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Related rates of an angle

  1. Sep 19, 2013 #1
    1. The problem statement, all variables and given/known data
    A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when 200ft of the string has been let out?


    2. Relevant equations

    -


    3. The attempt at a solution
    I've tried two different methods to come up with the answer but i'm not sure which one is correct.

    In one case, I've got:

    cosθ = x/200

    And in the other one:

    tanθ = 100/x

    However, when I try to derive both sides of both equations in order to get the result, I found out that the values of dθ/dt are actually different for each equation. For example, in the first case:

    -sinθ (dθ/dt) = 1/200 (dx/dt)

    Clearing this out makes the final result as (-2/25)

    On the other hand:

    sec2 (dθ/dt) = -(dx/dt) / 3

    Clearing this out makes the final result as (-1/50)

    Is there a reason for this incongruity or have I made a mistake in my calculations? Thank you very much guys.
     
  2. jcsd
  3. Sep 19, 2013 #2

    Simon Bridge

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    You should be able to work out which, if either, approaches are correct by writing down your reasoning.
    How did you arrive at those equations?
     
  4. Sep 19, 2013 #3
    Firstly, I made a triangle with angles θ, 90-θ and 90. Then, i put in the sides as 100, 100√3 and 200. So i just set up the equations:

    cosθ = x/200

    tanθ = 100/x
     
  5. Sep 20, 2013 #4

    Simon Bridge

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    Yes OK - those were your math steps - as far as they go ... but not the reasoning behind them.
    It's the reasoning that is important - why did you set up that particular triangle? Why would you expect it to help you find the answer you are looking for?

    It looks like you reasoned that the string was almost straight so you could model it as the hypotenuse of a right-angle triangle. The height of the kite forms the opposite side, which is given as a constant: y=100ft. You are looking for something (what?) when the hypotenuse is: h=200ft.

    It is usually easier to see what's going on if you just keep the variables.

    The horizontal distance is x in your equation? Thus ##\small x=h\cos\theta = y/\tan\theta## - which you got.
    Of course you can find the angle from the third trig relation, ##\small y=h\sin\theta ##, or by solving the simultaneous equations, but that's not what you are asked to find is it?

    How are you factoring in the speed of the kite? Hint: ##\small v=dx/dt (=8\text{ft/s})##
     
  6. Sep 20, 2013 #5

    HallsofIvy

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    "tangent" is defined as "opposite leg over near leg". But the length of string in this problem is the hypotenuse of the triangle, isn't it?
     
  7. Sep 22, 2013 #6

    Dick

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    This was bugging me for a while. I kept doing it over and over again and getting the same as you. But the length of the string, 200 ft, is not a constant if the height is constant. It depends on x. That's your mistake. The cos form is wrong because you can't take 200 to be constant, it's sqrt(100^2+x^2), isn't it? The tan form is ok because it doesn't involve assuming the length is a constant. Now I feel stupid.
     
    Last edited: Sep 22, 2013
  8. Sep 22, 2013 #7

    Simon Bridge

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    @fogfajarash: how are you getting on?

    @dick: you are probably correct about where OP (fogavjarash) made the mistake.
    The cosine form is OK as ling as you realize that the hypotenuse changes with time - to get the relationship between the rates, you have to use the chain-rule, probably with Pythagoras or another trig function.

    The idea was to try to get OP to realize this without actually spelling it out :)
     
  9. Sep 22, 2013 #8

    Dick

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    I know I'm correct. I worked it out that way. I'm just annoyed it took me so long to figure out the problem. Sorry if I gave away too much.
     
  10. Sep 22, 2013 #9
    Thank you guys. I couldn't figure out the problem on my own and I was getting kind of desperate.
     
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