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Math Help - Applications of circular functions.

  1. #1
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    Applications of circular functions.(is this to hard because its a graphing question)

    1) A bicycle tire with radius 33cm rotates at a rate of 5 revolutions per second. The top of the valve stem is located 7cm above the surface of the road.

    a) Draw a graph showing the height of the top valve stem above ground level in terms of time. Assume that the valve stem is initially at its lowest position.(i just need to know where to put some stuff and how to get it)

    b) Write the equation of a cosine function that describes the height of the top of the valve stem, y, as a function of the time, x. Use a negative value for A and a phase shift of 0.

    c) What is the height of the valve stem above the ground 10.11 seconds after it is at its lowest point?
    Last edited by Dave19; July 8th 2008 at 06:43 PM.
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  2. #2
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    a) This problem is all about breaking it down. Draw a picture of the wheel on the ground. Relative to the axle, the top of the valve stem just makes a circle as the wheel rotates. The center of the circle is at the axle at 33cm and the radius is 33-7 = 26 cm. At 0 seconds the height is 7. If it rotates at 5 times per second, then at 1/5=0.2 seconds the wheel and bike stem are back at the original position. The height is given by an offset sine curve. So draw a sine curve that oscillates between y=7 and y=33+26=59 and does one full up and down every 0.2 seconds.

    Make sense?

    b) So we just have to shoe horn a cosine curve to look like what we just drew. A cosine curve starts at 1 at time 0 and goes down to -1 and back up to 1. cos(2*pi*t) does this full cycle every 1 second. Let's start with the frequency. we want the oscillation every 0.2. cos(2*pi*5 t) (5 is the frequency). Ok, but this curve starts at the top and goes down. The directions said to not phase shift. So -cos(2*pi*5*t) is now doing the right thing. But it is not oscillating between the right numbers. Lets make the amplitude (radius) correct. -26*cos(2*pi*5*t) Now it is bouncing up and down by the right amount. Just at the wrong height. The center of our cos curve oscillation is 0, where it should be 33. So
    33-26cos(2*pi*5*t)

    I did this just from reasoning, but you make have been starting from a formula like A cos(2*pi*F*(t+S)) + C. If so, sorry for the long drawn out explanation.

    c) well it is at it's lowest point at 0, so just plug in 10.11 for t into the formula above.
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