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Math Help - Differentiation word problem

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    Differentiation word problem

    A runner sprints around a circular track of radius 100m at a constant speed of 7 m/s. The runner's friend is standing at a distance of 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m ?
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    Grand Panjandrum
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    Quote Originally Posted by cityismine View Post
    A runner sprints around a circular track of radius 100m at a constant speed of 7 m/s. The runner's friend is standing at a distance of 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m ?
    Set the geometry up so that the runners position at time t is:

    p(t)=(100 \cos(0.07 t), 100 \sin (0.07 t))

    That is we take their position at time t=0 to be p(0)=(100,0), the circle is centred at the origin, and we assume the runner is running anti-clockwise.

    We may place the friend at any convienient point 200 m from the centre of the circle, so lets put them at (200,0).

    So now your problem is to find:

     <br />
\frac{d}{dt} \sqrt{(p_1(t)-200)^2+p_2(t)^2}<br />

    when \sqrt{(p_1(t)-200)^2+p_2(t)^2}=200

    RonL
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  3. #3
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    Quote Originally Posted by cityismine View Post
    A runner sprints around a circular track of radius 100m at a constant speed of 7 m/s. The runner's friend is standing at a distance of 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m ?
    If D = distance in meters between the two friends at any time t in seconds, then we are looking for
    dD/dt = rate of change of the D, in m/sec.

    We draw the figure based on the givens. (I do not know how to draw in computers; neither do I know how to use Latex for easy reading of math expressions.)

    It is a triangle whose 3 sides are 100m, 200m and D. The center of the circle is at the intersection of the 100m and 200m sides.
    Let us call the central angle as theta.
    The D side of the triangle is opposite the angle theta.

    Connecting the givens and the D, we will use the Law of Cosines,
    D^2 = 100^2 +200^2 -2(100)(200)cos(theta)
    D^2 = 50,000 -40,000cos(theta) ------------------(1)

    Since we are looking for dD/dt, we differentiate (1) with respect to time t,
    2D*dD/dt = 40,000sin(theta) *dtheta/dt ------------(2)

    In this Eq.2, when D=200m, we don't know theta and dtheta/dt, so we solve for them first.

    For theta, use the Eq.1,
    200^2 = 50,000 -40,000cos(theta)
    theta = arccos(10,000 /40,000) = 75.5224878 degrees ----**

    For dtheta/dt, use the rotational speed of the central angle theta,
    w = dtheta/dt = (tangential velocity)/(radius) = (7m/sec)/100m = 0.07 rad/sec -------------------**

    Hence, substituting all those in (2),
    2(200)*dD/dt = 40,000sin(75.5224878 deg) *(0.07)
    400dD/dt = 2711.088
    dD/dt = 2711.088 /400 = 6.778 m/sec

    Therefore, at the instant when the distance between the two friends is 200 meters, they are separating at the rate of 6.778 m/sec. -----answer.
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