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Related rates and a lamp

  1. Sep 7, 2005 #1
    A lamp is located on the ground 10 m from a building. A man 1.8 m tall walks from the light toward the building at a rate of 1.5 m s⁻¹. What is the rate at which the man's shadow on the wall is shortening when he is 3.2 m from the building ? Give your answer correct to two decimal places. [0.58 m/s]
    I tried using similar triangles first to find the height of the lamp but nothing came out of that.
    let h be the height of the lamp
    {h \over{x+10}} = {1.8 \over {x+3.2}}[/tex]

    h = { {1.8x+18} \over {x + 3.2}}

    How is the velocity of the man related to the height of the shadow? Or, what else should I find?

    Also, shouldn't the shadow of the man on the wall become taller as he approaches the building?
  2. jcsd
  3. Sep 7, 2005 #2


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    What is x? The distance the man is from the wall or the distance the man is from the lamp?

    First rule for a problem like this is to draw a picture and label it carefully. I took x as the distance the man is from the lamp, because that gave simpler distances. Drawing a line from the lamp to the top of the man's head to the wall (i.e. the light ray that gives the top of the shadow) I see two similar right triangles. the smaller has the man as opposite side, so height 1.7m and base x. The larger has the wall as opposite side. Taking h as the height of the shadow, the height of this right triangle is h and the base is 10m. Since they are similar triangles, [tex]\frac{h}{10}= \frac{1.8}{x}[/tex].

    You appear to have tried the same thing but have "10+ x" and "3.2+ x". I see no reason to add the distance from the lamp to the wall, especially since the man is standing between the lamp and the wall. If your x is the distance from the man to the wall (rather than the lamp) then the formula would be [itex]\frac{h}{10}= \frac{1.8}{10-x}[/itex].

    I finally got around to looking at your picture. You have the lamp on the opposite side of the wall from the man!? How is the light getting through the wall? Where is it casting the shadow?

    Also, the "3.2 m" is (related to) the specific value of x at which you want to find the rate of change. It has nothing to do with the general formula connecting h and x.

    Once you have a general "static" formula for two quantities, you can find a formula relating their "rates of change" by differentiating with respect to t (using the chain rule since there is no t in the formula itself).
    With my formula [itex]\frac{h}{10}= \frac{1.8}{x}[/itex], the derivative of [itex]\frac{h}{10}[/itex] with respect to x is [itex]\frac{1}{10}\frac{dh}{dt}[/itex] and the derivative of [itex]\frac{1.8}{x}= 1.8x^{-1}[/itex] is [itex]-1.8x^{-2}\frac{dx}{dt}= \frac{-1.8}{x^2}\frac{dx}{dt}[/itex].
    That is, from [itex]\frac{h}{10}= \frac{1.8}{x}[/itex] we get [itex]\frac{1}{10}\frac{dh}{dt}=\frac{-1.8}{x^2}\frac{dx}{dt}[/itex]. Now, set x= 10- 3.2= 6.8 (since my x is the distance from the lamp to the man, not from the wall to the man), set [itex]\frac{dx}{dt}= 1.5 m/s[/itex] and solve for [itex]\frac{dh}{dt}[\itex].

    And, yes, the man's shadow gets smaller as he walks toward the wall. When he is right at the wall, his shadow is exactly his height. When he is very close to the lamp, he is blocking a lot of the light and his shadow is huge.
    Last edited: Sep 7, 2005
  4. Sep 7, 2005 #3
    Many thanks for your help.

    I assumed that the lamp had a height and that's why I got everything wrong. Now I understand...

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