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Homework Help: Please help me solve a multivariable limit question

  1. Apr 20, 2010 #1

    CR9

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

    Find limit as (r, θ)----> (0, pi/2) for the function:

    r= (w secθ)/(secθ+tanθ)^(Vs/Vr)

    Both w and Vs/Vr are constants in this question


    3. The attempt at a solution

    I tried with L'hopital but it didnt turn well as when I differentiate secθ, I got ln (secθ + tan θ)

    which when θ approaches 0 still gives me infinite..

    Please help me....

    I've tried so long to get the answer, but nothing seems to work and I have to pass this up tommorow.
     
  2. jcsd
  3. Apr 20, 2010 #2

    tiny-tim

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    Hi CR9! :smile:

    (have a pi: π :wink:)
    (you mean "as θ----> π/2" ?)

    Try multiplying top and bottom by cosθ :wink:
     
  4. Apr 20, 2010 #3

    CR9

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    Hi tiny Tim,

    Thanks for the reply, yea I mean ( "as θ----> π/2" ?)

    Multiplying cos top and bottom would cancel off the sec on top, but how do i times cos inside the denominator? It has power of Vs/Vr

    Please advice.

    Thanks
     
  5. Apr 20, 2010 #4

    tiny-tim

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    Hi CR9! :smile:

    (secθ+tanθ)Vs/Vr = (secθ+tanθ)(secθ+tanθ)Vs/Vr - 1 :wink:
     
  6. Apr 20, 2010 #5

    CR9

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    Hi Tiny Tim,

    Thanks again for quick reply. You rock!

    okay, so after multiplying cos top and bottom and expanding sec+ tan at the bottom as your previous post;

    I got:

    r/w=1/(1+sinθ )((secθ+tanθ)^(Vs/Vr-1) )
     
  7. Apr 20, 2010 #6

    CR9

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    But there are too many unknowns, how can I solve this in order to find a value for Vs/Vr.

    I need to find the value for Vs/Vr and then find the angle.

    Please advice, tim
     
  8. Apr 20, 2010 #7

    tiny-tim

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    Hi CR9! :smile:
    or even 1/(1+sinθ)Vs/Vr(secθ)Vs/Vr - 1
    no problemo … Vs/Vr is a constant, and θ -> π/2

    (so, for example, the (1+sinθ) at the beginning obviously –> 2)

    Deal with the three cases separately: Vs/Vr > = or < 1 :wink:
     
  9. Apr 20, 2010 #8

    CR9

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    Hi Tiny Tim,
    I was waiting for you online just now...

    Anyway, Im still stuck at r/w=1/(1+sinθ )((secθ+tanθ)^(Vs/Vr-1)

    What do i do from here?
     
  10. Apr 20, 2010 #9

    tiny-tim

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    Hi CR9! :smile:

    I don't understand why you're stuck. :confused:

    What does (1+sinθ ) tend to as θ -> π/2 ?

    And what does (secθ+tanθ) tend to?
     
  11. Apr 20, 2010 #10

    CR9

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    as θ -> π/2,
    (1+sin θ) approaches 2
    (sec θ + tan θ) approaches infinite?

    then from
    r/w=1/(1+sinθ )((secθ+tanθ)^(Vs/Vr-1)

    it will be
    r/w= 1/(2)(infinite)^(vs/vr)-1

    What do I do with this?

    Please help :(
     
  12. Apr 20, 2010 #11

    tiny-tim

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    Well, what's ∞Vs/Vr-1 ?
     
  13. Apr 20, 2010 #12

    CR9

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    I dont know, Vs/vr could = 0 and it will become 1....

    I dont have values for Vs/Vr, r and w...

    Please help
     
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