Math Problem Statement

sketch the curves in polar coordinates

Sketch the curves in polar coordinates: (ii) r = 6sinθ (iii) r = -1 - cosθ.

Solution

Let's examine the two polar equations and analyze how to sketch them.

(ii) r=6sinθr = 6 \sin \theta

This is the polar equation of a limacon with a sine term. Here's how it works:

  • When θ=0\theta = 0, r=6sin(0)=0r = 6 \sin(0) = 0, so the curve passes through the origin.
  • When θ=π2\theta = \frac{\pi}{2}, r=6sin(π2)=6r = 6 \sin\left(\frac{\pi}{2}\right) = 6, so the point (r,θ)=(6,π2)(r, \theta) = (6, \frac{\pi}{2}) is plotted on the positive y-axis.
  • When θ=π\theta = \pi, r=6sin(π)=0r = 6 \sin(\pi) = 0, so it returns to the origin.
  • When θ=3π2\theta = \frac{3\pi}{2}, r=6sin(3π2)=6r = 6 \sin\left(\frac{3\pi}{2}\right) = -6, which indicates a point in the negative y-direction.

This gives a limacon shape without an inner loop, symmetric with respect to the vertical axis.

(iii) r=1cosθr = -1 - \cos \theta

This is also a polar equation, and it generates a limacon with a negative rr and cosine term. Here's the interpretation:

  • When θ=0\theta = 0, r=1cos(0)=2r = -1 - \cos(0) = -2, so the point is 2 units to the left of the origin.
  • When θ=π2\theta = \frac{\pi}{2}, r=1cos(π2)=1r = -1 - \cos\left(\frac{\pi}{2}\right) = -1, the point is 1 unit below the origin.
  • When θ=π\theta = \pi, r=1cos(π)=0r = -1 - \cos(\pi) = 0, meaning the curve passes through the origin.
  • When θ=3π2\theta = \frac{3\pi}{2}, r=1cos(3π2)=1r = -1 - \cos\left(\frac{3\pi}{2}\right) = -1, a point below the origin.

This results in a limacon with an inner loop, symmetric with respect to the horizontal axis.

Would you like me to generate an exact sketch of these curves for you?


Here are 5 related questions:

  1. How does the graph of r=asinθr = a \sin \theta differ from r=acosθr = a \cos \theta?
  2. What happens to the curve when you adjust the coefficient in r=ksinθr = k \sin \theta?
  3. How do we determine the symmetry of a polar curve based on the equation?
  4. What effect does a negative coefficient have in polar equations like r=abcosθr = -a - b \cos \theta?
  5. Can you explain how to convert a polar equation to Cartesian coordinates?

Tip: When plotting polar curves, always start by determining key points for special angles like θ=0,π2,π,3π2\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} to understand the general shape of the graph.

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Math Problem Analysis

Mathematical Concepts

Polar Coordinates
Limacon Curves
Trigonometric Functions

Formulas

r = 6sin(θ)
r = -1 - cos(θ)

Theorems

Symmetry of polar curves
Behavior of limacons based on trigonometric equations

Suitable Grade Level

Grades 10-12