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Interesting rate of change question

  1. Mar 14, 2006 #1
    Two sides of a triangle have lengths a = 5cm and b = 10cm, and the included angle is [tex]\theta = \frac{\pi}{3}.[/tex] If a is increasing at a rate of 2cm/s, b is decreasing at a rate of 1cm/s and [tex]\theta[/tex] remains constant, at what rate os the third side changing? Is it increasing or decreasing?

    Just wondering the best way to go about this question. I could calculate values of third side c after one second, then two etc, and figure out the rate of change. But we've been doing partial differentiation and differentials and so on. So do i say [tex] c^2 = a^2 + b^2 - 2ab\cos C [/tex] and take partial derivatives getting something like
    [tex] (2a + 2b\cos C)da + (2b + 2a\cos C)db + (2ab\sin C)dc [/tex]

    Then plug some form of values into that?
     
    Last edited: Mar 14, 2006
  2. jcsd
  3. Mar 14, 2006 #2

    HallsofIvy

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    Something like that but not that itself!

    First remember that you started with an equation so, taking the differential of both sides, you should still have an equation. Also where did that (2ab)dc come from?
     
  4. Mar 14, 2006 #3
    Sorry, the sin C hadn't shown up. Is the equation:

    [tex] dc^2 = (2a + 2b\cos C)da + (2b + 2a\cos C)db + (2ab\sin C)dC [/tex]

    Or is it the same but instead of dc^2, its just dc?

    Assuiming this is correct (a shakey assumption no doubt), do i plug in the initial values ie. a = 5, b = 10, C = 60. And would da = 2, db = 1 and dC = 0?
     
  5. Mar 14, 2006 #4
    i may be wrong in this aspect but ill give it a shot

    shouldntu just tak the derivative with respect to time?? So then ur looking for dc/dt...
    differentiate wrt time and see what u get...

    also u can find c using the coniditions given to u, that is a,b, and theta
     
  6. Mar 14, 2006 #5
    i may be wrong in this aspect but ill give it a shot

    shouldntu just tak the derivative with respect to time?? So then ur looking for dc/dt...
    differentiate wrt time and see what u get...

    also u can find c using the coniditions given to u, that is a,b, and theta
     
  7. Mar 14, 2006 #6

    StatusX

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    Instead of leaving it in terms of infinitessimals, why don't you differentiate both sides with respect to t? It's the same basic idea, but then the answer becomes clear immediately. Also, you have a sign mistake in your da and db terms.
     
  8. Mar 14, 2006 #7
    Ok thanks, but how can i differentiate wrt if there are no t terms in the equation, or do i say da/dc = 2, db/dc = 1, then what does dC/dc equal? 0?
     
  9. Mar 14, 2006 #8
    you need to differentiate implicitly

    so then for example to differentiate a variable x^2 wrt to t you get
    [tex] 2x \frac{dx}{dt} [/tex]
    differentiate the function itself then the variable wrt time
     
  10. Mar 14, 2006 #9
    Ok, so if i have the equation
    [tex] c^2 = a^2 + b^2 - 2ab\cos C [/tex]

    do i take the square root before i begin differentiating?
     
  11. Mar 14, 2006 #10
    Still can;t do this. I cant differentiate with respect to time. For the first two terms i get [tex] 2a\frac{da}{dt} + 2b\frac{db}{dt} \left then \left ??? \left is \left it \left 2ab\sin C\frac{dC}{dt}? [/tex]
     
  12. Mar 14, 2006 #11

    StatusX

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    Use the product rule.
     
  13. Mar 14, 2006 #12
    since ur angle is CONSTANT, the value of C (or theta, whatever) reamins CONSTANT

    however you need to apply product rule because its a product of a and b who are both dependant on time
     
  14. Mar 15, 2006 #13
    But using the product rule isnt it just:

    u = -2ab, du/dt = 0
    v = cosC dv/dt = -sinC

    giving udv/dt = vdu/dt = -2ab-sinC + cosC.0 = 2absinC?

    Using this i have

    [tex] \frac{dc^2}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt} + 2ab\sin C
    = 2 \times 5 \times 2 + 2 \times 10 \times 1 + 2 \times 5 \times 10 \times .866 = 126.6 [/tex]

    therefore [tex] \frac{dc2}{dt} = \sqrt{126.6} = 11.25cm/s ?[/tex]

    Seems too big and i know its probably wrong.
     
    Last edited: Mar 15, 2006
  15. Mar 15, 2006 #14

    HallsofIvy

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    People have been trying to tell you to use the product rule for the "ab" part. Since C is a constant, you don't have to differentiate with respect to C.

    [tex]c^2= a^2+ b^2- 2ab cos (C)[/tex]

    Since C is constant, the derivative of just the -2ab cos(C) part is

    [tex]-2b cos(C)\frac{da}{dt}- 2a cos(C)\frac{da}{dt}[/tex]

    And you don't want [itex]\frac{dc^2}{dt}[/itex], you want [itex]\frac{dc}{dt}[/itex]- but they are closely related.
     
    Last edited: Mar 15, 2006
  16. Mar 16, 2006 #15
    Ah sorry, of course! I was tired, wasn't thinking straight. Hand in time has passed now, but i was determined to get to the bottom of this. I can see now how to do it. Thanks very much!
     
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