Math Problem Statement

Convert polar coordinates to Cartesian and sketch a polar curve: a. (3, 45°), b. (2, -120°), c. (-1, 270°), d. (-2, -60°), and the curve r = 1 - 2sin(θ).

Solution

The image contains two math problems:

  1. The first problem seems to be asking for the Cartesian coordinates corresponding to polar coordinates of different points. Specifically, it lists:

    • a.(3,45)a. (3, 45^\circ)
    • b.(2,120)b. (2, -120^\circ)
    • c.(1,270)c. (-1, 270^\circ)
    • d.(2,60)d. (-2, -60^\circ)
  2. The second problem asks to sketch the curve defined by the polar equation: r=12sinθr = 1 - 2 \sin \theta and then convert this equation into Cartesian coordinates.


Solving these problems:

1. Convert Polar to Cartesian Coordinates:

To convert polar coordinates (r,θ)(r, \theta) to Cartesian coordinates (x,y)(x, y), we use the formulas: x=rcosθx = r \cos \theta y=rsinθy = r \sin \theta

Let's calculate for each point:

  • a. (3, 45°): x=3cos45=3×12=322.12x = 3 \cos 45^\circ = 3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}} \approx 2.12 y=3sin45=3×12=322.12y = 3 \sin 45^\circ = 3 \times \frac{1}{\sqrt{2}} = \frac{3}{\sqrt{2}} \approx 2.12 Cartesian coordinates: (2.12,2.12)(2.12, 2.12)

  • b. (2, -120°): x=2cos(120)=2×(12)=1x = 2 \cos (-120^\circ) = 2 \times (-\frac{1}{2}) = -1 y=2sin(120)=2×(32)=31.73y = 2 \sin (-120^\circ) = 2 \times \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3} \approx -1.73 Cartesian coordinates: (1,1.73)(-1, -1.73)

  • c. (-1, 270°): x=1cos270=1×0=0x = -1 \cos 270^\circ = -1 \times 0 = 0 y=1sin270=1×(1)=1y = -1 \sin 270^\circ = -1 \times (-1) = 1 Cartesian coordinates: (0,1)(0, 1)

  • d. (-2, -60°): x=2cos(60)=2×12=1x = -2 \cos (-60^\circ) = -2 \times \frac{1}{2} = -1 y=2sin(60)=2×(32)=31.73y = -2 \sin (-60^\circ) = -2 \times \left(-\frac{\sqrt{3}}{2}\right) = \sqrt{3} \approx 1.73 Cartesian coordinates: (1,1.73)(-1, 1.73)

2. Sketch and Convert the Polar Equation r=12sinθr = 1 - 2 \sin \theta to Cartesian Form:

To convert the polar equation to Cartesian coordinates, use the following relationships: r=x2+y2,sinθ=yr,cosθ=xrr = \sqrt{x^2 + y^2}, \quad \sin \theta = \frac{y}{r}, \quad \cos \theta = \frac{x}{r}

Given: r=12sinθr = 1 - 2 \sin \theta Substitute sinθ=yr\sin \theta = \frac{y}{r}: r=12×yrr = 1 - 2 \times \frac{y}{r} Multiply both sides by rr: r2=r2yr^2 = r - 2y Substitute r2=x2+y2r^2 = x^2 + y^2: x2+y2=r2yx^2 + y^2 = r - 2y Since r=x2+y2r = \sqrt{x^2 + y^2}, replace rr with x2+y2\sqrt{x^2 + y^2}: x2+y2=x2+y22yx^2 + y^2 = \sqrt{x^2 + y^2} - 2y This is the equation of the curve in Cartesian coordinates. It is a bit complex to sketch, but it resembles a limaçon.

Let me know if you would like more details or have any questions!


Here are some related questions:

  1. How do you determine which trigonometric function to use when converting polar coordinates?
  2. What other types of polar curves are common, besides limaçons?
  3. What is the significance of the negative radius in polar coordinates?
  4. How can we graph polar equations using a Cartesian grid?
  5. What happens to the graph of r=12sinθr = 1 - 2 \sin \theta if you change the coefficients?

Tip: Always double-check the sign of the angle when converting from polar to Cartesian coordinates, as it impacts the direction in which the point is plotted.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Polar coordinates
Cartesian coordinates
Trigonometry
Graphing curves

Formulas

x = r * cos(θ)
y = r * sin(θ)
r = sqrt(x^2 + y^2)
sin(θ) = y / r

Theorems

Polar to Cartesian conversion
Polar equations in Cartesian form

Suitable Grade Level

Grades 10-12