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Homework Help: Related Rates.

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data

    A spotlight on the ground shines on a wall 12 m away. If a man 2m tall walks from the spotlight toward the building at a speed of 1.6m/s, how fast is the length of his shadow on the building decreasing when he is 4m from the building?

    2. Relevant equations

    x2 + y2 = z2

    3. The attempt at a solution

    So the x and y values are decreasing, the x at 1.6m/s which is dx/dt with respect to time. We want to find out what dy/dt is which is how fast the shadow is decreasing.

    so using x2 + y2 = z2

    plug in 12m for x

    144 + y2 = z2

    implicit differentiate that equation and I get

    2y*(dy/dt) = 2z*(dz/dt)

    solve for (dy/dt)

    (dy/dt) = (2z*(dz/dt))/2y

    now I believe we plug in 2m for y at that point which would be

    (dy/dt) = (2z*(dz/dt))/4

    this is where I start to get lost. I don't know what to plug in for z, or dz/dt to get dy/dt
  2. jcsd
  3. Oct 12, 2009 #2


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    rather than jumping into pythagoras, i would start by thinking about similar triangles and what you know in each
  4. Oct 12, 2009 #3
    quick question about the question, I copied the exact question, the question isn't saying that the shadow at the beginning is 12 m high right? It's saying that the distance between the light and the building is 12 m apart?
  5. Oct 12, 2009 #4


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    yeah teh wya i read it anyway, so to get what i was saying:

    draw the light man & wall, then draw a line form the light through the top of the man, then look at where it hits the wall, this will be the top of the shadow... now think about those similar triangles
  6. Oct 12, 2009 #5
    Okay, so I drew the picture and I got the height of the shadow where the man is 4 m away to be 3m. with 2/8 = y/12
  7. Oct 13, 2009 #6
    Okay, I was looking at it again and here's what I've come up with.

    So y/2 = 12/(12-x)

    I solved for y

    y = 24/(12-x)

    Then I differentiated it

    dy/dt = -24(12-x)-2 * -1(dx/dt)


    dy/dt = (24*(dx/dt))/(12-x)2

    then plug in dx/dt which is 1.6
    and plug in x which is 4

    dy/dt = 38.4/64 m/s

    This seems like a mistake for some reason?
    Last edited: Oct 13, 2009
  8. Oct 13, 2009 #7


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    how come?

    note that dx/dt is negative, so will be dy/dt
  9. Oct 13, 2009 #8
    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?
  10. Oct 13, 2009 #9


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    that looks ok to me as well, note its the same numerical answer, just formulated differently
  11. Oct 13, 2009 #10
    That wouldn't change my answer would it?
  12. Oct 13, 2009 #11


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    well the way you defined y, it make the rate of change negative, but the magnitude is correct...

    so your number should be negative, ie the shadow is shrinking in size as the man nears the wall
  13. Oct 14, 2009 #12
    So I tried doing the problem from scratch again, and this time I received a different answer :P.

    y/2 = 12/(12-x)

    Differentiated it right away

    12(dy/dt) - x(dy/dt) + y(dx/dt) = 0
    12(dy/dt) - x(dy/dt) = -y(dx/dt)
    dy/dt = -y(dx/dt)/12-x
    Then I received 4.8/8 as the rate of change.
  14. Oct 14, 2009 #13
    Haha, nevermind, I just divided both of the numbers and got .6, so they're both the same answer.
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