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Math Help

M

MrCodeDude

Lifer
Convert from Polar to Rectangular Form

I don't know how to do the restrictions for the circle. I created a picture for my problem.

Thanks

Link
-- mrcodedude
 
pi = 180 degrees
pi/2 = 90 degrees
pi/4 = 45 degrees

Is that basically what you need?

Edit: whoops, never mind..let's see..
 
Equation of circle is x^2+y^2=25

Then restrict x and y appropriately..somehow.

Your picture looks like pacman btw 🙂
 
Can the circle be in parametric form? If so, wouldn't it just be...

x = 5*cos(t), y = 5*sin(t), pi/6 <= t <= 5pi/3

Don't yell at me if this is stupid, I'm not sure on whether the terminology "rectangular form" allows for parametric representation or not.
 
Originally posted by: a7berwill
Can the circle be in parametric form? If so, wouldn't it just be...

x = 5*cos(t), y = 5*sin(t), pi/6 <= t <= 5pi/3

Don't yell at me if this is stupid, I'm not sure on whether the terminology "rectangular form" allows for parametric representation or not.
Click to expand...

I think you're onto something.
 
Hmmm I was about to say that I have no idea what the answer is (and I still dont) but that it looks A LOT like pacman...............but then I noticed the filename😛
 
y = +sqrt(25 - x^2) {x = -5 .. 5*sqrt(3)/2}
y = -sqrt(25 - x^2) {x = -5 .. 5/2]
y = x/sqrt(3) {x = 0 .. 5*sqrt(3)/2}
y = -sqrt(3)*x {x = 0 .. 5/2}

This probably isn't right, as I'm tired.
 
Originally posted by: silverpig
y = +sqrt(25 - x^2) {x = -5 .. 5*sqrt(3)/2}
y = -sqrt(25 - x^2) {x = -5 .. 5/2]
y = x/sqrt(3) {x = 0 .. 5*sqrt(3)/2}
y = -sqrt(3)*x {x = 0 .. 5/2}

This probably isn't right, as I'm tired.
Click to expand...

I think you are close but I dont' think these two are correct:
y = x/sqrt(3) {x = 0 .. 5*sqrt(3)/2}
y = -sqrt(3)*x {x = 0 .. 5/2}

because at x=0, y doesn't equal 0. It's been to long since I've done this stuff but I think you are on the right track.
 
Originally posted by: JetBlack69
Originally posted by: silverpig
y = +sqrt(25 - x^2) {x = -5 .. 5*sqrt(3)/2}
y = -sqrt(25 - x^2) {x = -5 .. 5/2]
y = x/sqrt(3) {x = 0 .. 5*sqrt(3)/2}
y = -sqrt(3)*x {x = 0 .. 5/2}

This probably isn't right, as I'm tired.
Click to expand...

I think you are close but I dont' think these two are correct:
y = x/sqrt(3) {x = 0 .. 5*sqrt(3)/2}
y = -sqrt(3)*x {x = 0 .. 5/2}

because at x=0, y doesn't equal 0. It's been to long since I've done this stuff but I think you are on the right track.
Click to expand...

Sure it does. There are 4 equations there describing 4 parts of pacman. The first one listed describes the curved part above the x-axis. The second one describes the curved part below the x-axis. The third one represents the straight part that makes the upper part of his mouth, and the last one makes the bottom straight part of his mouth (the theta = 60 degrees and theta = 300 degrees from r=0 to 5 respectively)
 
Originally posted by: silverpig
Originally posted by: JetBlack69
Originally posted by: silverpig
y = +sqrt(25 - x^2) {x = -5 .. 5*sqrt(3)/2}
y = -sqrt(25 - x^2) {x = -5 .. 5/2]
y = x/sqrt(3) {x = 0 .. 5*sqrt(3)/2}
y = -sqrt(3)*x {x = 0 .. 5/2}

This probably isn't right, as I'm tired.
Click to expand...

I think you are close but I dont' think these two are correct:
y = x/sqrt(3) {x = 0 .. 5*sqrt(3)/2}
y = -sqrt(3)*x {x = 0 .. 5/2}

because at x=0, y doesn't equal 0. It's been to long since I've done this stuff but I think you are on the right track.
Click to expand...

Sure it does. There are 4 equations there describing 4 parts of pacman. The first one listed describes the curved part above the x-axis. The second one describes the curved part below the x-axis. The third one represents the straight part that makes the upper part of his mouth, and the last one makes the bottom straight part of his mouth (the theta = 60 degrees and theta = 300 degrees from r=0 to 5 respectively)
Click to expand...

Oh, ok I didn't know you were describing the part of the mouth as a function of f(x).
 
Theda is not a Greek letter. TheTa is, however.



Edit: Man... what a misopportune time for a typo. God I'm stupid. Heh.
 
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