Problem 236 - 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. $$ r=1-\sin \theta ;\left(\frac{1}{2}, \frac{\pi}{6}\right) $$

Step-by-Step Solution

Answer

The slope of the tangent line at \(\theta = \frac{\pi}{6}\) is \(\frac{\sqrt{3} - 1}{\sqrt{3} - 3}\).

Step 1: Identify the Problem

We need to find the slope of the tangent line to the polar curve given by the equation \( r = 1 - \sin \theta \) at the point \( \left( \frac{1}{2}, \frac{\pi}{6} \right) \).

Step 2: Convert Polar to Parametric Equations

The parametric equations are given by:\[ x = r \cos \theta = (1 - \sin \theta) \cos \theta \]\[ y = r \sin \theta = (1 - \sin \theta) \sin \theta \].

Step 3: Compute Derivatives of Parametric Equations

Find the derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\): \[ \frac{dx}{d\theta} = \frac{d}{d\theta}[(1 - \sin \theta) \cos \theta] = -\cos^2 \theta + \sin \theta \cdot \cos \theta \]\[ \frac{dy}{d\theta} = \frac{d}{d\theta}[(1 - \sin \theta) \sin \theta] = \sin \theta \cdot \cos \theta - \sin^2 \theta \].

Step 4: Apply the Slope Formula for Parametric Curves

The slope \(m\) of the tangent line to a parametric curve is given by \( m = \frac{dy/d\theta}{dx/d\theta} \).

Step 5: Evaluate the Slope at \(\theta = \frac{\pi}{6}\)

Substitute \(\theta = \frac{\pi}{6}\) into \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) and find the slope:For \(\frac{dx}{d\theta}\):\[ \frac{dx}{d\theta} = -\cos^2 \frac{\pi}{6} + \sin \frac{\pi}{6} \cdot \cos \frac{\pi}{6} = -\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = -\frac{3}{4} + \frac{\sqrt{3}}{4} \]For \(\frac{dy}{d\theta}\):\[ \frac{dy}{d\theta} = \sin \frac{\pi}{6} \cdot \cos \frac{\pi}{6} - \sin^2 \frac{\pi}{6} = \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right)^2 = \frac{\sqrt{3}}{4} - \frac{1}{4} \]Thus, the slope is:\[ m = \frac{\frac{\sqrt{3}}{4} - \frac{1}{4}}{-\frac{3}{4} + \frac{\sqrt{3}}{4}} \].

Step 6: Simplify the Slope Expression

Simplify the expression:\[ m = \frac{\frac{\sqrt{3} - 1}{4}}{\frac{\sqrt{3} - 3}{4}} = \frac{\sqrt{3} - 1}{\sqrt{3} - 3} \].

Key Concepts

Parametric Equations

When dealing with polar coordinates, equations can often be converted into parametric form for easier analysis. This conversion involves transforming the polar equation, which is expressed in terms of the radius and angle \( r = f(\theta) \), into a set of parametric equations in terms of \( x \) and \( y \). For example, if \( r = 1 - \sin \theta \), the corresponding parametric equations are derived by substituting \( r \) with its terms:

• \( x = r \cos \theta = (1 - \sin \theta) \cos \theta \)
• \( y = r \sin \theta = (1 - \sin \theta) \sin \theta \)

This powerful technique allows us to examine the curve's behavior using calculus by considering the variables \( x \) and \( y \) that depend on \( \theta \). Parametric equations are essential when we need to compute properties of curves, such as tangent lines, in different coordinate systems.

Slope of Tangent Line

The slope of a tangent line gives us the rate of change of a curve, which is crucial in understanding the curve's behavior at a particular point. For curves described with parametric equations, the slope of the tangent line can be found using the formula: \[ m = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \] Here, \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \) are the derivatives of \( y \) and \( x \) with respect to \( \theta \). For the given polar curve \( r = 1 - \sin \theta \), at the point \( (\frac{1}{2}, \frac{\pi}{6}) \), the derivatives are evaluated to provide the required slope:

• \( \frac{dx}{d\theta} = -\cos^2 \frac{\pi}{6} + \sin \frac{\pi}{6} \cdot \cos \frac{\pi}{6} = -\frac{3}{4} + \frac{\sqrt{3}}{4} \)
• \( \frac{dy}{d\theta} = \sin \frac{\pi}{6} \cdot \cos \frac{\pi}{6} - \sin^2 \frac{\pi}{6} = \frac{\sqrt{3}}{4} - \frac{1}{4} \)

Calculating the slope involves more than just math; it's about understanding how changes in \( \theta \) affect the direction and steepness of the curve at that exact point.

Derivatives

Derivatives are a fundamental tool in calculus that help us understand how functions change. In the context of parametric equations derived from polar coordinates, derivatives with respect to \( \theta \) are used to examine the rates of change for both \( x \) and \( y \) with respect to \( \theta \), the angle component. For a polar equation \( r = 1 - \sin \theta \), the derivatives are calculated to understand how each coordinate changes:

• \( \frac{dx}{d\theta} \) is derived from \( x = (1 - \sin \theta) \cos \theta \), resulting in \( -\cos^2 \theta + \sin \theta \cdot \cos \theta \).
• \( \frac{dy}{d\theta} \) comes from \( y = (1 - \sin \theta) \sin \theta \), giving \( \sin \theta \cdot \cos \theta - \sin^2 \theta \).

These derivatives reveal not only how \( x \) and \( y \) change individually, but also how they collectively define the curve’s tangent in different directions of \( \theta \). Mastering derivatives in various coordinate systems facilitates advanced studies in motion, physics, and engineering.