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Math Help - Calculus again!!!

  1. #1
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    Calculus again!!!

    A ship is sailing due north at 12 kn when it observes another ship dead at a distance of 15 n. mi. The second ship is sailing due east at 9 kn. What is the closest distance of approach of the two ships?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Raina View Post
    A ship is sailing due north at 12 kn when it observes another ship dead at a distance of 15 n. mi. The second ship is sailing due east at 9 kn. What is the closest distance of approach of the two ships?
    immagine these two boats on a coordinate system. The first boat is at
    (0,-15) and the is at the origin.

    the coordinate of the 1st boat is given by the equation

    (0,12t-15)

    and the coordinate of the 2nd is

    (9t,0)

    We can find the distance between using the pythagorean theorem.

    d^2=(0-9t)^2+((12t-15)-0)^2

    d=\sqrt{225t^2-360t+225}

    d=3\sqrt{25t^2-40t+25}

    taking the derivative we get

    d'=\frac{3(25t-20)}{\sqrt{25t^2-40t+25}}

    setting the numerator equal to zero we get

    0=3(25t-20) \iff 0=25t-20 \iff 20 =25t \iff t=\frac{4}{5}

    d=3\sqrt{25(\frac{4}{5})^2-40(\frac{4}{5})+25}

    d=3\sqrt{16-32+25}=3\sqrt{9}=9

    Yeah!!

    I hope this helps
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