Math Help

[MrCodeDude]

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

 

[MrCodeDude]

Sorry, the arrows should be going from the X axis, not the Y axis.
-- mrcodedude

 

[Anubis]

42. the answer is 42

 

[MrCodeDude]

Originally posted by: TheEvil1
42. the answer is 42

OMG THX..
-- mrcodedude

 

[Syringer]

pi = 180 degrees
pi/2 = 90 degrees
pi/4 = 45 degrees

Is that basically what you need?

Edit: whoops, never mind..let's see..

 

[MrCodeDude]

I know that, that's radians. But how do I make cut-off restrictions for the circle with 60° and 300° ?
-- mrcodedude

 

[theNEOone]

couldn't hold this one back

 

[Syringer]

Equation of circle is x^2+y^2=25

Then restrict x and y appropriately..somehow.

Your picture looks like pacman btw

 

[Anubis]

Originally posted by: theNEOone
couldn't hold this one back

0wn3d

 

[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.

 

[Syringer]

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.

I think you're onto something.

 

[yoda291]

I'm repressing calculus right now...so I can't help.

 

[InverseOfNeo]

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

 

[MrCodeDude]

Yeah, we had to model something using circles and stuff. I chose PacMan because he seemed easy enough
-- mrcodedude

 

[JetBlack69]

doesn't:
x=r*cos(@)
y=r*sin(@)
r^2=x^2 + y^2
where @ is the theda and r is the radius?

 

[a7berwill]

Originally posted by: JetBlack69
doesn't:
x=r*cos(@)
y=r*sin(@)
r^2=x^2 + y^2
where @ is the theda and r is the radius?

Correct.

 

[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.

 

[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.

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.

 

[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.

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.

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)

 

[JetBlack69]

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.

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.

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)

Oh, ok I didn't know you were describing the part of the mouth as a function of f(x).

 

[dighn]

for purple: y = +-(x*sqrt(3)), x = 0 to 2.5

for red: y = +-sqrt(25-x^2), x = -5 to 2.5

 

[raptor13]

Theda is not a Greek letter. TheTa is, however.



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

 

[dighn]

Originally posted by: raptor13
Theda is not a Greek letter. TheT is, however.

lol