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Math Help - Related Rate

  1. #1
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    Related Rate

    A lighthouse is .8 miles away from shore. The light rotates 6 times a minute. How fast in MPH is the spot of light moving along the shore when it is .6 miles away?


    So I did the problem and I'm not sure but I think I got 1/.6. is that right?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Related Rate

    Let's let D be the perpendicular distance from the lighthouse to the shore (a straight line) and x be the distance of the spot of light from the point on the shore intersected by the distance line segment. Let \theta be the angle subtended by the distance line segment and the beam of light. Let \omega=\frac{d\theta}{dt}. We may then state:

    \tan(\theta)=\frac{x}{D}

    Differentiating with respect to time t, we find:

    \omega\sec^2(\theta)=\frac{1}{D}\cdot\frac{dx}{dt}

    Solving for \frac{dx}{dt} and using \sec(\theta)=\frac{\sqrt{x^2+D^2}}{D} we have:

    \frac{dx}{dt}=\frac{\omega(x^2+D^2)}{D}

    For this problem, we are given:

    \omega=6\cdot2\pi\frac{\text{rad}}{\text{min}} \cdot\frac{60\text{ min}}{1\text{ hr}}=720\pi\,\frac{1}{\text{hr}}

    x=\frac{3}{5}\,\text{mi}

    D=\frac{4}{5}\,\text{mi}

    Now, plug 'n' chug.
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