Introduction to Polar Coordinates

Understanding curves in polar form

🎯 Introduction to Polar Coordinates

What Are Polar Coordinates?

Instead of locating a point by (x,y)(x, y) (Cartesian), use:

  • rr = distance from origin (radius)
  • θ\theta = angle from positive x-axis (measured counterclockwise)

Polar notation: (r,θ)(r, \theta)

💡 Key Idea: Every point can be described by how far and in what direction from the origin!

Converting Between Polar and Cartesian

From Polar to Cartesian:

x=rcos⁡θx = r\cos\theta y=rsin⁡θy = r\sin\theta

Derivation: Draw a right triangle from origin to point.

From Cartesian to Polar:

r=x2+y2r = \sqrt{x^2 + y^2} tan⁡θ=yx\tan\theta = \frac{y}{x} (so θ=arctan⁡yx\theta = \arctan\frac{y}{x}, with quadrant adjustment)

Alternative for θ\theta:

\arctan(y/x) & \text{if } x > 0 \\ \arctan(y/x) + \pi & \text{if } x < 0 \\ \pi/2 & \text{if } x = 0, y > 0 \\ -\pi/2 & \text{if } x = 0, y < 0 \end{cases}$$ --- ## Example 1: Polar to Cartesian Convert $(r, \theta) = (5, \frac{\pi}{3})$ to Cartesian coordinates. **Step 1: Apply formulas** $$x = r\cos\theta = 5\cos\frac{\pi}{3} = 5 \cdot \frac{1}{2} = \frac{5}{2}$$ $$y = r\sin\theta = 5\sin\frac{\pi}{3} = 5 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}$$ **Answer**: $\left(\frac{5}{2}, \frac{5\sqrt{3}}{2}\right)$ --- ## Example 2: Cartesian to Polar Convert $(x, y) = (3, 4)$ to polar coordinates. **Step 1: Find $r$** $$r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ --- **Step 2: Find $\theta$** $$\tan\theta = \frac{4}{3}$$ $$\theta = \arctan\frac{4}{3} \approx 0.927 \text{ radians} \approx 53.1°$$ **Answer**: $(r, \theta) = (5, \arctan\frac{4}{3})$ or approximately $(5, 0.927)$ --- ## Non-Uniqueness of Polar Coordinates **Important**: Unlike Cartesian, polar coordinates are **not unique**! The point $(r, \theta)$ is the same as: - $(r, \theta + 2\pi n)$ for any integer $n$ - $(-r, \theta + \pi)$ (negative radius means opposite direction) **Example**: $(2, \frac{\pi}{4})$ is the same as: - $(2, \frac{9\pi}{4})$ (add $2\pi$) - $(-2, \frac{5\pi}{4})$ (negative $r$, add $\pi$) --- ## Polar Equations An equation in the form $r = f(\theta)$ describes a curve in polar coordinates. **Example**: $r = 2$ is a circle of radius 2 centered at origin. --- ## Common Polar Curves ### Circle centered at origin, radius $a$: $$r = a$$ --- ### Circle passing through origin: $$r = 2a\cos\theta$$ (circle with diameter on x-axis) $$r = 2a\sin\theta$$ (circle with diameter on y-axis) --- ### Line through origin: $$\theta = c$$ (constant angle) This is a ray from the origin! --- ### Spiral (Archimedean): $$r = a\theta$$ As $\theta$ increases, $r$ increases linearly. --- ## Example 3: Graphing $r = 1 + \cos\theta$ This is a **cardioid** (heart-shaped curve). **Make a table**: | $\theta$ | $\cos\theta$ | $r = 1 + \cos\theta$ | |-----------|----------------|------------------------| | 0 | 1 | 2 | | $\pi/3$ | 1/2 | 3/2 | | $\pi/2$ | 0 | 1 | | $2\pi/3$ | -1/2 | 1/2 | | $\pi$ | -1 | 0 | | $4\pi/3$ | -1/2 | 1/2 | | $3\pi/2$ | 0 | 1 | | $5\pi/3$ | 1/2 | 3/2 | | $2\pi$ | 1 | 2 | --- **Key features**: - Maximum $r = 2$ at $\theta = 0$ - Minimum $r = 0$ at $\theta = \pi$ (cusp at origin) - Symmetric about x-axis (since $\cos(-\theta) = \cos\theta$) --- ## Rose Curves ### $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$ - If $n$ is odd: $n$ petals - If $n$ is even: $2n$ petals - Each petal has length $a$ **Example**: $r = 3\cos(2\theta)$ has 4 petals of length 3. **Example**: $r = 2\sin(5\theta)$ has 5 petals of length 2. --- ## Limaçons ### $r = a \pm b\cos\theta$ or $r = a \pm b\sin\theta$ Shape depends on