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Math Help - Related Rates

  1. #1
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    Related Rates

    The shadow casts by a man standing 1 m from a lamppost is 1.2 m long. If the man is 1.8 m tall and walks away from the lamppost at a speed of 120 m/min, at what rate is the shadow lengthening after 5 seconds?

    This is my work:

    I used Pythagorean theorm
    a^2 + b^2 = c^2
    differentiating
    2a da/dt + 2b db/dt = 2c dc/dt
    2(1) da/dt + 2(1.8) db/dt = 2(1/12* -120) dc/dt
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    The shadow casts by a man standing 1 m from a lamppost is 1.2 m long. If the man is 1.8 m tall and walks away from the lamppost at a speed of 120 m/min, at what rate is the shadow lengthening after 5 seconds?

    This is my work:

    I used Pythagorean theorm
    a^2 + b^2 = c^2
    differentiating
    2a da/dt + 2b db/dt = 2c dc/dt
    Related Rates Example
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  3. #3
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    Ok, so I ended up getting 14.4 m/min as my answer.. but the answer on the worksheet says 144 m/min..

    Here's how I got my answer: we want to find how the length of the shadow is changing with time.. so s shadow is 1.2 m, and at 5 sec or 1/12 min, then (1.2 ) (1/12) = 14.4 m/min

    ?
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  4. #4
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    Quote Originally Posted by skeske1234 View Post
    Ok, so I ended up getting 14.4 m/min as my answer.. but the answer on the worksheet says 144 m/min..

    Here's how I got my answer: we want to find how the length of the shadow is changing with time.. so s shadow is 1.2 m, and at 5 sec or 1/12 min, then (1.2 ) (1/12) = 14.4 m/min

    ?
    the 5 seconds doesn't matter.

    ratio ... (distance from post)/(shadow length)

    \frac{x}{s} = \frac{1}{1.2}

    s = 1.2 x

    \frac{ds}{dt} = 1.2 \cdot \frac{dx}{dt}

    \frac{ds}{dt} = 1.2(120) = 144 m/min
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