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Compression and Conversion Help (page 2)

By — McGraw-Hill Professional
Updated on Aug 30, 2011

Practice 1

Provide an example of a graphical object that can be represented as a function in polar coordinates, but not in Cartesian coordinates.

Solution 1

In polar coordinates, let θ represent the independent variable, and let r represent the dependent variable. Then when we talk about a function f , we can say that r = f(θ) . A simple function of θ in polar coordinates is a constant function such as this:

f (θ) = 3

Because f (θ) is just another way of denoting r , the radius, this function tells us that r = 3. This is a circle with a radius of 3 units.

In Cartesian xy -coordinates, the equation of the circle with radius of 3 units is more complicated:

x 2 + y 2 = 9

(Note that 9 = 3 2 , the square of the radius.) If we let x be the independent variable and y be the dependent variable, we can rearrange the equation of the circle to get:

y = ±(9 – x 2 ) 1/2

If we say that y = g ( x ) where g is a function of x in this case, we are mistaken. There are values of x (the independent variable) that produce two values of y (the dependent variable). For example, when x = 0, y = ±3. If we want to say that g is a relation, that’s fine, but we cannot call it a function.

Practice 2

Consider the point ( θ 0 , r 0 ) = (135°, 2) in polar coordinates. What is the ( x 0 , y 0 ) representation of this point in Cartesian coordinates?

Solution 2

Use the conversion formulas above:

x 0 = r 0 cos θ 0

y 0 = r 0 sin θ 0

Plugging in the numbers gives us these values, accurate to three decimal places:

x 0 = 2 cos 135° = 2 × (–0.707) = –1.414

y 0 = 2 sin 135° = 2 × 0.707 = 1.414

Thus, ( x 0 , y 0 ) = (–1.414, 1.414).

Practice problems for these concepts can be found at:  Polar Coordinates Practice Test

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