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Write a function that describes this situation (pics coming...) Two circles, radius 1, are drawn so their centers are 3 in. apart. Points marked on the circles rotate at a speed of 1 rotation every 4 seconds. Find a function that describes the distance (d) between the two marked points at any time (t).

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I know it will be periodic, and can calculate for integral values of t (0,1,2,3,4,etc) but can't figure out how to find when t isn't a whole number.
@ParthKohli @agentx5 can you help?
Gear ratios? I still trying to understand this question the way it's worded, but if that's the case it's just a matter of the ratios of the radii, like you would for pulleys or gears.

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Other answers:

Ah wait cycloids...
no, both are rotating clockwise. (prolate, nested function)
Which of course is going to make it look a whole lot like a sine function
yeah, but a cycloid only describes 1 point. I'm looking for distance between the 2 points.
I know that, which is why it'll end up being two of those together making it a sine function when I think about it visually. If they were touching it would be double amplitude and half period.
The tricky part is that they start out with an initial distance that also changes with respect to time
Or does it... Hmm, might just be additive.
this is what i've figured out: \[d(0)= 1\]\[d(1) = \sqrt{13}\]\[d(2) = 5\]\[d(3)= \sqrt{13}\]
They are pulleys, not gears in this case. |dw:1343667494092:dw|
at time t=.5, |dw:1343667478687:dw| so we get roughly a triangle:|dw:1343667609895:dw| by pyth. theorem, \[d=\sqrt{2+2.514} \approx 2.124\]
t(.5) = t(3.5) right?
Input your angular frequency and your initial condition (+1 inch), phase shift is zero here, for the formula see in this link. Your function should match my sketch above. When it does you can plug in any value for t and you'll have your distance at time t. Local max will be 5 every time, local min will be 1 every time. Also review:
Standard form for this parametric is: \[y(t,x) = (A) \sin (\omega t - kx + \phi) + D\] In this case double A for amplitude Phi is zero D = 1
all you need to find is the angular frequency and see how the wave number will cancel out This formula can be used to construct any wave formula, but you'll need to figure out a few facts about the wave. At the moment I'm on a bit of a time crunch today so leave you to the number crunching :-D
@agentx5 @ganeshie8 I figured out d as a function of the angle: \[d(\theta) = \sqrt{(2 \sin \theta)^2 + (3 - 2 \cos \theta)^2}\]Tips on getting it as a function of time? it: \[d(t) = \sqrt{(2 \sin (\frac{\pi x}{2}))^2 + (3-2 \cos (\frac{\pi x}{2}))^2}\]
Angular frequency relates the rate of the angular rotation per time (i.e.: degrees per second, or radians per second, or rpm). That's is how you shall have to relate time. I've tried finding something comparable to help you out but everything I've found has to do with energy, mechanical advantage, and torque, which doesn't help much here...
That actually looks plausible as an answer, but double check the phase shifts for sin and cosine. The net function should be similar to what I described initially, plot it on a calculator on on Wolf or something and see what it looks like. is my preferable graphing tool...and it works!
Highest distance should be when they're both "pointing" away, and shortest distance should be when they are "pointing" towards. I would test the horizontal and vertical cases as test angles to make sure your function works (that's 4 test cases). ;-D
@across is there an easy way to figure this out?

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