Math Problem Statement

Ubah ke sistem koordinat kartesius, persamaan: a. r cos θ = 6, b. r^2 - 8r cos θ - 4r sin θ + 11 = 0

Solution

The problem asks to convert polar equations into Cartesian coordinates. Let's address both parts:

Part (a) rcosθ=6r \cos \theta = 6:

In polar coordinates, we know: rcosθ=xr \cos \theta = x Thus, the equation becomes: x=6x = 6 This is the Cartesian form of the equation, which represents a vertical line at x=6x = 6.

Part (b) r28rcosθ4rsinθ+11=0r^2 - 8r \cos \theta - 4r \sin \theta + 11 = 0:

Using the following polar to Cartesian conversions: r2=x2+y2,rcosθ=x,rsinθ=yr^2 = x^2 + y^2, \quad r \cos \theta = x, \quad r \sin \theta = y We substitute these into the given equation: r28rcosθ4rsinθ+11=0r^2 - 8r \cos \theta - 4r \sin \theta + 11 = 0 becomes (x2+y2)8x4y+11=0(x^2 + y^2) - 8x - 4y + 11 = 0 This is the Cartesian form of the given equation.

Would you like further clarification or details on these steps?

Here are 5 questions to extend this discussion:

  1. How do we derive the polar to Cartesian coordinate conversion formulas?
  2. Can you visualize the geometric meaning of these equations in Cartesian coordinates?
  3. What is the relationship between radial distance rr and θ\theta in polar coordinates?
  4. How can we apply these conversion techniques to more complex polar equations?
  5. How would the equation change if we shift or rotate the coordinate system?

Tip: When converting polar to Cartesian equations, remember the key conversions for rr and θ\theta: r2=x2+y2r^2 = x^2 + y^2, rcosθ=xr \cos \theta = x, and rsinθ=yr \sin \theta = y.

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
Equation Conversion

Formulas

r cos θ = x
r sin θ = y
r^2 = x^2 + y^2

Theorems

Polar to Cartesian Coordinate Conversion

Suitable Grade Level

Grades 10-12