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

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

    Differentiating with respect to time $\displaystyle t$, we find:

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

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

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

    For this problem, we are given:

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

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

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

    Now, plug 'n' chug.
    Thanks from camjenson
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