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What do you think? Give us your opinion. Anonymous comments allowed.
User avatar #4 - coolcalx (02/19/2013) [-]
I'm confused as **** . the straight sides of the square should give you a linear graph, shouldn't they? why are they curved like x^3 graphs?
#12 to #4 - anon (02/19/2013) [-]
You're stupid, imagaine the x doesn't change while the y does.
User avatar #10 to #4 - insaniacfedde (02/19/2013) [-]
because they'd have to alter speed im order to keep up with the circle (I'ts less evenly shaped)
#9 to #4 - peikkonaama has deleted their comment [-]
#5 to #4 - puredeliciousness (02/19/2013) [-]
I think it has something to do with the middle of the sides of the square being closer to the center point than the corners. But I could be wrong.
User avatar #7 to #5 - coolcalx (02/19/2013) [-]
then again, maybe you're right. I may be overthinking it.
User avatar #6 to #5 - coolcalx (02/19/2013) [-]
I just took a Cal B test earlier today, so I'm thinking derivatives. shouldn't... shouldn't it just be a y=x line?
User avatar #13 to #6 - anniethreeone (02/19/2013) [-]
It would be linear if the point tracing the square had a constant linear velocity, but it's the angular velocity, so the graph is actually dy/dtheta -- how much the vertical coordinate changes for a change in degrees around the centre of rotation.
User avatar #15 to #13 - anniethreeone (02/19/2013) [-]
Corrections: it's constant angular velocity and the graph is y(theta). The slope of the line is dy/dtheta which is why it's curved rather than constant.
User avatar #20 to #15 - coolcalx (02/19/2013) [-]
that makes sense. thank you.
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