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Math Help - Road and Railroad Related Rates Problem

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    Road and Railroad Related Rates Problem

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    Re: Road and Railroad Related Rates Problem

    Quote Originally Posted by Chaim View Post
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    You have the right idea, but when t=0 you train is negative 15 units from the crossing still. The way you set up the equations, the trains crosses when t=10.


    So ten minutes later from then is when t=20 in your model

    P.S. I didn't check your algebra or calculus.
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    Re: Road and Railroad Related Rates Problem

    Quote Originally Posted by TheEmptySet View Post
    You have the right idea, but when t=0 you train is negative 15 units from the crossing still. The way you set up the equations, the trains crosses when t=10.


    So ten minutes later from then is when t=20 in your model

    P.S. I didn't check your algebra or calculus.
    So when t = 0, it is negative 15. Was that right when I did 90(t - 10/60).
    Cause I'm kind of still confused how my answer is still pretty far off from the actual answer :O
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    Re: Road and Railroad Related Rates Problem

    Quote Originally Posted by Chaim View Post
    So when t = 0, it is negative 15. Was that right when I did 90(t - 10/60).
    Cause I'm kind of still confused how my answer is still pretty far off from the actual answer :O
    What you have is a right triangle

    c^2=a^2+b^2

    What you want to find is the rate of change of the hypotenuse.

    Then taking the derivative with respect to time gives

    2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}

    So we need to know a,b and c.

    The car has traveled 12 miles (a=12) and the train has traveled 9 miles (b=9) and by the Pythagorean theorem c=15.

    Plugging all this in gives

    2(15)\frac{dc}{dt}=2(12)(60)+2(9)(90) \iff 30\frac{dc}{dt}=3060 \iff \frac{dc}{dt}=102
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    Re: Road and Railroad Related Rates Problem

    Quote Originally Posted by TheEmptySet View Post
    What you have is a right triangle

    c^2=a^2+b^2

    What you want to find is the rate of change of the hypotenuse.

    Then taking the derivative with respect to time gives

    2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}

    So we need to know a,b and c.

    The car has traveled 12 miles (a=12) and the train has traveled 9 miles (b=9) and by the Pythagorean theorem c=15.

    Plugging all this in gives

    2(15)\frac{dc}{dt}=2(12)(60)+2(9)(90) \iff 30\frac{dc}{dt}=3060 \iff \frac{dc}{dt}=102
    Thanks for replying and showing how you got the answer
    But just wondering, how did you get 12 and 9 :O
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