ratio $\frac{a}{b}$: - $\frac{a}{b} < 1$: **Inner loop** - $\frac{a}{b} = 1$: **Cardioid** (heart shape) - $1 < \frac{a}{b} < 2$: **Dimpled** (indented) - $\frac{a}{b} \geq 2$: **Convex** (oval) --- ## Lemniscates ### $r^2 = a^2\cos(2\theta)$ or $r^2 = a^2\sin(2\theta)$ **Figure-eight** or infinity symbol shape! **Example**: $r^2 = 4\cos(2\theta)$ Note: $r = \pm 2\sqrt{\cos(2\theta)}$ (exists only when $\cos(2\theta) \geq 0$) --- ## Symmetry in Polar Graphs ### Symmetric about x-axis: If replacing $\theta$ with $-\theta$ gives same equation. **Test**: Does $r(\theta) = r(-\theta)$? **Example**: $r = 1 + \cos\theta$ (since $\cos(-\theta) = \cos\theta$) --- ### Symmetric about y-axis: If replacing $\theta$ with $\pi - \theta$ gives same equation. **Test**: Does $r(\theta) = r(\pi - \theta)$? **Example**: $r = 1 + \sin\theta$ (since $\sin(\pi - \theta) = \sin\theta$) --- ### Symmetric about origin: If replacing $r$ with $-r$ OR $\theta$ with $\theta + \pi$ gives same equation. **Test**: Does $-r(\theta) = r(\theta)$ or $r(\theta) = r(\theta + \pi)$? **Example**: $r = \sin(2\theta)$ --- ## Converting Polar Equations to Cartesian Use substitutions: - $r = \sqrt{x^2 + y^2}$ - $\cos\theta = \frac{x}{r} = \frac{x}{\sqrt{x^2 + y^2}}$ - $\sin\theta = \frac{y}{r} = \frac{y}{\sqrt{x^2 + y^2}}$ --- ## Example 4: Convert $r = 4\sin\theta$ to Cartesian **Step 1: Multiply both sides by $r$** $$r^2 = 4r\sin\theta$$ --- **Step 2: Substitute** $$x^2 + y^2 = 4y$$ --- **Step 3: Complete the square** $$x^2 + y^2 - 4y = 0$$ $$x^2 + (y^2 - 4y + 4) = 4$$ $$x^2 + (y - 2)^2 = 4$$ **This is a circle** with center $(0, 2)$ and radius 2! --- ## Converting Cartesian to Polar Use substitutions: - $x = r\cos\theta$ - $y = r\sin\theta$ - $x^2 + y^2 = r^2$ --- ## Example 5: Convert $y = x$ to Polar **Step 1: Substitute** $$r\sin\theta = r\cos\theta$$ --- **Step 2: Simplify** (assuming $r \neq 0$) $$\sin\theta = \cos\theta$$ $$\tan\theta = 1$$ $$\theta = \frac{\pi}{4}$$ (or $\theta = \frac{5\pi}{4}$) **Answer**: $\theta = \frac{\pi}{4}$ (a ray from origin at 45°) --- ## ⚠️ Common Mistakes ### Mistake 1: Forgetting Quadrant When using $\arctan\frac{y}{x}$, check which quadrant the point is in! Calculator gives $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, but point might be in quadrant II or III. --- ### Mistake 2: Negative Radius In polar, $r$ can be negative! It means go in the opposite direction. $(−2, 0)$ is the same as $(2, \pi)$ (both represent point $(-2, 0)$ in Cartesian). --- ### Mistake 3: Not Checking Domain For $r = \sqrt{\cos\theta}$, we need $\cos\theta \geq 0$! The curve only exists where the expression under the square root is non-negative. --- ### Mistake 4: Missing Multiplicity $r^2 = 4$ gives $r = \pm 2$ This is TWO circles (though they overlap), not one! --- ## Polar Grid Polar graphs use circular grid: - Concentric circles for different $r$ values - Radial lines for different $\theta$ values **Tip**: When plotting, mark angles at multiples of $\frac{\pi}{6}$ or $\frac{\pi}{4}$ for common points. --- ## 📝 Practice Strategy 1. **For plotting**: Make a table of $\theta$ and $r$ values 2. **Use special angles**: $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, \ldots$ 3. **Check symmetry** before plotting full curve 4. **Find where $r = 0$**: These are points at the origin 5. **Find maximum/minimum $r$**: These are furthest/closest points 6. **For conversion**: Use the standard formulas, simplify 7. **Sketch before calculating** - builds intuition!

📚 Practice Problems

1Problem 1medium

❓ Question:

Convert the Cartesian equation x2+(y−3)2=9x^2 + (y-3)^2 = 9 to polar form.

