# Thread: Polar coordinates stuff - urgent

1. ## Polar coordinates stuff - urgent

Hey guys, i'm stuck on these 3 probs. Polar stuff is hard for me to grasp, cuz i've been doing rectangular my whole math life... any and all help much appreciated. The second one is a cool kind that I usually like to do, but ...

1. Convert the polar equation r = acos(x) + bsin(x) to rectangular form. Show that the resulting equation represents a circle by completing the square on x and y. State the center and radius.

2. A forest ranger at observation post A sights a fire at location C in the direction N65 degrees E. Another ranger at observation post B, 8 kilometers directly east of A, sights the same fire at N32.2 degrees W. How far is the fire from observation post A? Draw and label a triangle using the given info. Estimate to 2 decimal places.

3. Sketch the graph of (x - 2)^2 + y^2 = 4. then find an equation of the graph in polar form.

2. Originally Posted by leviathanwave
1. Convert the polar equation r = acos(x) + bsin(x) to rectangular form. Show that the resulting equation represents a circle by completing the square on x and y. State the center and radius.
1. r = acos(x) + bsin(x)
I don't like the form of this equation, because it has cos(x) and sin(x), but we're trying to convert this polar equation into rectangular form (which contains x's and y's), so the angle needs to be something other than x. Let's use u instead.
r = acos(u) + bsin(u)

Now, I assume you know that the radius r can be considered the hypotonuse of a right triangle where the 2 legs are x and y and the angle is u. Knowing this, we can use a few identities to convert this equation:
cos(u) = x/r -> rcos(u) = x
sin(u) = y/r -> rsin(u) = y
x^2 + y^2 = r^2

Let's take the equation r = acos(u) + bsin(u), and multiply both sides by r:
r^2 = arcos(u) + brsin(u) ... look at what we have: r^2, rcos(u), rsin(u). We can substitute the above relations to get:
x^2 + y^2 = ax + by
x^2 - ax + y^2 - by = 0

Now by completing the square, we get:
[x^2 - ax + (a/2)^2] + [y^2 - by + (b/2)^2] = (a/2)^2 + (b/2)^2

(x - a/2)^2 + (y - b/2)^2 = (a^2 + b^2)/4
Center is (a/2,b/2)

3. Originally Posted by ecMathGeek

(x - a/2)^2 + (y - b/2)^2 = (a^2 + b^2)/4
Center is (a/2,b/2)
You did everything correct.
However, the problem should have said,
"a,b are not both zero".

Because if they are you have that,

Center is (a/2,b/2)=(0,0)
What type of circle is that?

4. Originally Posted by leviathanwave
3. Sketch the graph of (x - 2)^2 + y^2 = 4. then find an equation of the graph in polar form.
I'm skipping 2) for now, as it looks like it might take a bit more work (and I'm lazy )

3. (x-2)^2 + y^2 = 4
This one is easy if we just use the relationships:
x = rcos(u)
y = rsin(u)

(rcos(u) - 2)^2 + (rsin(u))^2 = 4
r^2(cos(u))^2 - 4rcos(u) + 4 + r^2(sin(u))^2 = 4 ... subtract 4 from both sides and factor r^2 from r^2(cos(u))^2 and r^2(sin(u))^2:
r^2[(cos(u))^2 + (sin(u))^2] - 4rcos(u) = 0
r^2 = 4rcos(u)

5. Originally Posted by ThePerfectHacker
You did everything correct.
However, the problem should have said,
"a,b are not both zero".

Because if they are you have that,

Center is (a/2,b/2)=(0,0)
What type of circle is that?
lol, ty.

AND ... I just noticed this:
if a and b are 0, then
r = 0cos(u) + 0sin(u) = 0

6. ## Thanks!

thanks a lot guys. I'm reading up all i can on info to help with number 2 now, heh. Feel free to help out if u can

Thanks again

7. ## Is this right?

I did problem #2 and got 5.07km for my final answer. Is it right?