Mathematics
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OpenStudy (anonymous):
How do I convert the rectangular coordinates (5,7) to polar coordinates?
13 years ago
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OpenStudy (anonymous):
What are rectangular and polar coordinates?
13 years ago
OpenStudy (anonymous):
x = r cos(theta)
y = r sin(theta)
13 years ago
OpenStudy (anonymous):
hm?
13 years ago
OpenStudy (anonymous):
(x,y)
x = r cos(theta)
y = r sin(theta)
13 years ago
OpenStudy (anonymous):
x^2 + y^2 = r^2
13 years ago
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OpenStudy (anonymous):
what is r and theta
13 years ago
OpenStudy (anonymous):
theta is the angle
you should have learned it by now if you are doing polar coordinates
13 years ago
OpenStudy (anonymous):
Well, I'm a freshman and this is just on an academic team test, so I don't know any of it
13 years ago
OpenStudy (anonymous):
|dw:1347584313553:dw|
13 years ago
OpenStudy (anonymous):
ah, that's nice
13 years ago
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OpenStudy (anonymous):
How the hell did you draw that perfectly? mine was completely disgusting...
13 years ago
OpenStudy (anonymous):
so rectangular coordinates just means cartesian coordinates?
13 years ago
OpenStudy (anonymous):
yes
13 years ago
OpenStudy (anonymous):
so, what next?
13 years ago
OpenStudy (anonymous):
theta = arctan(y/x)
r= x/cos(theta)
13 years ago
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OpenStudy (anonymous):
what is arctan?
13 years ago
OpenStudy (anonymous):
inverse operation of tangent
13 years ago
OpenStudy (anonymous):
tan^-1 like on my calculator?
13 years ago
OpenStudy (anonymous):
[INV] [TAN] on your calculator
13 years ago
OpenStudy (anonymous):
probably.
13 years ago
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OpenStudy (anonymous):
Dont use the calculator..learn it
13 years ago
OpenStudy (anonymous):
you using Google's calculator? Then yes it's tan^-1
13 years ago
OpenStudy (anonymous):
no, my ti-84
13 years ago
OpenStudy (anonymous):
2nd tan = tan^-1
13 years ago
OpenStudy (anonymous):
I think that's labeled tan^-1 also... don't remember for sure...
13 years ago
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OpenStudy (anonymous):
yeah
13 years ago
OpenStudy (anonymous):
so would doing arctan(7/5) give me the actual angle?
13 years ago
OpenStudy (anonymous):
yes
13 years ago
OpenStudy (anonymous):
can you explain to me why I use arctan please?
13 years ago
OpenStudy (anonymous):
oh wait, I think i see
13 years ago
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OpenStudy (anonymous):
tangent is opp/adj which is 7/5 which gives you the actual tangent, and then convert it to the angle?
13 years ago
OpenStudy (anonymous):
tan(theta) = opp/adj = y/x
tan^-1(tan(theta)) = tan^-1 (y/x)
theta = tan^-1(y/x)
13 years ago
OpenStudy (anonymous):
yes, exactly.
13 years ago
OpenStudy (anonymous):
sweet
13 years ago
OpenStudy (anonymous):
and then we have our angle which is about 54.46 degrees
13 years ago
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OpenStudy (anonymous):
yes
13 years ago
OpenStudy (anonymous):
now we need to find r which is sqrt(25+49)?
13 years ago
OpenStudy (anonymous):
you can...
13 years ago
OpenStudy (anonymous):
or r= x/cos(theta)
or r= y/sin(theta)
13 years ago
OpenStudy (anonymous):
can you explain to me why the 3 are the same thing?
13 years ago
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OpenStudy (anonymous):
better to use the sqrt. it's valid for all quadrants, since r is usually reckoned positive (except in some quirky cases). ...since this is 1st quadrant you can use the 'shortcuts' I gave...
13 years ago
OpenStudy (anonymous):
ok, would the way we found theta always work or are their exceptions too?
13 years ago
OpenStudy (anonymous):
there*
13 years ago
OpenStudy (anonymous):
Sort of... calculators give values for theta (when finding it by arctan) that are 1st or 4th quadrants...
13 years ago
OpenStudy (anonymous):
|dw:1347585395916:dw|
13 years ago