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Related rates

  1. Oct 1, 2007 #1
    1. The problem statement, all variables and given/known data
    "On a certain clock the minute had is 4in long, and the hour hand is 3in long. How fast is the distance between the tips of the hands changing at 9 o'clock?"

    2. Relevant equations
    - a[tex]^{2}[/tex] + b[tex]^{2}[/tex] = c[tex]^{2}[/tex]
    - Law of Cosines?

    3. The attempt at a solution
    Ok i drew a clock, and the 3-4-5 triangle positioned at 9:00. The distance between the 2 tips is 5in, but how would I find the rate of change between them? If the distance is marked by x, then I'm solving for dx/d[tex]\Theta[/tex], right? For every 6 degrees the minute hand moves, the hour hand moves one. I'm thoroughly confused........
  2. jcsd
  3. Oct 2, 2007 #2


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    I would think of this as two concentric circles with radii R and r (R > r). The inner circle is stationary, but the outer circle is rotating. At t = 0 two points (one on each circle) lie on a straight line through the origin so their distance is d = R - r. At t = [itex]\epsilon[/itex], the outer circle has rotated some, so the distance between the two points has increased. Find (d([itex]\epsilon[/itex]) - d(0))/[itex]\epsilon[/itex] as [itex]\epsilon[/itex] --> 0.
  4. Oct 3, 2007 #3


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    Welcome to PF!

    What I would do is the following. There are two lines of different lengths (r and R) rotating about a common origin at different frequencies. If the tips of the lines are points a and b, I'd write equations that represent the x,y locations of a and b (with respect to t) using initial conditions such that the postions of the lines are at 9:00 at t=0. I'd then write an expression for the distance between points a and b, take the derivative and evaluate it at t=0. Chain rule involved.
    Last edited: Oct 3, 2007
  5. Oct 3, 2007 #4


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    In general, the angle formed by the hands is not a right angle (it happens to be at 9:00 but not immediately after). Your second thought was correct: you can use the cosine law to write the distance between the tips of the hands as a function of the angle between them.
  6. Oct 3, 2007 #5

    D H

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    Why is the chain rule needed here? Defining [itex]\hat x[/itex] as the 12:00 position, [itex]\hat y[/itex] as the 3:00 position, and [itex]\theta_m[/itex] as the angle between the 12:00 position and the minute hand measured clockwise, then the tip minute hand is at

    [tex]\vec m = 5\text{in}(\cos \theta_m \hat x + \sin \theta_m \hat y)[/tex]

    The minute hand makes one revolution per hour. Thus [itex]\dot \theta_m = 2\pi/\text{hr}[/tex]. Differentiating the expression for [itex]\vec m[/itex],

    [tex]\frac{d}{dt}\vec m = 10\pi\text{in}/\text{hr}(-\sin \theta_m \hat x + \cos \theta_m \hat y)[/tex]

    You can treat the hour hand similarly. Taking the difference between these two velocity vectors yields the tip-to-tip velocity. Apply the specific values for [itex]\theta_m[/itex] and [itex]\theta_h[/itex], take the magnitude, and voila, you have the distance rate of change.

    The above is not correct. Correction is imminent.
    Last edited: Oct 3, 2007
  7. Oct 3, 2007 #6

    D H

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    I should know better. The magnitude of a velocity vector does not equal to the time derivative of the magnitude underlying displacement vector.

    Suppose [itex]\vec d[/itex] is some displacement vector (e.g., the vector from the end of the hour hand to the end of the minute hand). Denote [itex]d=||\vec d||[/itex] as the magnitude of this displacement vector. For an inner product metric space,

    [tex]d^2 \equiv \vec d \cdot \vec d[/itex]

    Differentiating with respect to time,

    [tex]2d\dot d = 2\vec d \cdot \dot {\vec d}[/itex]

    Solving for the time derivative of the magnitude of the displacement vector,

    [tex]\dot d = \frac{\vec d \cdot \dot {\vec d}}{d}[/itex]

    The magnitude of the velocity vector, [tex]\sqrt{\dot {\vec d} \cdot \dot {\vec d}}[/tex] is obviously not the same as the time derivative of the magnitude of the displacement vector.
    Last edited: Oct 3, 2007
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