Problem 60 - Calculus Early Transcendentals

Question

\(57-62\) Find the slope of the tangent line to the given polar curve at the point specified by the value of \(\theta\) . $$r=\cos (\theta / 3), \quad \theta=\pi$$

Step-by-Step Solution

Answer

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

Step 1: Understand the Polar Equation

The polar curve given is \(r = \cos(\theta / 3)\). This means that for a given angle \(\theta\), the radius \(r\) is determined by the cosine of a third of that angle.

Step 2: Convert to Cartesian Coordinates

Use the formulas \(x = r \cdot \cos(\theta)\) and \(y = r \cdot \sin(\theta)\) to convert the polar equation to Cartesian coordinates. Here \(r\) is \(\cos(\theta / 3)\). For \(\theta = \pi\), \(r = \cos(\pi / 3) = 1/2\). Thus, \(x = \frac{1}{2} \cdot (-1) = -\frac{1}{2}\) and \(y = \frac{1}{2} \cdot 0 = 0\).

Step 3: Differentiate to Find Slopes

To find the slope of the tangent line in polar coordinates, calculate \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\). Using the chain rule, this requires differentiating \(x(\theta) = r(\theta) \cdot \cos(\theta)\) and \(y(\theta) = r(\theta) \cdot \sin(\theta)\).

Step 4: Calculate \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \)

First, differentiate \(x(\theta) = \cos(\theta/3) \cdot \cos(\theta)\) and \(y(\theta) = \cos(\theta/3) \cdot \sin(\theta)\) with respect to \(\theta\). For \(y\), use the product rule:\[ \frac{dy}{d\theta} = \frac{d}{d\theta}(\cos(\theta/3)) \cdot \sin(\theta) + \cos(\theta/3) \cdot \frac{d}{d\theta}(\sin(\theta)) \]\[= -\frac{1}{3}\sin(\theta/3)\sin(\theta) + \cos(\theta/3)\cos(\theta)\]Similarly for \(x\):\[\frac{dx}{d\theta} = \frac{d}{d\theta}(\cos(\theta/3)) \cdot \cos(\theta) + \cos(\theta/3) \cdot \frac{d}{d\theta}(\cos(\theta)) \]\[= -\frac{1}{3}\sin(\theta/3)\cos(\theta) - \cos(\theta/3)\sin(\theta)\]

Step 5: Evaluate Derivatives at \( \theta = \pi \)

Substitute \(\theta = \pi\) into \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\):\[ \frac{dy}{d\theta} = -\frac{1}{3}\sin(\pi/3)\cdot 0 + \cos(\pi/3)\cdot (-1) = -\frac{1}{2} \]\[ \frac{dx}{d\theta} = -\frac{1}{3}\sin(\pi/3)\cdot(-1) - \cos(\pi/3)\cdot 0 = \frac{\sqrt{3}}{6} \]

Step 6: Find the Slope of the Tangent Line

The slope of the tangent line to the polar curve at \(\theta = \pi\) is given by \( \frac{dy/d\theta}{dx/d\theta} \).Substitute the derivatives found in Step 5:\[ \text{slope} = \frac{-1/2}{\sqrt{3}/6} = -3/\sqrt{3} = -\sqrt{3} \].

Step 7: Simplify the Final Slope

Simplify the expression for the slope:\[ \frac{-3}{\sqrt{3}} = -\sqrt{3} \]This is the slope of the tangent line at the point where \(\theta = \pi\).

Key Concepts

Polar Coordinates

Polar coordinates are a way of representing points in a plane using a distance and an angle. Instead of the standard Cartesian system which uses

• x-coordinates (horizontal distance from the origin)
• y-coordinates (vertical distance from the origin)

polar coordinates use:

• r for the radial distance from the origin
• \( \theta \) for the angle measured from the positive x-axis

This system is particularly useful for circular and spiral structures, where describing points in Cartesian coordinates is more complex. To understand our exercise fully, imagine the angle \( \theta \) spinning around from the positive x-axis, and the radius \( r \) extending or contracting based on the function \( r = \cos(\theta/3) \). It describes various positions of the point in relation to the origin.

Derivatives

In mathematics, a derivative represents the rate of change of a function with respect to a variable. When dealing with polar coordinates, finding the slope of the tangent line means calculating derivatives. Here we're interested in finding the derivative of the path (or curve) of the polar function with respect to \( \theta \). This involves differentiating both the x and y components viewed as functions of \( \theta \).

For the curve \( r = \cos(\theta/3) \), we convert it into two separate functions:

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

By differentiating these with respect to \( \theta \), we obtain expressions for \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), which are vital in calculating the slope of the tangent at any given point.

Product Rule

The product rule is a crucial differentiation technique used when differentiating the product of two functions. If you have two functions u(θ) and v(θ), their derivative is given by:
\[ \frac{d}{d\theta} [u(\theta) \, v(\theta)] = u'(\theta) \, v(\theta) + u(\theta) \, v'(\theta) \]
In our exercise, this rule is applied to differentiate the Cartesian form of the curve. For the y-component, \( y(\theta) = \cos(\theta/3) \cdot \sin(\theta) \), apply the product rule:

• Derive \( u'(\theta) = \frac{d}{d\theta}(\cos(\theta/3)) \)
• Derive \( v'(\theta) = \frac{d}{d\theta}(\sin(\theta)) \)
• Apply the product rule for complete differentiation

Using this process, you can separately derive the x and y-component derivatives needed to find the overall slope.

Polar to Cartesian Conversion

Converting from polar to Cartesian coordinates simplifies the analysis of curves, especially when dealing with derivatives. The conversion is done using the formulas:

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

To apply this conversion, plug the polar equations directly in. For our example, with \( r = \cos(\theta/3) \), we calculate \( x \) and \( y \) given specific angle values, like \( \theta = \pi \). This method translates the abstract mathematical function into physical coordinates, making it easier to perform further analysis like finding the tangent slope. The results are then tangible points in the Cartesian plane, ready for derivative application.