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Related Rates - Sort of confused

  1. Dec 4, 2007 #1
    [SOLVED] Related Rates - Sort of confused

    1. The problem statement, all variables and given/known data

    There are two buildings. One 20' high and other 40' high with a 60' distance between them. A person walking between them creates a theta with the buildings (see pic)

    PICTURE: http://allyoucanupload.webshots.com/v/2003959771064888674

    Now i am asked to find maximum theta possible if the person is walking 4ft/s to the left.

    2. Relevant equations

    ?

    3. The attempt at a solution

    I am kind of lost in coming up with the relation between theta and the two buildings. I know once i get the relation in terms of theta, i can take the derivative and find when the derivative = 0 and see if thats a max.

    But any help with finding this equation would be greatly appreciated.

    Thanks!
     
  2. jcsd
  3. Dec 4, 2007 #2

    HallsofIvy

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    Are you sure that you have stated the problem correctly? The "maximum theta" will occur at a specific position no matter how fast the person is moving.

    Let x be the distance the person is from the building on the right, in feet. Then the distance from the building on the left is 60- x.
    Let [itex]\phi[/itex] be the angle the line from the top of the building on the left to the person makes with the ground. Then [itex]tan(\phi)= 20/(60-x)[/itex] so [itex]\phi= arctan(20/(60-x))[/itex].
    Let [itex]\psi[/itex] be the angle the line from the top of the building on the right to the person makes with the ground. Then [itex]tan(\psi)= 40/x[/itex] so [itex]\psi= arctan(40/x).

    Of course, [itex]\theta[/itex] is just [itex]\pi- \phi- \psi= \pi- arctan(20/(60-x))- arctan(40/x)[/itex]. To find the x that gives the maximum value for [itex]\theta[/itex], differentiate that and set equal to 0.

    Again, I don't see that how fast the person is walking has anything to do with that.
     
  4. Dec 4, 2007 #3
    o ok thx!
     
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