polar equation \(r=f(\theta)\) is now written in parametric form.\(r=4 \cos \theta ;\left(2, \frac{\pi}{3}\right)\)

Short answer

The slope of the tangent line at \(\left(2, \frac{\pi}{3}\right)\) is \(-\frac{\sqrt{3}}{3}\).

Step-by-step solution

Convert Polar to Parametric Form

Given the polar equation \( r = 4\cos\theta \), we need to express it in terms of \(x\) and \(y\). By the formulas, \(x = 4\cos\theta \cos\theta = 4\cos^2\theta\) and \(y = 4\cos\theta \sin\theta\). So our parametric equations are \(x(\theta) = 4\cos^2\theta\) and \(y(\theta) = 4\cos\theta \sin\theta\).

Differentiate Parametric Equations

Differentiate \(x(\theta) = 4\cos^2\theta\) and \(y(\theta) = 4\cos\theta\sin\theta\) with respect to \(\theta\). We find \(\frac{dx}{d\theta} = 4 \, \times \, 2\cos\theta(-\sin\theta) = -8\cos\theta\sin\theta\), and \(\frac{dy}{d\theta} = 4\left(-\sin^2\theta + \cos^2\theta\right) = 4(\cos^2\theta - \sin^2\theta)\).

Find Slope of Tangent Line

The slope of the tangent line to the polar curve is given by \(\frac{dy/d\theta}{dx/d\theta}\). Thus, we calculate \(\frac{4(\cos^2\theta - \sin^2\theta)}{-8\cos\theta\sin\theta}\). This simplifies to \(\frac{-\tan\theta + \cot\theta}{2}\).

Plug in the Given Point

To find the slope at point \((2, \frac{\pi}{3})\), we need to plug \(\theta = \frac{\pi}{3}\) into the slope formula. Evaluate: \(\cos\frac{\pi}{3} = \frac{1}{2}\) and \(\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\). Substitute these into \(\frac{-\tan\theta + \cot\theta}{2}\) to get the slope.

Calculate Slope at Given Point

Calculate the actual slope: \(\tan\frac{\pi}{3} = \sqrt{3}\) and \(\cot\frac{\pi}{3} = \frac{1}{\sqrt{3}}\). Thus, the slope becomes \(\frac{-\sqrt{3} + \frac{1}{\sqrt{3}}}{2} = \frac{-\frac{2\sqrt{3}}{3}}{2} = \frac{-\sqrt{3}}{3}\). This is the slope of the tangent line at the given polar coordinate.

Key concepts

Understanding Parametric Equations in Polar Coordinates

Parametric equations let us describe a curve using a parameter, often denoted as \( \theta \) in polar coordinates. In polar settings where we have \( r = f(\theta) \), our goal is to express \( r \) in terms of \( x \) and \( y \) to study the curve's behavior with familiar Cartesian parameters.
To convert a polar equation like \( r = 4\cos\theta \) into parametric form, we use:

• \( x = r \cos\theta = 4\cos\theta \cos\theta = 4\cos^2\theta \)
• \( y = r \sin\theta = 4\cos\theta \sin\theta \)

This transformation is essential for tasks like calculating slopes or integrating, where having coordinates in \( x, y \) form simplifies the problem.
Grasping parametric equations helps bridge the gap between polar and Cartesian systems, offering a different perspective for visualizing and analyzing curves.

Differentiation of Parametric Equations

Differentiation of parametric equations using the parameter \( \theta \) is a key step in exploring the properties of a curve in polar coordinates. From the derived parametric equations \( x(\theta) = 4\cos^2\theta \) and \( y(\theta) = 4\cos\theta \sin\theta \), we differentiate each with respect to \( \theta \).
Knowing the differentiation formulas helps:

• \( \frac{dx}{d\theta} = -8\cos\theta\sin\theta \)
• \( \frac{dy}{d\theta} = 4(\cos^2\theta - \sin^2\theta) \)

These derivatives are fundamental when needing to analyze the rate at which \( x \) and \( y \) change concerning \( \theta \). This insight lays down the groundwork to solve for the slope of a tangent line on the polar curve, highlighting differentiation's role in understanding motion and change.

Tangent Line to a Polar Curve

The concept of a tangent line is crucial for understanding the behavior of a curve at a specific point. In polar coordinates, finding the slope of a tangent line involves differentiating the parametric equations and forming a ratio of these derivatives. From our exercise, the slope \( m \) is given by:

• \( m = \frac{dy/d\theta}{dx/d\theta} = \frac{4(\cos^2\theta - \sin^2\theta)}{-8\cos\theta\sin\theta} \)

This simplifies the tangent line's slope formula to \( \frac{-\tan\theta + \cot\theta}{2} \).
To find this slope at a specific polar coordinate, such as \((2, \frac{\pi}{3})\), we substitute \( \theta = \frac{\pi}{3} \), using trigonometric values for cosine and sine. This entire process shows how tangent lines help us understand how steep or flat a curve is at a particular point, essential for applications in physics and geometry.