Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Related rates solution

  1. Mar 20, 2006 #1
    "A street light is mounted at the top of a 15-ft-tall pole. A man 6ft tall walks away from the pole with a speed of 5ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?"
    This is how I did it:
    [tex]\frac{15}{6} = \frac{x+y}{y}[/tex]
    [tex]0 = \frac{y(\frac{dy}{dx} + \frac{dy}{dt}) - (x + y)\frac{dy}{dt}}{y^2}[/tex]
    [tex]y\frac{dx}{dt} - x\frac{dy}{dt} = 0[/tex]
    [tex]\frac{dy}{dt} = \frac{y}{x} \frac {dx}{dt}[/tex]
    [tex]\frac{dy}{dt} = \frac{6}{9}(5)[/tex]
    [tex]\frac{dy}{dt} = \frac{30}{9} = \frac{10}{3} ft/s[/tex]
    then, since the guy is also moving at 5ft/s, add 5 to dy/dt, and the answer is:
    \frac{25}{3} ft/s
    This is the right answer. However, my solution manual does it another way. They find y = (2/3)x, and then say that the tip of the shadow moves at a rate of:
    [tex]\frac{d}{dt}(x + y) = \frac{d}{dt}(\frac {5}{3} x) = \frac {5}{3} (5) = \frac {25}{3} ft/s.
    Are both ways correct? I was trying to explain my solution to someone, but now I'm confused as to whether my solution is correct.

    According to my way, why don't you simply stop at
    \frac{10}{3} ft/s
    I'm thinking that this is the rate of the shadow moving relative to the man. But to find the rate of the shadow relative to the street light (or the earth), you must add the speed of the man. Is my reasoning correct??
    Last edited: Mar 20, 2006
  2. jcsd
  3. Mar 20, 2006 #2
    sorry, i'm new to latex... I'm trying to fix this at the moment
    I think that's good. I couldn't get new lines in there for some reason...
    Last edited: Mar 20, 2006
  4. Mar 21, 2006 #3


    User Avatar
    Science Advisor

    I have a complaint about the way you did this: You have "x" and "y" in your formulas but there is no "x" or "y" mentioned in the problem. Always start "word problems" by stating what your variables represent in the problem (perhaps by drawing a picture and labeling the picture)! If I were your teacher, I would warn you one or two times about that and then start deducting points.

    I assume that x is the distance the man is from the light pole and y is the length of his shadow. What you have done is correct. (If it were me I would have mutltiplied through by the denominators to clear the fractions: 6x+ 6y= 15 so 6x= 7y. I hate fractions and I hate the quotient rule even more!)

    Yes, both methods are correct. The problem asked for the rate at which the tip of the shadow is moving away from the light post. The text does that by finding the rate of change of (x+ y) which is the total distance from the light post to the tip of the shadow. You determined the rate of change of y, the length of the shadow, and then added on the rate at which the distance of the person from the light post, x, is changing.

    In simple terms, the text calculated d(x+y)/dt and you calculated
    dx/dt+ dy/dt. They are, of course, the same.
  5. Mar 21, 2006 #4
    Sorry about those variables. You should know that I was not copying exactly what I wrote on my paper. I was thinking too fast. I did draw a diagram and did state what the variables represented. I did make x the distance from teh pole to the man, and y the length of the shadow.

    Thanks for confirming my answer!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook