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Related rates waliking away from light towards building

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    a spotlight won the ground shines on a wall 12 m away if am man 2 m tall walks from the spotlight towards the building at a speed of 1.6 m/s how fast is the length of his shadow on the building decreasing when he si 4 m from the building?

    2. Relevant equations
    using relative triangles


    3. The attempt at a solution
    trying to learn related rates as well, this is what i got tell me if i am wrong here...


    draw triangle ABC with A being the light, B being the base of building, and C being top of shaddow/building.

    the second triangle is formend with the man and the light, using ADE, D being the mans feet and E being his head at 2 m height.

    using the 2 meter horizontal from the mans height we have two relative triangles.

    call the range from the light (line AD) x and call the building/shaddow (line BC) y

    using the two triangles we can infer that 2/x=y/12 or xy=24

    if we then differentiate relative to time 0=dx/dt * y + dy/dt * x

    we then plug in 1.6 for dx/dt and 8 for x(range from the light not the building); and y=3 (using xy=24 @ x=8) and solve for dy/dt= -.6m/s which should be negative.

    Is this correct?
     
  2. jcsd
  3. Oct 13, 2009 #2

    lanedance

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    Homework Helper

    wasn't this in the other posting?
     
  4. Oct 13, 2009 #3
    yes i had already posted this, i have a new posting out titled related rates kite
     
  5. Oct 13, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that is correct.
     
  6. Oct 13, 2009 #5

    HallsofIvy

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    Please don't post the same thing more than once!
     
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