Chapter 1: Problem 236
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. $$ r=1-\sin \theta ;\left(\frac{1}{2}, \frac{\pi}{6}\right) $$
Short Answer
Step by step solution
Identify the Problem
Convert Polar to Parametric Equations
Compute Derivatives of Parametric Equations
Apply the Slope Formula for Parametric Curves
Evaluate the Slope at \(\theta = \frac{\pi}{6}\)
Simplify the Slope Expression
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
- \( x = r \cos \theta = (1 - \sin \theta) \cos \theta \)
- \( y = r \sin \theta = (1 - \sin \theta) \sin \theta \)
Slope of Tangent Line
- \( \frac{dx}{d\theta} = -\cos^2 \frac{\pi}{6} + \sin \frac{\pi}{6} \cdot \cos \frac{\pi}{6} = -\frac{3}{4} + \frac{\sqrt{3}}{4} \)
- \( \frac{dy}{d\theta} = \sin \frac{\pi}{6} \cdot \cos \frac{\pi}{6} - \sin^2 \frac{\pi}{6} = \frac{\sqrt{3}}{4} - \frac{1}{4} \)
Derivatives
- \( \frac{dx}{d\theta} \) 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 \( \sin \theta \cdot \cos \theta - \sin^2 \theta \).