Mathematics
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OpenStudy (babynini):
a) sketch curve
b) find a rectangular coordinate equation for the curve by eliminating the perimeter
x=sec(t), y=tan(t), 0 = t < pi/2
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OpenStudy (babynini):
I simplified x into
x= 1/cos(t)
and y = sin(t)/cos(t)
then y = [sin(t)][1/cos(t)]
OpenStudy (babynini):
then plugged x into y and got
y=sint(x)
OpenStudy (babynini):
eem..I got confused on that second part there because there's lots of ='s xD
yeah the first part is easy
OpenStudy (loser66):
ok, line by line y^2 = tan^2 t , ok?
OpenStudy (babynini):
right
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OpenStudy (babynini):
yup
OpenStudy (babynini):
yeah.aah I see.
OpenStudy (loser66):
ok, last line is obvious. right?
OpenStudy (babynini):
Yep!!
OpenStudy (babynini):
well the
sec^2t= x^2 makes sense
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OpenStudy (babynini):
how you got to x^2-1=y^2 I don't know.
OpenStudy (loser66):
how cannot?
y^2 +1 = x^2
just -1 both sides
OpenStudy (babynini):
Oooh sorry, i had written it wrong on my paper. k.
OpenStudy (babynini):
it could also be written x^2-y^2=1 right?
OpenStudy (loser66):
sure
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OpenStudy (babynini):
then we just plug in numbers for x and find y.
OpenStudy (babynini):
Kk's thanks! :)
OpenStudy (loser66):
hihihi... tip for a)
x^2 -y^2 =1 is a hyperbola with the center at origin
OpenStudy (babynini):
Oo that is helpful.
OpenStudy (babynini):
Hold up, how do I find the perimeter?
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OpenStudy (babynini):
like they give me 0 </= t </= pi/2
but I have to do that for
x and y