Problem 246 - Calculus Volume 3

Question

Find the slope of the tangent line to the given polar curve at the point given by the value of \(\theta\). $$ r=3 \cos \theta, \theta=\frac{\pi}{3} $$

Step-by-Step Solution

Answer

The slope of the tangent line is \( \frac{\sqrt{3}}{3} \).

Step 1: Calculate the Derivatives of r with Respect to θ

First, we differentiate the polar equation given: \( r = 3 \cos \theta \).* The derivative \( \frac{dr}{d\theta} \) is obtained using standard differentiation rules for cosine:\[ \frac{dr}{d\theta} = -3 \sin \theta. \]

Step 2: Compute x and y in Cartesian Coordinates

Convert the polar coordinates to Cartesian using the formulas:* \( x = r \cos \theta \)* \( y = r \sin \theta \)Substitute \( r = 3 \cos \theta \) into these equations:* \( x = (3 \cos \theta) \cos \theta = 3 \cos^2 \theta \)* \( y = (3 \cos \theta) \sin \theta = 3 \cos \theta \sin \theta \)

Step 3: Differentiate x and y with respect to θ

To find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), differentiate the Cartesian coordinates:* For \( x = 3 \cos^2 \theta \): \[ \frac{dx}{d\theta} = 6 \cos \theta (-\sin \theta) = -6 \cos \theta \sin \theta \]* For \( y = 3 \cos \theta \sin \theta \): \[ \frac{dy}{d\theta} = 3 [\cos \theta \cos \theta + (-\sin \theta)(\sin \theta)] = 3 \cos^2 \theta - 3 \sin^2 \theta \]

Step 4: Apply the Formula for Slope of the Tangent Line

The slope of the tangent line \( \frac{dy}{dx} \) in terms of \( \theta \) is given by:\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \]Substitute \( \frac{dy}{d\theta} = 3 \cos^2 \theta - 3 \sin^2 \theta \) and \( \frac{dx}{d\theta} = -6 \cos \theta \sin \theta \):\[ \frac{dy}{dx} = \frac{3 \cos^2 \theta - 3 \sin^2 \theta}{-6 \cos \theta \sin \theta} \]

Step 5: Substitute the Given θ Value

Once the derivative in terms of \( \theta \) is established, we substitute \( \theta = \frac{\pi}{3} \) into the equation:* \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \)* \( \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \)Substituting these into the expression:\[ \frac{dy}{dx} = \frac{3 \left( \frac{1}{2} \right)^2 - 3 \left( \frac{\sqrt{3}}{2} \right)^2}{-6 \left( \frac{1}{2} \right) \left( \frac{\sqrt{3}}{2} \right)} = \frac{3/4 - 3(3/4)}{-6 \times (1/2) \times (\sqrt{3}/2)} = \frac{3/4 - 9/4}{-3\sqrt{3}/2} \]

Step 6: Calculate the Final Slope Value

Compute the value to find:\[ \frac{dy}{dx} = \frac{-6/4}{-3\sqrt{3}/2} = \frac{-3/2}{-3\sqrt{3}/2} = \frac{1}{\sqrt{3}} \]Simplifying, we rationalize the denominator:\[ \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3} \] which is the slope of the tangent line at the given \( \theta = \frac{\pi}{3} \).

Key Concepts

Tangent Line

When dealing with polar coordinates, the concept of a tangent line involves finding the slope of a line that just touches the curve at a given point. Just like in Cartesian coordinates, where a tangent line gives you the slope of a curve at a specific point, in polar coordinates, you can translate this idea using differentiation and conversion to Cartesian coordinates.
To determine the slope of the tangent line to a polar curve, you need to first represent the polar equation in terms of Cartesian coordinates. The reason is that slopes are more naturally found using Cartesian coordinates, since slopes relate horizontal and vertical changes.
This means converting the polar coordinates \( r \) and \( \theta \) to \( x \) and \( y \) using the formulas:

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

Once you have \( x \) and \( y \), you differentiate them with respect to \( \theta \) to find \( \frac{dy}{dx} \). This quotient gives us the slope of the tangent line at a particular point \( \theta \) on the curve.

Differentiation

Differentiation is a fundamental tool in calculus used to find rates of change. In the context of polar coordinates, you often need to differentiate functions to understand how one variable affects another. For example, if you have a polar equation for a curve, such as \( r = 3 \cos \theta \), you can differentiate it to examine how \( r \) changes with \( \theta \).
To differentiate \( r = 3 \cos \theta \), apply the standard rule for differentiating cosine functions. The derivative becomes \( \frac{dr}{d\theta} = -3 \sin \theta \), indicating how the radius changes as the angle increases.
Once you have these derivatives, converting to Cartesian coordinates allows you to leverage differentiation again. Specifically, find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) after substituting \( r=3 \cos \theta \) into \( x \) and \( y \). Differentiation here helps shadow the polar complexity into more straightforward Cartesian relationships. This serves as a bridge to calculate the final desired slope \( \frac{dy}{dx} \).
In summary, differentiation allows us to unlock the behavior of curves, offering insights into tangential motions and changes in any coordinate system.

Cartesian Coordinates

Cartesian coordinates are a familiar system for graphing mathematical problems, aiding in visualization and making calculations straightforward. Curves can initially be defined in other systems, such as polar coordinates, which use the radius and angle \( \theta \). However, to analyze these curves, especially for understanding slopes and tangents, you often convert to Cartesian coordinates.
In Cartesian systems, coordinates are given by \( x \) and \( y \), with formulas to convert from polar:

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

This conversion is crucial when investigating curves defined in polar form. Cartesian equations paint the line with x and y axes, helping calculate crucial aspects like the slope of a tangent using techniques such as differentiation.
Upon transforming to Cartesian coordinates, further differentiation provides rates of change (i.e., \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \)). You then use these rates to calculate the slope \( \frac{dy}{dx} \) needed for the tangent line.
This interplay between coordinate systems can initially seem complex, but each system's strength becomes apparent as they simplify the process of calculating important properties of curves and shapes.