darkknight:
precalc
Shadow:
@dude You remember this stuff?
darkknight:
@dude plz help
darkknight:
@dude!
hep:
What is the question
hep:
Dark knight
darkknight:
Question is right there 1c
DanielleLove:
you want to just solve 1c?
darkknight:
yes
darkknight:
I think I got it! So we know the center is at (0,237.42725) we can use the formulas to model the function of x and y respectively 233.42725(cos((2pi/5)t-.06283) 233.42725(sin((2pi/5)t-.06283) at 6 sec she is at the point (85.93, 454.4626) the center is at (0,237.42725) I got a slope of 2.5722722! someone check my work plz
DanielleLove:
did you check this? A professor already explained it online just with different numbers https://sites.math.washington.edu/~ep2/classes/120/problems/Ch16part1
hep:
Well from the way that you typed it I can see where you get your numbers but i also agree with what @DanielleLove posted
darkknight:
No, I didn't check that
darkknight:
radius is 233.42725 17600ft/min)/(2pi(r)ft/rev = 12RPM 24pir=17600
hep:
Yes
darkknight:
It takes 1.3 sec to get to the top 1Rev/5sec 1.3/5 = .26 of circumference = 1144/3 1144/3= pi/180)(theta)(233.42725) =93.6 - 90 to get to x-axis. She is 3.6 degrees away from the x-axis -3.6pi/180 = -.06283
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