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Related rates(shadow problem)

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    A sandbag is dropped from a balloon at a height of 60m when the angle of elevation to the sun is 30 degrees. Find the rate at which the shadow is at a height of 35 meters


    2. Relevant equations
    the position of the sandbag is given by s(t)=60-4.9t^2


    3. The attempt at a solution
    i tried to find the relation between the shadow's path and the position of the sandbag, but my efforts were futile.
     
  2. jcsd
  3. Feb 10, 2009 #2

    Dick

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    The point below the sandbag, the position of the sandbag and the position of the shadow make a right triangle. Use trig.
     
  4. Feb 10, 2009 #3
    i know that but i need help trying to solve it
    i don't know how to relate the shadow to something else
     
  5. Feb 10, 2009 #4

    Dick

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    One leg of your triangle has length s(t). You want to figure out the length of the other leg. One of the angles in your triangle is 30 degrees.
     
  6. Feb 10, 2009 #5
    i got the hypotenuse to be 70 and b to be 61 since a would be 35 since the problem asks when the height is 35
    is this right and if so, how would i go from there im still having trouble
     
  7. Feb 10, 2009 #6

    Dick

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    What's the relation between the vertical leg (s(t)) and the horizontal leg? Which trig function would be good to use? One of them is the ratio between the two legs.
     
  8. Feb 10, 2009 #7
    im sorry if the problem doesn't make sense
    s(t)=60-4.9t^2 is the instantaneous position of the sandbag in the air, not a leg.
    I think that's what makes this problem so confusing
    so to find the sides of the triangle, i did sin(30)/35= the hypotenuse and tan(30)/35=the base of the triangle since the problem says when the sandbag is 35 feet in the air, giving us the 35 to use.
     
  9. Feb 10, 2009 #8

    Dick

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    s(t) is the position of the sandbag at time t. It's also the length of the vertical leg at time t. If h(t) is the length of the horizontal leg at time t, then yes, s(t)/h(t)=tan(30 degrees). So s(t)=tan(30 degrees)*h(t). The rates are then related by d/dt(s(t))=d/dt(h(t))*tan(30 degrees). The vertical rate is equal to the horizontal rate times tan(30 degrees). You need to find the vertical rate when s(t)=35 and then solve for the horizontal rate. The other lengths aren't important.
     
  10. Feb 10, 2009 #9
    okay so would 60-4.9t^2=xtan(30) where x=horizontal length so if you differentiate that you would get -9.8t=(dx/dt)tan(30)+xsec^2(30)
    is this right and if so where would i go from here
    im sorry for being a pain but i really dont understand this problem
     
  11. Feb 10, 2009 #10

    Dick

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    Don't worry about being a pain, but no, you don't understand the problem. But you are getting closer. You have to figure out what the rate -9.8*t is at the point where 60-4.9*t^2=35. That means you have to solve the latter equation for t and plug it into the first equation to get the vertical rate. Once you have that, (vertical rate)=(horizontal rate)*tan(30). You don't have to differentiate the trig function, it's a constant!
     
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