Polar Graphing

Grade 12 · trigonometry · 88 practice problems · read aloud

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Polar Graphing: A Different Coordinate System

Polar graphing represents points using a distance from a central point (the pole, or origin) and an angle from a fixed direction (the polar axis, usually the positive x-axis). Instead of (x, y), we use (r, θ). This is incredibly useful for modeling circular, spiral, and rotational patterns found in nature and engineering, which are often cumbersome in rectangular (Cartesian) coordinates.

How to Graph a Polar Equation

  1. Understand the Equation: Identify the relationship between the radius r and the angle θ.
  2. Create a Table of Values: Choose key values for θ (e.g., 0, π/6, π/4, π/2, π) and calculate the corresponding r values.
  3. Plot the Points: For each (r, θ) pair, move counter-clockwise to the angle θ, then move outward a distance of r (if r is negative, move in the opposite direction).
  4. Connect the Points: Draw a smooth curve through the plotted points to reveal the graph.

Visual Examples

Example 1: The Circle r = 4

This equation states that the radius is always 4, regardless of the angle θ. As θ sweeps from 0 to 2π, you are always 4 units from the pole, creating a perfect circle with a radius of 4.

Example 2: The Cardioid r = 2 + 2sinθ

Let's build a table for this heart-shaped curve:

  • θ = 0°: r = 2 + 2(0) = 2
  • θ = 90° (π/2): r = 2 + 2(1) = 4
  • θ = 180° (π): r = 2 + 2(0) = 2
  • θ = 270° (3π/2): r = 2 + 2(-1) = 0

Plotting these points reveals a curve with a cusp at the bottom and a loop on top.

Common Mistakes to Avoid 🚫

  • Misplotting Negative r: When r is negative, you plot the point at the angle opposite to θ. The coordinate ( -r, θ ) is the same point as ( r, θ + π ).
  • Using Degrees Instead of Radians: In advanced math, polar equations almost always use radians. Ensure your calculator is in radian mode.
  • Not Using Enough Points: Polar graphs can have loops and dimples. Use a sufficient number of θ values (especially between 0 and 2π) to capture the true shape.

Tips & Tricks

  • Symmetry is Your Friend: Look for symmetry to cut your work in half.
    • If the equation has cos(θ), it's symmetric about the polar axis (x-axis).
    • If it has sin(θ), it's symmetric about the line θ = π/2 (y-axis).
  • Know Your Classics: Memorize the general forms for common graphs like circles (r=a), limaçons (r=a±b sin/cos θ), and roses (r=a sin/cos(nθ)).
  • Use Technology: Use a graphing calculator or online tool to check your hand-drawn sketches.

How to Practice

Start by graphing basic equations like r = a and r = θ (the spiral). Then, move on to more complex forms like limaçons and rose curves. Practice converting between polar and rectangular coordinates to deepen your understanding of the relationship between the two systems. Finally, try to identify the polar equation just by looking at its graph, and vice-versa.

Practice problems

6 of the 88, worked through step by step — try them before opening the answer.

1 ∫(x²e^x) dx from 0 to 1 = ?

Hint: Consider using integration by parts, which follows the formula ∫u dv = uv - ∫v du. Try selecting u and dv strategically to simplify the integral.

Show the answer

Answer: e-2

  1. Use integration by parts. Let u = x² and dv = e^x dx
  2. Then du = 2x dx and v = e^x
  3. Apply integration by parts: ∫x²e^x dx = x²e^x - ∫2xe^x dx
  4. Apply integration by parts again to ∫2xe^x dx. Let u = 2x and dv = e^x dx
  5. Then du = 2 dx and v = e^x
  6. ∫2xe^x dx = 2xe^x - ∫2e^x dx = 2xe^x - 2e^x
  7. Substitute back: ∫x²e^x dx = x²e^x - (2xe^x - 2e^x) = x²e^x - 2xe^x + 2e^x
  8. Evaluate from 0 to 1: [1²e¹ - 2(1)e¹ + 2e¹] - [0²e⁰ - 2(0)e⁰ + 2e⁰]
  9. = [e - 2e + 2e] - [0 - 0 + 2(1)]
  10. = [e] - [2] = e - 2

The answer is e-2.

2 ∫(x²e^(x)) dx from 0 to 1 = ?

Hint: Consider using integration by parts, where one function becomes simpler when differentiated and the other remains manageable when integrated.

