Question

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

Short answer

The slope is \(\frac{1}{\sqrt{3}}\).

Step-by-step solution

Convert Polar to Cartesian Coordinates

Given the polar equation \(r = 3 \cos \theta\), we need to express this in Cartesian coordinates. We know that \(x = r \cos \theta\) and \(y = r \sin \theta\). We can substitute \(r = 3 \cos \theta\) into these equations to obtain Cartesian expressions: \(x = 3 \cos \theta \cdot \cos \theta = 3 \cos^2 \theta\) and \(y = 3 \cos \theta \cdot \sin \theta = 3 \cos \theta \sin \theta\).

Find Derivatives with respect to \(\theta\)

To find the slope, we first find the derivatives of \(x\) and \(y\) with respect to \(\theta\). The derivatives are \(\frac{dx}{d\theta} = \frac{d}{d\theta}(3 \cos^2 \theta) = -6 \cos \theta \sin \theta\) and \(\frac{dy}{d\theta} = \frac{d}{d\theta}(3 \cos \theta \sin \theta) = 3 \cos^2 \theta - 3 \sin^2 \theta\).

Use Derivatives to Find Slope of Tangent

The slope \(m\) of the tangent line is given by \(\frac{dy}{dx}\). We use the chain rule: \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3 \cos^2 \theta - 3 \sin^2 \theta}{-6 \cos \theta \sin \theta}\).

Substitute the Given \(\theta\) Value

Substitute \(\theta = \frac{\pi}{3}\) into the expression for the slope \(\frac{dy}{dx}\). Compute \(\cos \frac{\pi}{3} = \frac{1}{2}\) and \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\), resulting in \(\frac{dy}{dx} = \frac{3 \left(\frac{1}{2}\right)^2 - 3 \left(\frac{\sqrt{3}}{2}\right)^2}{-6 \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right)}\).

Simplify to Find the Slope

Simplify the expression from Step 4: \(\frac{dy}{dx} = \frac{3 \cdot \frac{1}{4} - 3 \cdot \frac{3}{4}}{-6 \cdot \frac{\sqrt{3}}{4}} = \frac{\frac{3}{4} - \frac{9}{4}}{-\frac{3\sqrt{3}}{2}}\). This simplifies to \(\frac{-\frac{6}{4}}{-\frac{3\sqrt{3}}{2}} = \frac{\frac{-3}{2}}{-\frac{3\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}\).

Finalize the Slope

Thus, the slope of the tangent line at \(\theta = \frac{\pi}{3}\) is \(\frac{1}{\sqrt{3}}\).

Key concepts

Polar to Cartesian Conversion

Converting from polar to Cartesian coordinates is a crucial skill when working with polar curves. In polar coordinates, a point is defined by a radius, r, and an angle, \(\theta\). However, in Cartesian coordinates, a point is defined by the coordinates \( (x, y) \). The conversion formulas are quite simple:

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

In the given problem, the polar equation is \( r = 3 \cos \theta \). By substituting this expression for \( r \) into the conversion formulas, we find:

• \( x = 3 \cos \theta \cdot \cos \theta = 3 \cos^2 \theta \)
• \( y = 3 \cos \theta \cdot \sin \theta = 3 \cos \theta \sin \theta \)

Now, the polar curve is effectively expressed in terms of \( x \) and \( y \), making it easier to apply calculus operations to find derivatives.

Derivative Calculation

Calculating derivatives with respect to \(\theta\) allows us to determine how \(x\) and \(y\) change as the angle changes along the curve. This is a vital step in finding the slope of the tangent line. To find these derivatives, we differentiate the Cartesian expressions with respect to \(\theta\):

• For \( x = 3 \cos^2 \theta \), use the derivative formula for \(\cos^2 \theta\): \(\frac{dx}{d\theta} = \frac{d}{d\theta}(3 \cos^2 \theta) = -6 \cos \theta \sin \theta \).
• For \( y = 3 \cos \theta \sin \theta \), apply the product rule: \(\frac{dy}{d\theta} = \frac{d}{d\theta}(3 \cos \theta \sin \theta) = 3 \cos^2 \theta - 3 \sin^2 \theta \).

These derivatives show how quickly the coordinates change with respect to changes in \(\theta\), providing the necessary information to find the slope of the tangent.

Chain Rule

The chain rule in calculus is a fundamental technique used to compute the derivative of a function with respect to one variable by relating it to derivatives with respect to another variable. When dealing with polar coordinates converted to Cartesian, we use the chain rule to find the slope \(\frac{dy}{dx}\), which involves both \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\):

\[\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3 \cos^2 \theta - 3 \sin^2 \theta}{-6 \cos \theta \sin \theta}\]

This expression represents how the \(y\) coordinate changes with respect to the \(x\) coordinate as we vary \(\theta\). Through careful application of the chain rule, we translate the radial changes into a familiar Cartesian form, allowing for straightforward calculation of tangent slopes.

Trigonometric Functions

In the solution, trigonometric functions \(\cos \theta\) and \(\sin \theta\) play a crucial role. Understanding their values at key angles is essential for solving such problems.


When substituting \(\theta = \frac{\pi}{3}\):

• \(\cos \frac{\pi}{3} = \frac{1}{2}\)
• \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\)

These trigonometric values are then substituted into the derivative expression from earlier:

\[\frac{dy}{dx} = \frac{3 \left(\frac{1}{2}\right)^2 - 3 \left(\frac{\sqrt{3}}{2}\right)^2}{-6 \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right)} = \frac{1}{\sqrt{3}}\]

Understanding how these functions behave at various angles enables us to calculate expressions easily, such as the slope of the tangent line at specific points on the curve.