Problem 61 - 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 \(\theta\) . $$r=\cos 2 \theta, \quad \theta=\pi / 4$$

Step-by-Step Solution

Answer

The slope of the tangent line is 1.

Step 1: Understand the Polar Curve

The equation of the polar curve given is \( r = \cos 2\theta \). We need to find the slope of the tangent line at \( \theta = \frac{\pi}{4} \). Understanding the meaning of \( r \) and \( \theta \) in the context of polar coordinates is crucial: \( r \) is the distance from the origin, and \( \theta \) is the angle from the positive x-axis.

Step 2: Convert Polar to Cartesian Coordinates

To find slopes, it is often easier in Cartesian coordinates. Use the formulas \( x = r \cos \theta \) and \( y = r \sin \theta \), where \( r = \cos 2\theta \). So: \[x = \cos 2\theta \cdot \cos \theta \]\[y = \cos 2\theta \cdot \sin \theta \]

Step 3: Differentiate with Respect to \( \theta \)

Find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \):- Differentiate \( x(\theta) = \cos 2\theta \cdot \cos \theta \) using the product and chain rule:\[ \frac{dx}{d\theta} = -2\sin 2\theta \cdot \cos \theta - \cos 2\theta \cdot \sin \theta \]- Differentiate \( y(\theta) = \cos 2\theta \cdot \sin \theta \):\[ \frac{dy}{d\theta} = -2\sin 2\theta \cdot \sin \theta + \cos 2\theta \cdot \cos \theta \]

Step 4: Find the Slope of the Tangent Line

The slope of the tangent line in polar coordinates is given by \( \frac{dy/d\theta}{dx/d\theta} \). So, calculate the following:\[\frac{dy/d\theta}{dx/d\theta} = \frac{-2\sin 2\theta \cdot \sin \theta + \cos 2\theta \cdot \cos \theta}{-2\sin 2\theta \cdot \cos \theta - \cos 2\theta \cdot \sin \theta}\]

Step 5: Substitute \( \theta = \frac{\pi}{4} \) and Simplify

Substitute \( \theta = \frac{\pi}{4} \) in the expressions:\( \sin \theta = \cos \theta = \frac{1}{\sqrt{2}} \), \(\sin 2\theta = 1\), and \(\cos 2\theta = 0\).Calculate:\[\frac{dy/d\theta}{dx/d\theta} = \frac{-2 \cdot 1 \cdot \frac{1}{\sqrt{2}} + 0 \cdot \frac{1}{\sqrt{2}}}{-2 \cdot 1 \cdot \frac{1}{\sqrt{2}} - 0 \cdot \frac{1}{\sqrt{2}}} = \frac{-\sqrt{2}}{-\sqrt{2}} = 1\]

Step 6: Conclusion

After simplifying, the slope of the tangent line to the polar curve \( r = \cos 2\theta \) at \( \theta = \frac{\pi}{4} \) is 1.

Key Concepts

Understanding Tangent Lines

A tangent line is a straight line that touches a curve at a single point without crossing it. This line shows the direction of the curve at that point, and its slope can tell us how steep the curve is there. In polar coordinates, where points are represented with an angle \( \theta \) and a radius \( r \), finding the tangent line involves converting these into Cartesian coordinates (\( x \) and \( y \)).

In simple terms, imagine you're tracing along a curve on a graph. The tangent line is what your finger would follow if the curve suddenly straightened out at a particular point. It's crucial for understanding the immediate rate of change of the curve. For the exercise provided, the goal was to find such a line for the polar curve \( r = \cos 2\theta \) at \( \theta = \frac{\pi}{4} \).

• The tangent's slope is a measure of how steep the line is.
• Finding the slope involves mathematical transformations and calculations.

Role of Differentiation

Differentiation is a mathematical process that helps us find how a function changes at any given point, which is essential for determining slopes of tangent lines. In the context of our exercise, differentiation allows us to calculate how the \( x \) and \( y \) positions change as \( \theta \) changes.

By going through the process of differentiation, you effectively dissect a curve to understand its behavior more deeply. It's like examining the DNA of the curve to see what makes it grow and change.

• For the exercise, differentiation was used to find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \).
• These derivatives were used to compute the slope of the tangent line.

Understanding Cartesian Coordinates

Cartesian coordinates are a system that uses two numbers, \( x \) and \( y \), to describe points on a flat surface. This contrasts with polar coordinates, which use distances from the origin and angles from the positive x-axis. Converting from polar to Cartesian coordinates can make some calculations, like finding slopes, easier.

In the context of this exercise, we used the formulas:
\[ x = r \cos \theta \]
\[ y = r \sin \theta \]
These help translate the circular movement described in polar coordinates into the linear grid of Cartesian coordinates. It's like switching from describing a circle's circumference (polar) to its width and height (Cartesian).

• Cartesian coordinates give a straightforward way to analyze changes.
• They help break down complex curves into linear components.

Applying the Chain Rule

The chain rule is a technique in calculus used to differentiate composite functions. It's especially useful when dealing with functions of functions, as it lets us navigate through layers of calculations. In this exercise, applying the chain rule was necessary for tackling expressions like \( \cos 2\theta \cdot \cos \theta \) and \( \cos 2\theta \cdot \sin \theta \), which are products of functions that themselves depend on \( \theta \).

Consider the chain rule as a way to peel the layers of an onion, with each layer representing a different function or operation. You handle one layer at a time, passing your findings to deeper layers till you find what's at the core.

• The chain rule allowed us to systematically find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \).
• It helped simplify complex expressions into manageable steps.