Question

For the following exercises, find the slope of a tangent line to a polar curve \(r=f(\theta) .\) Let \(x=r \cos \theta=f(\theta) \cos \theta\) and \(y=r \sin \theta=f(\theta) \sin \theta\), so the polar equation \(r=f(\theta)\) is now written in parametric form.\(r=4+\sin \theta ;\left(3, \frac{3 \pi}{2}\right)\)

Short answer

The slope of the tangent line is 0.

Step-by-step solution

Understand the given polar curve

We start with the polar equation given as \( r = 4 + \sin \theta \). This equation defines the polar curve whose tangent slope we need to find.

Convert to parametric equations

Using the relations \( x = r \cos \theta \) and \( y = r \sin \theta \), we convert the polar equation into parametric form. Substitute \( r = 4 + \sin \theta \) to get:\[ x = (4 + \sin \theta) \cos \theta \]\[ y = (4 + \sin \theta) \sin \theta \]

Find derivatives dx/dθ and dy/dθ

Differentiate the parametric equations with respect to \( \theta \). For \( x \):\[ \frac{dx}{d\theta} = \frac{d}{d\theta} \left((4 + \sin \theta) \cos \theta \right) \]\[ = \cos \theta \cdot \frac{d}{d\theta} (4 + \sin \theta) + (4 + \sin \theta) \cdot \frac{d}{d\theta} (\cos \theta) \]\[ = \cos \theta \cdot \cos \theta - (4 + \sin \theta) \sin \theta \]\[ = \cos^2 \theta - (4 \sin \theta + \sin^2 \theta) \]For \( y \):\[ \frac{dy}{d\theta} = \frac{d}{d\theta} \left((4 + \sin \theta) \sin \theta \right) \]\[ = \sin \theta \cdot \cos \theta + (4 + \sin \theta) \cdot \cos \theta \]\[ = 4 \cos \theta + 2 \sin \theta \cos \theta \]

Calculate tangent slope

The tangent line slope is given by \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \). Plug in the derivatives from Step 3:\[ \frac{dy}{dx} = \frac{4 \cos \theta + 2 \sin \theta \cos \theta}{\cos^2 \theta - (4 \sin \theta + \sin^2 \theta)} \]

Evaluate at the given point (3, \(\frac{3\pi}{2}\))

The polar point \( \left(3, \frac{3\pi}{2}\right) \) corresponds to the angle \( \theta = \frac{3\pi}{2} \). Substitute this \( \theta \) value into the slope expression:Substitute \( \cos(\frac{3\pi}{2}) = 0 \), \( \sin(\frac{3\pi}{2}) = -1 \):\[ \frac{dy}{dx} = \frac{4 \cdot 0 + 2 \cdot (-1) \cdot 0}{0^2 - (4 \cdot (-1) + (-1)^2)} \]\[ = \frac{0}{-3 - 4} = 0 \]

Conclusion: Find the slope

The slope of the tangent line to the polar curve at the given point \( \left(3, \frac{3\pi}{2}\right) \) is 0. This means the tangent at that point is a horizontal line.

Key concepts

Polar Coordinates

Polar coordinates offer an alternative way of describing points on a plane compared to the more familiar Cartesian coordinates. In polar coordinates, each point is defined by an angle, \(\theta\), and a radius, \(r\). This is particularly useful when working with curves that are naturally circular or spiral in form. For instance, the polar equation \(r = f(\theta)\) represents a curve by describing how the distance from the origin (radius) changes as you sweep around the circle (angle).
A key advantage of using polar coordinates over Cartesian coordinates is their ability to simplify the expression of curves that have symmetry about the origin. This is especially visible in our exercise, where the polar equation \(r = 4 + \sin \theta\) provides a straightforward way to understand the curve's form and behavior.

Parametric Equations

Parametric equations are a way to represent a curve by expressing the coordinates of its points as functions of a variable, often denoted as \(t\) or \(\theta\) in the case of polar coordinates. In this exercise, converting polar coordinates to parametric equations involves creating two separate equations, one for \(x\) and one for \(y\).
Specifically, from \(r = f(\theta)\), we set \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). For the given curve \(r = 4 + \sin \theta\), the parametric form becomes \(x = (4 + \sin \theta) \cos \theta\) and \(y = (4 + \sin \theta) \sin \theta\).
The importance of parametric equations lies in their ability to represent complex curves that are difficult to describe using only the standard \(y = f(x)\) form. They also facilitate the process of finding derivatives and analyzing the behavior of curves.

Derivatives

Derivatives are fundamental in calculus as they provide information about the rate of change of functions. In the context of this exercise, deriving \(x\) and \(y\) with respect to \(\theta\) tells us how these coordinates change as we move along the curve.
We differentiate the parametric equations to obtain \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\). For our equations, the derivatives are:

• \(\frac{dx}{d\theta} = \cos^2 \theta - (4 \sin \theta + \sin^2 \theta)\)
• \(\frac{dy}{d\theta} = 4 \cos \theta + 2 \sin \theta \cos \theta\)

Understanding these derivatives is crucial for analyzing the curve's behavior, particularly when finding the tangent line's slope.

Tangent Slope Calculation

The slope of a tangent line to a curve at a particular point is a measure of the curve's steepness and direction at that point. In polar coordinates, finding this slope involves using the derivatives of the parametric equations.
The formula \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\) gives the slope of the tangent line. In this exercise, substituting the calculated derivatives provides the slope expression:

• \(\frac{dy}{dx} = \frac{4 \cos \theta + 2 \sin \theta \cos \theta}{\cos^2 \theta - (4 \sin \theta + \sin^2 \theta)}\)

At the specific point \((3, \frac{3\pi}{2})\), we substitute \(\theta = \frac{3\pi}{2}\) (where \(\cos(\frac{3\pi}{2}) = 0\) and \(\sin(\frac{3\pi}{2}) = -1\)) into this formula. This computation reveals that the slope is 0, indicating the tangent at this point is a horizontal line.
Understanding how to calculate the tangent slope is essential for tasks like sketching curves and analyzing their geometric properties.