AP Calculus BC 9.7 Defining Polar Coordinates and Differentiating in Polar Form Study Notes - New Syllabus
AP Calculus BC 9.7 Defining Polar Coordinates and Differentiating in Polar Form Study Notes- New syllabus
AP Calculus BC 9.7 Defining Polar Coordinates and Differentiating in Polar Form Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Recognizing opportunities to apply derivative rules can simplify differentiation.
Key Concepts:
- Defining Polar Coordinates and Differentiating in Polar Form
Defining Polar Coordinates and Differentiating in Polar Form
Defining Polar Coordinates and Differentiating in Polar Form
Polar coordinates provide a way to describe points in the plane using a distance from the origin and an angle from the positive x-axis, instead of using Cartesian coordinates \((x, y)\).
- In polar form, a point \(P\) in the plane is represented as \((r, \theta)\):
- \(r\) = radial distance from the origin (can be positive or negative)
- \(\theta\) = angle measured counterclockwise from the positive x-axis (in radians or degrees)
- The relationship between Cartesian and polar coordinates:
- In polar form, a point \(P\) in the plane is represented as \((r, \theta)\):
$ x = r \cos\theta, \quad y = r \sin\theta $
- Also, $ r = \sqrt{x^{2} + y^{2}}, \quad \theta = \tan^{-1}\left(\dfrac{y}{x}\right) $
Polar Curves
A polar curve is given by an equation of the form \(r = f(\theta)\). As \(\theta\) varies, the function \(f(\theta)\) describes how far the point is from the origin.
Differentiation in Polar Form
For a polar curve \(r = f(\theta)\), the Cartesian coordinates are:
$ x(\theta) = r(\theta) \cos\theta, \quad y(\theta) = r(\theta) \sin\theta $
By the chain rule, the derivatives are:
$ \frac{dx}{d\theta} = r'(\theta) \cos\theta – r(\theta) \sin\theta $ $ \frac{dy}{d\theta} = r'(\theta) \sin\theta + r(\theta) \cos\theta $
The slope of the tangent line is:
$ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} $
Tangents Parallel or Perpendicular to Axes
- Tangent is horizontal when \(\dfrac{dy}{d\theta} = 0\) and \(\dfrac{dx}{d\theta} \neq 0\).
- Tangent is vertical when \(\dfrac{dx}{d\theta} = 0\) and \(\dfrac{dy}{d\theta} \neq 0\).
Example
Convert the point \((r, \theta) = (4, \pi/3)\) into Cartesian coordinates.
▶️ Answer/Explanation
Using the formulas:
$ x = r \cos\theta = 4 \cos\left(\frac{\pi}{3}\right) = 4 \cdot \frac{1}{2} = 2 $ $ y = r \sin\theta = 4 \sin\left(\frac{\pi}{3}\right) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} $
So the Cartesian coordinates are \((2, 2\sqrt{3})\).
Example
For the polar curve \(r = 2 + \sin\theta\), find \(\dfrac{dy}{dx}\) at \(\theta = \dfrac{\pi}{2}\).
▶️ Answer/Explanation
First, \(r'(\theta) = \cos\theta\).
At \(\theta = \pi/2\):
\(r = 2 + \sin\left(\frac{\pi}{2}\right) = 2 + 1 = 3\)
\(r’ = \cos\left(\frac{\pi}{2}\right) = 0\)
\(\frac{dx}{d\theta} = r’ \cos\theta – r \sin\theta = 0 \cdot 0 – 3 \cdot 1 = -3 $
\(\frac{dy}{d\theta} = r’ \sin\theta + r \cos\theta = 0 \cdot 1 + 3 \cdot 0 = 0 $
Slope: $ \frac{dy}{dx} = \frac{0}{-3} = 0 $
The tangent line is horizontal at \(\theta = \pi/2\).
Example
For the polar curve \(r = 1 – \cos\theta\), determine the values of \(\theta\) where the tangent is vertical.
▶️ Answer/Explanation
First, \(r'(\theta) = \sin\theta\).
$ \frac{dx}{d\theta} = r’ \cos\theta – r \sin\theta $
Substitute \(r = 1 – \cos\theta\) and \(r’ = \sin\theta\):
\(\frac{dx}{d\theta} = \sin\theta \cos\theta – (1 – \cos\theta)\sin\theta\)
Simplify: $ \frac{dx}{d\theta} = \sin\theta\cos\theta – \sin\theta + \sin\theta\cos\theta = 2\sin\theta\cos\theta – \sin\theta $
Factor: $ \frac{dx}{d\theta} = \sin\theta (2\cos\theta – 1) $
For vertical tangent: \(\frac{dx}{d\theta} = 0\) and \(\frac{dy}{d\theta} \neq 0\).
\(\frac{dx}{d\theta} = 0\) when \(\sin\theta = 0\) or \(\cos\theta = \frac12\).
Thus, \(\theta = 0, \pi, \frac{\pi}{3}, \frac{5\pi}{3}\), provided \(\frac{dy}{d\theta} \neq 0\) at those points.