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Homework Help: What's wrong with my similar triangle for a related rate question?

  1. Jul 18, 2012 #1
    1. The problem statement, all variables and given/known data

    (I don't know how to draw a picture here so I will just explain the dimensions of the similar triangles)

    A lightbulb on a floor shines at a 2m tall guy who is walking away from it towards a wall. As he gets closer to the wall, his shadow gets smaller and smaller on the wall. The lightbulb is 10m from the bottom of the wall. This makes a triangle where the base is 10m and the height of it is the height of the shadow (y). At what rate is the height of the shadow decreasing if the guy walks at a rate of 1.2m/s and is 7m (we will call x) from the lightbulb. This information creates a similar small triangle with a base of x = 7m (his distance from the lightbulb) and a height (his height) of 2m.

    3. The attempt at a solution

    So, in my notes my instructor found the related rate by using a similar triangle method of y/10 = 2/x When differentiated it looks like y' = -20/x^2(x') = -20/49(1.2) = -(24/49)m/s. I was fine with this, and it makes sense to me until I tried a different similar triangle method. I tried putting y/(x+3) = 2/7 which gives me the correct x and y dimensions of the triangles. But when I differentiate the equation, I don't get the right answer. Instead, I get:
    y = (2x)/7 + 3/7
    y' = (2(x'))/7 + 0
    y' = (2(1.2))/7
    y' = 12/35 which has the wrong sign and is a constant change - not an accelerated one.

    I never want to stop bashing my head into the wall until I understand why this doesn't work.
  2. jcsd
  3. Jul 18, 2012 #2
    so y/10=2/x, how do you arrive y' = -20/x^2(x') from that?
  4. Jul 18, 2012 #3
    What do you mean? This is correct isn't it?

    y/10 = 2/x
    y = 20/x
    y = 20x^(-1)
    y' = -20x^(-2)(1.2)
    y' = -20(7)^(-2)(1.2)
    y' = -24/49

    I am quite sure that this is the right answer.
  5. Jul 18, 2012 #4


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    What are the constants in the problem? What are the variables?

    The instantaneous (momentary) x value is 7m, and therefore 10 = x + 3 at that instant. Would this relationship always hold true? What happens when x = 5m or 8m? Remember, the 10m distance is the fixed one.

    Similarly, the ratio of 2/7 is true for that instant, but as the guy gets closer or further away from the bulb, the denominator will change, but the numerator remains constant (the guy does not change in height).
  6. Jul 18, 2012 #5


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    That part is fine.
  7. Jul 18, 2012 #6
    Oh, I misread the parenthese in your original post. It's correct.
  8. Jul 18, 2012 #7
    Ahhhh, I see what you're saying. Is there any general rule that I am breaking here because even though I see my mistake, it is not so obvious that I won't do it again. Or, should I just stick with what is constant when forming the formula (and of course differentiate lone constants to 0)?
  9. Jul 18, 2012 #8


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    The aim is to derive a relationship between y and x that holds no matter how the variables change. I don't know how to explain it any more simply than that.
  10. Jul 18, 2012 #9
    ok, thanks
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