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Related Rates help please!

  1. Nov 3, 2007 #1
    Related Rates!! help please!

    A and B are walking on straight paths that meet at right angles. A approaches at 2m/sec; B moves away from the intersection at 1m/sec. At what rate is the angle [tex]\vartheta[/tex] changing when A is 10m from the intersection and B is 20m from the intersection. Ans in degrees per second.


    attempted solving the question using tan ratio where:
    tan[tex]\vartheta[/tex] (t)= A/B= (10-2t)/(20+t)

    I know i have to take the derivative of this equation in order to get d[tex]\vartheta[/tex]
    /dt, but how?
     
  2. jcsd
  3. Nov 3, 2007 #2

    Dick

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    Just do it. Differentiate tan(theta(t)) and don't forget the chain rule. The chain rule gives you the dtheta(t)/dt part.
     
  4. Nov 3, 2007 #3
    thx for you response. ok, so

    sec^2(theta) (d(theta)/dt)= (B(dA/dt)-A(db/dt))/B^2

    where A=10-2t and B=20+t

    if i'm doing this right, then how would i solve for t?
     
  5. Nov 3, 2007 #4

    Dick

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    The way you've set it up, t=0, yes? All you need is theta and A and B.
     
  6. Nov 4, 2007 #5

    HallsofIvy

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    t is "when A is 10m from the intersection and B is 20m from the intersection." As Dick sad, since you cleverly used 10-2t and 20+t as the lengths of the sides, A= 10 and B= 20 when t= 0.
     
  7. Nov 6, 2007 #6
    hey, thought i would do this question randomly for some exam study, heres how i did it (not sure if its right, hopefully is though, lol).

    EDIT: sry stuffed up my working, here new working
     

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    Last edited: Nov 6, 2007
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