AP Calculus BC — Unit 9.7
Defining Polar Coordinates and Differentiating in Polar Form
Key Concepts Packet + Worked Examples + Interactive MCQ Practice (36 Questions)
Table of Contents
- Polar Coordinates: Points and Conversion
- Negative Radius and Multiple Representations
- Polar Curves as Parametric Curves
- Slope in Polar Form: \( \frac{dy}{dx} \)
- Horizontal/Vertical Tangents in Polar
- Calculator Expectations + Common Mistakes
- Unit 9.7 Multiple-Choice Practice (36 Questions)
- Answer Key
Unit 9.7 Key Concepts Packet
1) Polar Coordinates: Points and Conversion
Coordinate relationships
\(x=r\cos\theta,\quad y=r\sin\theta\)
\(r^2=x^2+y^2,\quad \tan\theta=\frac{y}{x}\) (with quadrant care)
\(r^2=x^2+y^2,\quad \tan\theta=\frac{y}{x}\) (with quadrant care)
- A polar point is \((r,\theta)\) (distance from origin, then angle from the positive \(x\)-axis).
- Converting polar → rectangular uses \(x=r\cos\theta,\ y=r\sin\theta\).
- Converting rectangular → polar uses \(r=\sqrt{x^2+y^2}\) and an angle \(\theta\) that matches the correct quadrant.
2) Negative Radius and Multiple Representations
Same point, different polar coordinates
\((r,\theta)\equiv (r,\theta+2\pi k)\) for any integer \(k\).
\((r,\theta)\equiv (-r,\theta+\pi)\) (negative radius identity).
\((r,\theta)\equiv (-r,\theta+\pi)\) (negative radius identity).
- This identity is a common AP “trap” when solving intersections or choosing bounds later.
- Always consider whether a curve is traced more than once for a given interval of \(\theta\).
3) Polar Curves as Parametric Curves
Key idea
A polar curve \(r=f(\theta)\) can be treated parametrically by letting \(t=\theta\):
\(x(\theta)=r(\theta)\cos\theta,\quad y(\theta)=r(\theta)\sin\theta\).
- This is why differentiation in polar uses the chain rule with derivatives in \(\theta\).
- In AP terms: polar differentiation is a direct application of parametric differentiation.
4) Slope in Polar Form: \( \frac{dy}{dx} \)
Core formula (must know)
If \(x=r\cos\theta,\ y=r\sin\theta\), then
\( \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \)
where
\( \displaystyle \frac{dx}{d\theta}=r'(\theta)\cos\theta-r(\theta)\sin\theta \),
\( \displaystyle \frac{dy}{d\theta}=r'(\theta)\sin\theta+r(\theta)\cos\theta \).
- Compute \(r'(\theta)\) first, then plug into the expressions for \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\).
- Finally divide: \( \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta} \).
5) Horizontal/Vertical Tangents in Polar
Tangent conditions
Horizontal tangent when \( \frac{dy}{d\theta}=0 \) and \( \frac{dx}{d\theta}\neq 0 \).
Vertical tangent when \( \frac{dx}{d\theta}=0 \) and \( \frac{dy}{d\theta}\neq 0 \).
Vertical tangent when \( \frac{dx}{d\theta}=0 \) and \( \frac{dy}{d\theta}\neq 0 \).
- You must check both conditions (a common AP mistake is to ignore the “not zero” requirement).
- Also check special points where \(r(\theta)=0\) (the curve passes through the pole/origin).
6) Calculator Expectations + Common Mistakes
- Expect to evaluate \( \frac{dy}{dx} \) at a specific \(\theta\) value (sometimes with a calculator).
- Trap: Using \( \frac{dy}{d\theta} \) and calling it slope. Slope is \( \frac{dy}{dx} \).
- Trap: Missing vertical tangents because you only solve \( \frac{dy}{d\theta}=0 \).
- Important: If you ever see raw backslashes instead of formatted math, it means MathJax did not typeset that section—this page forces typesetting on load and when solutions open.
Worked Examples + Notes
Example 1 — Convert polar to rectangular
Convert \((r,\theta)=(2,\,\frac{\pi}{3})\) to rectangular coordinates.
Step 1: Use \(x=r\cos\theta\), \(y=r\sin\theta\).
Step 2: \(x=2\cos(\frac{\pi}{3})=2\cdot\frac12=1\).
Step 3: \(y=2\sin(\frac{\pi}{3})=2\cdot\frac{\sqrt3}{2}=\sqrt3\).
Conclusion: Rectangular point is \((1,\sqrt3)\).
Example 2 — Use the negative radius identity
Find an equivalent polar coordinate for \((r,\theta)=(3,\,\frac{\pi}{6})\) with negative radius.
Step 1: Use \((r,\theta)\equiv(-r,\theta+\pi)\).
Step 2: \((3,\frac{\pi}{6})\equiv(-3,\frac{\pi}{6}+\pi)=(-3,\frac{7\pi}{6})\).
Conclusion: One equivalent coordinate is \((-3,\frac{7\pi}{6})\).
Example 3 — Find \( \frac{dy}{dx} \) for a polar curve
For \(r(\theta)=2\cos\theta\), find \( \frac{dy}{dx} \) at \(\theta=\frac{\pi}{3}\).
Step 1: Compute \(r'(\theta)=-2\sin\theta\).
Step 2: \( \frac{dx}{d\theta}=r'\cos\theta-r\sin\theta=(-2\sin\theta)\cos\theta-(2\cos\theta)\sin\theta=-4\sin\theta\cos\theta\).
Step 3: \( \frac{dy}{d\theta}=r'\sin\theta+r\cos\theta=(-2\sin\theta)\sin\theta+(2\cos\theta)\cos\theta=2(\cos^2\theta-\sin^2\theta)=2\cos(2\theta)\).
Step 4: \( \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{2\cos(2\theta)}{-4\sin\theta\cos\theta}=\frac{-\cos(2\theta)}{2\sin\theta\cos\theta}\).
Step 5: At \(\theta=\frac{\pi}{3}\), \(\cos(2\theta)=\cos(\frac{2\pi}{3})=-\frac12\), \(\sin\theta=\frac{\sqrt3}{2}\), \(\cos\theta=\frac12\).
Conclusion: \( \frac{dy}{dx}=\frac{-(-1/2)}{2(\sqrt3/2)(1/2)}=\frac{1/2}{\sqrt3/2}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}\).
Example 4 — Horizontal tangent in polar form
Let \(r(\theta)=1+\sin\theta\). Find \(\theta\) values where the curve has a horizontal tangent.
Step 1: \(r'(\theta)=\cos\theta\).
Step 2: \( \frac{dy}{d\theta}=r'\sin\theta+r\cos\theta=\cos\theta\sin\theta+(1+\sin\theta)\cos\theta=\cos\theta(\sin\theta+1+\sin\theta)=\cos\theta(1+2\sin\theta)\).
Step 3: Horizontal tangent requires \(dy/d\theta=0\) and \(dx/d\theta\neq 0\).
Step 4: Solve \(\cos\theta(1+2\sin\theta)=0\Rightarrow \cos\theta=0\) or \(\sin\theta=-\frac12\).
Conclusion: Candidate angles are \(\theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}\) (then confirm \(dx/d\theta\neq 0\) for each).
Note: This page forces MathJax typesetting on load and when solution panels open.
Unit 9.7 Multiple-Choice Practice (36 Questions)
Answer Key