💡 Show Solution

Step 1: Expand the equation

x2+y2−6y+9=9x^2 + y^2 - 6y + 9 = 9 x2+y2−6y=0x^2 + y^2 - 6y = 0

Step 2: Substitute polar equivalents

x2+y2=r2x^2 + y^2 = r^2 and y=rsin⁡θy = r\sin\theta:

r2−6rsin⁡θ=0r^2 - 6r\sin\theta = 0

Step 3: Factor

r(r−6sin⁡θ)=0r(r - 6\sin\theta) = 0

This gives r=0r = 0 (the origin) or:

r=6sin⁡θr = 6\sin\theta

Step 4: Simplify

Since r=0r = 0 is included in r=6sin⁡θr = 6\sin\theta (when θ=0\theta = 0 or π\pi), the polar equation is:

r=6sin⁡θr = 6\sin\theta

Answer: r=6sin⁡θr = 6\sin\theta

Note: This is a circle with diameter 6 on the y-axis, centered at (0,3)(0, 3).

2Problem 2easy

❓ Question:

a) Convert the polar point (4,π3)(4, \frac{\pi}{3}) to rectangular coordinates. b) Convert the rectangular point (−2,23)(-2, 2\sqrt{3}) to polar coordinates.

💡 Show Solution

Solution:

Part (a): Use formulas: x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta

x=4cos⁡π3=4⋅12=2x = 4\cos\frac{\pi}{3} = 4 \cdot \frac{1}{2} = 2

y=4sin⁡π3=4⋅32=23y = 4\sin\frac{\pi}{3} = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}

Rectangular: (2,23)(2, 2\sqrt{3})

Part (b): Use formulas: r=x2+y2r = \sqrt{x^2 + y^2}, tan⁡θ=yx\tan\theta = \frac{y}{x}

r=(−2)2+(23)2=4+12=16=4r = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4

tan⁡θ=23−2=−3\tan\theta = \frac{2\sqrt{3}}{-2} = -\sqrt{3}

The point is in Quadrant II (negative xx, positive yy).

θ=π−π3=2π3\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}

Polar: (4,2π3)(4, \frac{2\pi}{3}) or (−4,−π3)(-4, -\frac{\pi}{3})

3Problem 3hard

❓ Question:

Find all points of intersection of r=1r = 1 and r=2cos⁡θr = 2\cos\theta.

💡 Show Solution

Step 1: Set equations equal

1=2cos⁡θ1 = 2\cos\theta cos⁡θ=12\cos\theta = \frac{1}{2}

Step 2: Solve for θ\theta

In [0,2π)[0, 2\pi): θ=π3,θ=5π3\theta = \frac{\pi}{3}, \quad \theta = \frac{5\pi}{3}

Step 3: Find points

At θ=π3\theta = \frac{\pi}{3}: r=1r = 1 gives (1,π3)(1, \frac{\pi}{3})

At θ=5π3\theta = \frac{5\pi}{3}: r=1r = 1 gives (1,5π3)(1, \frac{5\pi}{3})

Step 4: Check for origin

For r=1r = 1: never passes through origin (r≠0r \neq 0)

For r=2cos⁡θr = 2\cos\theta: passes through origin when cos⁡θ=0\cos\theta = 0, at θ=π2,3π2\theta = \frac{\pi}{2}, \frac{3\pi}{2}

So both curves pass through origin, but at different angles!

The origin is also an intersection point (represented differently in each equation).

Answer: Three intersection points:

  • (1,π3)(1, \frac{\pi}{3})
  • (1,5π3)(1, \frac{5\pi}{3})
  • Origin (pole)

4Problem 4medium

❓ Question:

Identify and sketch r=2+2cos⁡θr = 2 + 2\cos\theta. Find the maximum value of rr.

💡 Show Solution

Step 1: Identify the curve type

This is a limaçon of the form r=a+bcos⁡θr = a + b\cos\theta with a=2,b=2a = 2, b = 2.

Since ab=22=1\frac{a}{b} = \frac{2}{2} = 1, this is a cardioid.

Step 2: Find maximum rr

r=2+2cos⁡θr = 2 + 2\cos\theta is maximized when cos⁡θ\cos\theta is maximized.

cos⁡θmax⁡=1\cos\theta_{\max} = 1 at θ=0\theta = 0

rmax⁡=2+2(1)=4r_{\max} = 2 + 2(1) = 4

Step 3: Find minimum rr

cos⁡θmin⁡=−1\cos\theta_{\min} = -1 at θ=π\theta = \pi

rmin⁡=2+2(−1)=0r_{\min} = 2 + 2(-1) = 0

So the curve passes through the origin at θ=π\theta = \pi.

Step 4: Check symmetry

Replace θ\theta with −θ-\theta: r(−θ)=2+2cos⁡(−θ)=2+2cos⁡θ=r(θ)r(-\theta) = 2 + 2\cos(-\theta) = 2 + 2\cos\theta = r(\theta)

Symmetric about the x-axis (polar axis).

Step 5: Key points

| θ\theta | rr | |-----------|-----| | 0 | 4 | | π/2\pi/2 | 2 | | π\pi | 0 | | 3π/23\pi/2 | 2 | | 2π2\pi | 4 |

Answer:

  • Curve: Cardioid
  • Maximum: r=4r = 4 at θ=0\theta = 0
  • Shape: Heart-shaped, pointing right, symmetric about x-axis
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