Problem 63 - Calculus Early Transcendentals

Question

\(63-68\) Find the points on the given curve where the tangent line is horizontal or vertical. $$r=3 \cos \theta$$

Step-by-Step Solution

Answer

Horizontal tangents at \(( \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}})\) and \((-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}})\). Vertical at \((3, 0)\), \((0, 0)\), and \((-3, 0)\).

Step 1: Convert to Cartesian Coordinates

Use the polar-to-Cartesian coordinate conversions. For the given polar equation \( r = 3 \cos \theta \), we have \( x = r \cos \theta = 3 \cos^2 \theta \) and \( y = r \sin \theta = 3 \cos \theta \sin \theta \).

Step 2: Expression for Tangent Slope

To find where the tangent line is horizontal or vertical, compute the derivative \( \frac{dy}{dx} \). By using the chain rule, this is determined by \( \frac{dy/d\theta}{dx/d\theta} \). Compute \( \frac{dy}{d\theta} = 3 \cos(2\theta) \) and \( \frac{dx}{d\theta} = -6 \cos \theta \sin \theta \).

Step 3: Horizontal Tangent Condition

For the tangent line to be horizontal, \( \frac{dy}{dx} = 0\). Set \( 3 \cos(2\theta) = 0 \) which results in \( \theta = \frac{\pi}{4}, \frac{3\pi}{4} \). Substitute these values into the original equation \( r = 3 \cos \theta \) to find the corresponding points: \(( \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}) \) and \( (-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}) \).

Step 4: Vertical Tangent Condition

For the tangent line to be vertical, \( \frac{dx}{d\theta} = 0\). Set \( -6 \cos \theta \sin \theta = 0\), which occurs at \( \theta = 0, \frac{\pi}{2}, \pi \). Substitute these values into \( r = 3 \cos \theta \) to find the corresponding points: \((3, 0)\), \((0, 0)\), and \((-3, 0)\).

Key Concepts

Tangent Line

A tangent line to a curve at a given point is a straight line that "just touches" the curve at that point without crossing it. This can be visualized as the line that best represents the curve's direction at a specific location.
In polar coordinates, a tangent line plays the same role as in Cartesian coordinates, but it's derived using polar-specific calculations. The slope of the tangent line can be found by differentiating the equation representing the curve, which gives us the instantaneous rate of change at that specific point.
In this exercise, we find tangent lines by converting polar equations to Cartesian to compute derivatives, enabling us to easily identify where these lines are horizontal or vertical.

Horizontal Tangent

A horizontal tangent line is a special case where the tangent line is parallel to the horizontal axis. For such lines, the slope is zero, indicating no vertical change as you move along the line.
To determine where this occurs on a curve defined by polar coordinates, we set the derivative \( \frac{dy}{dx} \) to zero. In our exercise, we derived the expression for the slope and set it to zero, leading us to specific angles \( \theta \) where this condition is met. By substituting these angles into the original equation, we find specific points on the curve where horizontal tangent lines exist. Points found in this process include \((\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}})\) and \((-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} )\).
These points represent locations where the curve "flattens out" momentarily.

Vertical Tangent

A vertical tangent line occurs at the point where the tangent line becomes perfectly vertical. This means the slope is infinitely steep, represented by an undefined slope in mathematical terms.
To find where the vertical tangent lines occur, we set \( \frac{dx}{d\theta} = 0 \) instead of \( \frac{dy}{dx} = 0 \).
In our problem, we solve the equation \(-6 \cos \theta \sin \theta = 0\). This solution gives specific angle values \( \theta = 0, \frac{\pi}{2}, \pi \).
Substituting these angles into the polar equation \( r = 3 \cos \theta \), the corresponding points are \((3, 0)\), \((0, 0)\), and \((-3, 0)\), indicating the exact locations where the curve exhibits vertical tangents.

Polar to Cartesian Conversion

Converting between polar and Cartesian coordinates is a critical step when dealing with curves described in polar form. This conversion is necessary to easily compute derivatives and slopes.
The basic formulas for conversion are:

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

For the given equation \( r = 3 \cos \theta \), we utilized these conversion techniques to convert into Cartesian coordinates, resulting in \(x = 3 \cos^2 \theta\) and \(y = 3 \cos \theta \sin \theta\).
By expressing the polar equation in terms of \(x\) and \(y\), we can use calculus-based strategies typical to Cartesian coordinates, allowing us to uncover the behavior of tangent lines on the curve.