Show the answer

Answer: e - 2

  1. Use integration by parts formula: ∫u dv = uv - ∫v du
  2. Let u = x², then du = 2x dx
  3. Let dv = e^(x) dx, then v = e^(x)
  4. Apply the formula: ∫x²e^(x) dx = x²e^(x) - ∫2xe^(x) dx
  5. Apply integration by parts again to ∫2xe^(x) dx
  6. Let u = 2x, then du = 2 dx
  7. Let dv = e^(x) dx, then v = e^(x)
  8. ∫2xe^(x) dx = 2xe^(x) - ∫2e^(x) dx = 2xe^(x) - 2e^(x)
  9. Substitute back: ∫x²e^(x) dx = x²e^(x) - [2xe^(x) - 2e^(x)] = x²e^(x) - 2xe^(x) + 2e^(x)
  10. Evaluate from 0 to 1: [1²e^(1) - 2(1)e^(1) + 2e^(1)] - [0²e^(0) - 2(0)e^(0) + 2e^(0)]
  11. Simplify: [e - 2e + 2e] - [0 - 0 + 2(1)] = [e] - [2] = e - 2

The answer is e - 2.

3 ∫(x²e^(x³)) dx from 0 to 1 = ?

Hint: Consider using substitution to simplify the integration of a product of polynomial and exponential functions.

Show the answer

Answer: (e-1)/3

  1. Let u = x³, then du = 3x² dx, so x² dx = du/3
  2. When x = 0, u = 0³ = 0
  3. When x = 1, u = 1³ = 1
  4. Substitute: ∫(x²e^(x³)) dx from 0 to 1 = ∫(e^u)(du/3) from 0 to 1
  5. = (1/3)∫(e^u) du from 0 to 1
  6. = (1/3)[e^u] from 0 to 1
  7. = (1/3)(e¹ - e⁰)
  8. = (1/3)(e - 1)
  9. = (e-1)/3

The answer is (e-1)/3.

4 ∫(x² + 2x + 1) dx from 0 to 3 = ?

Hint: Find the antiderivative of the polynomial, then evaluate it at the upper and lower limits of integration.

Show the answer

Answer: 21

  1. Find the antiderivative of x² + 2x + 1. The antiderivative of x² is (1/3)x³. The antiderivative of 2x is (2/2)x² = x². The antiderivative of 1 is x. So, the antiderivative F(x) = (1/3)x³ + x² + x.
  2. Evaluate F(x) from 0 to 3 using the Fundamental Theorem of Calculus: F(3) - F(0). F(3) = (1/3)(3)³ + (3)² + (3) = (1/3)(27) + 9 + 3 = 9 + 9 + 3 = 21. F(0) = (1/3)(0)³ + (0)² + (0) = 0 + 0 + 0 = 0.
  3. Calculate the definite integral: F(3) - F(0) = 21 - 0 = 21.

The answer is 21.

5 ∫(x² + 2x - 3) dx from 1 to 4 = ?

Hint: Find the antiderivative of the polynomial, then evaluate it at the upper and lower limits of integration.

Show the answer

Answer: 27

  1. Find the antiderivative of x² + 2x - 3. The antiderivative of x² is (1/3)x³. The antiderivative of 2x is (2/2)x² = x². The antiderivative of -3 is -3x. So the antiderivative F(x) = (1/3)x³ + x² - 3x.
  2. Evaluate F(4) - F(1). F(4) = (1/3)(4)³ + (4)² - 3(4) = (1/3)(64) + 16 - 12 = 64/3 + 4 = 64/3 + 12/3 = 76/3 F(1) = (1/3)(1)³ + (1)² - 3(1) = 1/3 + 1 - 3 = 1/3 - 2 = 1/3 - 6/3 = -5/3
  3. Calculate F(4) - F(1) = 76/3 - (-5/3) = 76/3 + 5/3 = 81/3 = 27

The answer is 27.

6 ∫(3x² + 2x - 1) dx from 0 to 2 = ?

Hint: To evaluate a definite integral, first find the antiderivative of the function, then apply the Fundamental Theorem of Calculus by evaluating at the upper and lower limits.

Show the answer

Answer: 10

  1. Find the antiderivative** We integrate term by term: ∫ 3x² dx = 3 * (x³ / 3) = x³ ∫ 2x dx = 2 * (x² / 2) = x² ∫ (-1) dx = -x So the antiderivative F(x) = x³ + x² - x --- **
  2. Apply the Fundamental Theorem of Calculus** The definite integral = F(2) - F(0) First, compute F(2): F(2) = (2)³ + (2)² - (2) = 8 + 4 - 2 = 10 Next, compute F(0): F(0) = (0)³ + (0)² - (0) = 0 --- **
  3. Subtract** F(2) - F(0) = 10 - 0 = 10 --- **Final Answer:** 10

Let's solve the definite integral step by step. We are given: ∫ from 0 to 2 of (3x² + 2x - 1) dx --- **

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