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Change in Angle of Elevation

  1. Oct 27, 2013 #1

    Qube

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

    A weather balloon is rising vertically at a speed of 10 meters per second. An observer is 40 meters away from the launch point of the balloon. the distance between the observer and the balloon in the air is 50 meters. How fast is the angle of elevation changing?

    2. Relevant equations

    Given the length of the hypotenuse is 50, and the bottom leg of the triangle is 40, it follows that the height of the balloon from its launch site is 30 meters.

    3. The attempt at a solution

    tan[itex]\theta[/itex] = opposite side "a" / adjacent side "b" = 30 / 40 = 3/4

    Taking the derivative of both sides, we get d[itex]\theta[/itex]/dt*sec^2([itex]\theta[/itex]) = (a'b - b'a) / b^2

    We are given a' is 10, and we know b' is 0 since the observer isn't moving on the ground. Therefore for the right hand side we get [10(40)]/(40^2). We divide this by sec^(theta) and get the answer right?
     
    Last edited: Oct 27, 2013
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  3. Oct 27, 2013 #2

    LCKurtz

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    No. You need to distinguish what the variables are and what is constant. The variables are the height of the balloon, which I will call ##h##, the hypotenuse, which I will call ##c##, and the angle ##\theta##. The distance along the ground is a constant ##40##. So, the equation with the variables would be ##\tan\theta = h/40## or ##h = 40\tan\theta##. You must differentiate this equation before you put in the instantaneous "snapshot" values of 50 and 10.
     
  4. Oct 27, 2013 #3

    Qube

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    I suppose it's dangerous to mix the variables and the "snapshot" values right? I think I get the same answer either method. Is my way simply wrong, in terms of method?
     
  5. Oct 27, 2013 #4

    LCKurtz

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    It's certainly not a good idea. Generally you don't want to use them until after you have the related rates equation. If you put them in too early you may differentiate that variable thinking its derivative is zero when it isn't. That is a very common mistake although you may not have made it.

    You didn't work yours all the way out and I didn't check it; maybe you do get the same answer. I get 4/25. Still, I have suggested a better way. You have treated both a and b in your equation as variables when differentiating when one is actually constant.
     
  6. Oct 27, 2013 #5

    Qube

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    I get 4/25 using either method. Thanks for clarifying! I'll avoid mixing up instantaneous values with the variables from now on. In calculus, it's always the small things that get you (from the fact that critical points aren't really critical points if they're outside the domain of f(x) to this!)
     
  7. Oct 27, 2013 #6

    Qube

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  8. Oct 27, 2013 #7

    LCKurtz

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    I wish everyone who posted images had as good handwriting as you do. You have it correct but I would have one more little nit-pick. You have written$$
    \frac{d\theta}{dt}\sec^2\theta = \frac 1 {40} a'$$ $$
    =\frac{10}{40}$$It would be better to write$$
    \frac{d\theta}{dt}\sec^2\theta = \frac 1 {40} \frac{da}{dt}$$ $$
    \frac{d\theta}{dt}\left(\frac 5 4 \right)^2 =\frac{10}{40}$$That would make it clear you are substituting the snapshot values into both sides of the related rates equation at the same time.
     
  9. Oct 27, 2013 #8

    Qube

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    Thank you! That was just me scribbling some work down; my actual handwriting tends to be neater :)!
     
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