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Need help! Related rates

  1. Mar 11, 2008 #1
    Need help! Related rates!!!

    1. The problem statement, all variables and given/known data
    I am stuck somewhere on a related rates problem, i think that i am missing something rather obvious, but i cannot figure out so far.

    -A street light is mounted at the top of the 15 ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole.



    2. Relevant equations



    3. The attempt at a solution
    Here it is what i did so far. I drew a triangle to describe the situation, i let one side of the triangle be 15 ft, i let x be the distance from the pole to the man, and i let y be the length of the shadow of the man. Then from the similarity of the triangles i got this relationship

    [tex]\frac{15}{x+y}=\frac{6}{y}[/tex] i tried to rearrange this a little and i came up with

    [tex]y=\frac{2}{3}x[/tex]

    From the data i also know that [tex]\frac{dx}{dt}=5[/tex]
    I also know that i should let [tex]x=40[/tex] after i come up with a relationship between the rate at which the man is moving and the rate of change of the tip of his shadow.

    Here it is where i am stuck, for if we just implicitly differentiate [tex]y=\frac{2}{3}x[/tex] with respect to time it does not work. What am i missing here??

    Any help would be really appreciated!!
     
    Last edited: Mar 11, 2008
  2. jcsd
  3. Mar 11, 2008 #2

    HallsofIvy

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    Why doesn't differentiating work? dy/dt= (2/3)dx/dt and you know that dx/dt= 5 ft/sec. It looks like dy/dt= 10/3 ft/sec. Why is that wrong?
     
  4. Mar 11, 2008 #3

    tiny-tim

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    Hi sutupidmath!

    You're asked for the speed of the tip of the shadow, which is x + y, and you've only worked out the "speed" of the length of the shadow - so you've got the shadow moving slower than the man, and presumably ending up behind him!

    Is that what happened to Peter Pan? :smile:
     
  5. Mar 11, 2008 #4
    This problem is from Stewart's, it tripped me up for a whole week. Makes a lot more sense to me now though, and has helped to solve other related rate problems ... so great problem.

    Should be set up as ...

    [tex]\frac{d}{dt}(x+y)=\frac{d}{dt}\left(\frac 5 3 x\right)[/tex]
     
    Last edited: Mar 11, 2008
  6. Mar 11, 2008 #5
    Yeah, i realized that what i actually should have changed in my approach is that i should have let x be the length of the shadow, (and not y) and let y be the distance of the man from the pole. But, the reason that i also got confused, is that we used nowhere the fact that the man is 40 ft from the pole when we are measuring the rate at which the tip of his shadow moves. In other words, even if we were not given this extra information, we still could have solved this problem? So why did they add this extra info, there? So is this extra information here given basically to trick us, or? For if it would not be given, i could get to the result way easier.

    Edit: I think that the rate of the tip of the man's shadow will be constant, right? so no matter how long from the pole the man moves, the rate of the tip of his shadow will still be 25 ft/s. For if this is not the case then, i still think we are missing something there, and we have to somehow incorporate the distance of the man from the pole...
     
    Last edited: Mar 11, 2008
  7. Mar 11, 2008 #6
    Because the answer at the end of the book is 25/3, and we used nowhere the fact that the man is 40 ft from the pole when we calculated the rate at which the tip of his shadow moves.!!
     
  8. Mar 11, 2008 #7

    HallsofIvy

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    Boy, I should have read that more closely! Yes, his shadow is lengthing by 10/3 ft./sec. and he is walking away from the light at 5 ft/sec so the tip of his shadow is moving away from the light at 10/3+ 5= 10/3+ 15/3= 25/3 ft/sec. You don't have to use the distance itself because the function involved is linear- the derivative with respect to time is a constant.
     
  9. Mar 11, 2008 #8

    I thought once to go exactly like you did it here, but i wasn't sure as to why aren't we considering the distance, so it thought this would be wrong.

    Thnx all of you!
     
    Last edited by a moderator: Mar 11, 2008
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