We can look at rectangular and polar coordinates as two different languages. We use them to describe the same things using different words. In the last section we learned how to translate points from one coordinate system to another.
Begin with a basic example, the right triangle. Rectangular and polar coordinates give different information about a right triangle.
From looking at the triangle we can see that these statements are true:

These transformation equations let us translate equations and inequalities between different coordinates.
To translate an equation or inequality from rectangular to polar coordinates, x becomes rcosθ and y becomes rsinθ. We can also replace x2 + y2 with r2.
Translating from polar to rectangular coordinates isn't quite as straightforward as going the other way. We can replace r2 with x2 + y2, rcosθ with x, and rsinθ with y.
However, sometimes we might need to do an extra step or two before we have any recognizable terms to replace.
Why can't we use one coordinate system and be done with it?
The answer is that some equations and inequalities look better in one system than the other. We know from practice that the equation
x2 + y2 = 36
describes a pizza with a 6 inch radius, but isn't
r = 6
a much nicer way to describe it?
Going the other way,
the polar inequality
r ≤ 2/cosθ
is a mess. If we translate into rectangular coordinates, we find
rcosθ ≤ 2
x≤ &2
which describes the part of the (x,y)-plane that lies on and to the left of the line x = 2:
If we're good at translating between coordinate systems, we can quickly find the simplest representation of a particular equation.
Practice:
Translate the equation x + y = 2x2 into polar coordinates.
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Replace {x} with {rcosθ} and replace {y} with {rsin θ}. x + y = 2 x2 rcos θ + r sin θ = 2 (rcos θ)2 rcos θ + rsinθ& = 2r2cos2θ One solution to this equation is r = 0. If r isn't zero, we can divide both sides of the equation by r to find cosθ + sinθ = 2rcos2θ. Since polar equations are usually written in the form r = f(θ) we rearrange this a bit more to find r = (cosθ + sinθ)/2cos2θ. | |
Translate the inequality x2 + y2 > 1 into polar coordinates.
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We can replace x2 + y2 with r to find the polar inequality r > 1. This describes all points that lie outside the unit circle:
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Translate the equation
rcosθ + rsinθ = 2
into rectangular coordinates. | |
This requires straightforward substitution. rcosθ + rsinθ = 2 x + y = 2. | |
Translate the inequality
into rectangular coordinates, assuming . | |
The cosθ piece is difficult to translate. But if we multiply both sides of the inequality by cos θ, we find rcosθ > 3 x>3. The restriction on the values of θ ensures that cosθ isn't negative, this way we don't accidentally multiply the inequality by a negative number. To go from polar to rectangular coordinates we need to rearrange the equation or inequality until we have terms that can be replaced with x, y, or x2 + y2. We may need to multiply both sides of an equation or inequality by r or by cosθ. We may need to square both sides. We may need to do other things. | |
- Translate each equation into polar coordinates.
- (x + 2)2 + (y + 2)2 = 4
- x + y2 = 0
- (x + y)2 = 4
Answer
Translate each equation into polar coordinates.
- We can square the terms first or replace x and y first. If we square the terms first, we find

Rearranging, and remembering that we can replace x2 + y2 with r2, we find

Since we have r2 and r terms, it's not reasonable to isolate r by itself.

If r≠ 0 then we can divide both sides by r to find

Then replace:

It's reasonable to separate out r2 in this equation. Remember that factoring r2 out from r2 leaves 1 behind.

- Translate each inequality into polar coordinates. %Graph the region being described by the inequality.
- y ≥ 5
- x + y≤ 3
- x2 + y2 < 100
Answer
Translate each inequality into polar coordinates. Graph the region being described by the inequality. Replace y to find

Replace x and y to find

Replace x2 + y2 with r2 to find

Therefore
- Translate each equation into rectangular coordinates.
- rcosθ + r2sin2θ = 4rsinθ
(hint: multiply through by r)- r = 4 (hint: square both sides.)
Answer
- This requires straightforward substitution.
rcosθ + r2sin2θ& = 4rsinθrcosθ + (rsinθ)2& = 4 rsinθx + y2 = 4y
Following the hint, multiply both sides by r:
sinθ = 5cosθ + frac2/rrsinθ = 5 rcosθ + 2
y = 5x + 2
Following the hint, square both sides and then substitute:
r = 4r2 = 16x2 + y2 = 16
- Translate each inequality into rectangular coordinates. Assume r is non-negative.%Graph the region being described by the inequality.
- sin θ < 1/r
- rcos2θ + rsin2θ = 1
- 5r2cos2θ + 10rcosθ + 5 -4r2sin2θ + 16rsinθ-16 = 20
Hint
substitute in x and y before trying to rearrange
Answer
\item Multiply both sides by r to find
sin θ &<& \frac{1}{r}rsinθ&<&1y&<&1
\item Factor out r on the left-hand side:
rcos2θ + rsin2θ& = &1r(cos2θ + sin2θ)& = &1r& = &1
Then square both sides:
r& = &1r2& = &1x2 + y2& = &1
\item Substitute in x and y first:
5r2cos2θ + 10rcosθ + 5 -4r2sin2θ + 16rsinθ-16& = &205({ rcosθ})2 + 10{ rcosθ} + 5-4({ rsinθ})2 + 16{ rsinθ}-16& = &205{ x}2 + 10{ x} + 5-4{ y}2 + 16{ y}-16& = &20.
Now we can factor to tidy things.
5x2 + 10x + 5-4y2 + 16y-16& = &205(x2 + 2x + 1)-4(y2-4y + 4)& = &205(x + 1)2-4(y-2)2& = &20.
Dividing both sides by 20, we see that this is the equation of a hyperbola:\frac{(x + 1)2}{4}-\frac{(y-2)2}{5} = 1.