Problem 238 - Calculus Volume 3

Question

Find the slope of a tangent line to a polar curve \(r=f(\theta)\). Let \(x=r \cos \theta=f(\theta) \cos \theta\) and \(y=r \sin \theta=f(\theta) \sin \theta, \quad\) so the polar equation \(r=f(\theta)\) is now written in parametric form. $$ r=8 \sin \theta ;\left(4, \frac{5 \pi}{6}\right) $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We are given a polar curve defined by the equation \( r = 8\sin\theta \). We need to find the slope of the tangent line at the specific point \( \left(4, \frac{5\pi}{6}\right) \). The polar coordinates have \( r = 4 \) and \( \theta = \frac{5\pi}{6} \).

Step 2: Convert Polar to Parametric Equations

The polar equation \( r = 8\sin\theta \) can be converted to parametric equations using the relations \(x = r\cos\theta\) and \(y = r\sin\theta\). Therefore, we have: \[ x = 8\sin\theta\cos\theta \]\[ y = 8\sin^2\theta \]

Step 3: Find Derivatives of Parametric Equations

To find the slope of the tangent line, compute \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \):- \( \frac{dx}{d\theta} = \frac{d}{d\theta}(8 \sin\theta \cos\theta) = 8(\cos^2\theta - \sin^2\theta) \).- \( \frac{dy}{d\theta} = \frac{d}{d\theta}(8 \sin^2\theta) = 16 \sin\theta \cos\theta \).

Step 4: Use Chain Rule to Find \( \frac{dy}{dx} \)

The slope of the tangent line is given by:\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{16 \sin\theta \cos\theta}{8 (\cos^2\theta - \sin^2\theta)} = \frac{2 \sin\theta \cos\theta}{\cos^2\theta - \sin^2\theta} \].

Step 5: Substitute the Given Point

Substitute \( \theta = \frac{5\pi}{6} \) into the expression for \( \frac{dy}{dx} \):- \( \sin \frac{5\pi}{6} = \frac{1}{2} \) and \( \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} \).- Substitute these values into the expression:\[ \frac{dy}{dx} = \frac{2 \times \frac{1}{2} \times -\frac{\sqrt{3}}{2}}{(-\frac{\sqrt{3}}{2})^2 - (\frac{1}{2})^2} = \frac{-\sqrt{3}/2}{3/4 - 1/4} = \frac{-\sqrt{3}}{2} \].

Step 6: Calculate the Slope

Simplify the expression to find the slope:\[ \frac{dy}{dx} = -\sqrt{3} \].

Key Concepts

Tangent Line Slope

Understanding the slope of a tangent line for a curve is a fundamental concept in calculus. It essentially describes how steep the line is at a particular point on the curve. This slope can help us understand the behavior and direction of the curve at that specific location.

For a polar curve like the one in the problem, calculating the slope isn't as straightforward as it is with Cartesian coordinates. This is because polar coordinates describe a point based on its distance from the origin and the angle from a reference direction, which makes the calculations a bit more involved.

When we want to find the slope of a tangent line to a polar curve, we convert polar equations to parametric ones, allowing us to use derivative techniques that are familiar from Cartesian coordinates. Ultimately, the slope of the tangent line is derived by combining derivatives of these parametric equations.

Parametric Equations

Parametric equations provide a way to express a mathematical relationship by using parameters, making it an invaluable tool for various kinds of problems, such as plotting curves. In our exercise, the polar equation is converted to parametric form based on trigonometric identities.

- For polar curves, this involves using the relationships:

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

These allow us to reformulate the polar equation into a form that corresponds to Cartesian coordinates with respect to parameter \( \theta \).

In the given problem, the conversion results in:

• \( x = 8 \sin \theta \cos \theta \)
• \( y = 8 \sin^2 \theta \)

This transformation facilitates derivative calculations and the eventual determination of the tangent line slope at any point on the curve.

Derivative Calculations

Calculating derivatives of parametric equations derived from polar coordinates is essential for analyzing curves and their slopes. The derivative \( \frac{dy}{dx} \), needed for determining tangent lines, arises through a series of steps using the chain rule from calculus.

Here's how we do it:

• First, find \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \) based on the parametric equations.
• Then, calculate \( \frac{dy}{dx} \) by dividing \( \frac{dy}{d\theta} \) by \( \frac{dx}{d\theta} \).

This process transforms our need for the slope of the tangent line into a more manageable calculation using derivatives.

In our exercise, the derivatives were calculated as follows:

• \( \frac{dx}{d\theta} = 8(\cos^2 \theta - \sin^2 \theta) \)
• \( \frac{dy}{d\theta} = 16 \sin \theta \cos \theta \)

The division leads to the expression \( \frac{dy}{dx} = \frac{2 \sin \theta \cos \theta}{\cos^2 \theta - \sin^2 \theta} \), and substituting specific values of \( \theta \) determines the slope at that point on the curve.