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Math Help - Rates of Decay

  1. #1
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    Rates of Decay

    In the question attached what is the answer to part b) i can do part a) and c) thanks
    Last edited by nath_quam; June 18th 2006 at 12:08 AM.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by nath_quam
    In the question attached what is the answer to part ii) i can do part 1 and 3 thanks
    you could see the \frac{dI}{dt}= -5(1-cosx)
    where x= \frac {\pi*t}{12}
    now, 1- cosx is always greater than or equal to zero.
    if 1- cosx is greater than zero., I is decreasing
    I is stationary if cosx =1 i.e. x=\frac{\pi}{2}
    or \frac {\pi t}{12}=\frac{\pi}{2}
    hence I is stationary at t=6
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  3. #3
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    What Malaygoel forgot to mention though it is implicit in what he wrote
    is that a differentiable function f(x) is decreasing at x if and only
    if f'(x)<0 and is stationary if and only if f'(x)=0

    RonL
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  4. #4
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    Stationary Points

    Thanks guys, for this question there must be other stationary points such as 18?? and so on
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by malaygoel
    you could see the \frac{dI}{dt}= -5(1-cosx)
    where x= \frac {\pi*t}{12}
    now, 1- cosx is always greater than or equal to zero.
    if 1- cosx is greater than zero., I is decreasing
    I is stationary if cosx =1 i.e. x=\frac{\pi}{2}
    or \frac {\pi t}{12}=\frac{\pi}{2}
    hence I is stationary at t=6
    \cos(x)=1 only when x is a multiple of 2\pi,
    so the stationary points are solutions of:

    <br />
\frac{\pi t}{12}=n \pi<br />
,

    or: t=12n, n=0,\ 1,\ ....

    RonL
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  6. #6
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    the question asks for when I is stationary, ain't we saying that at t = 12n the rate is stationary because for example at t = 12 -- I is decreasing still
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  7. #7
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    Quote Originally Posted by nath_quam
    the question asks for when I is stationary, ain't we saying that at t = 12n the rate is stationary because for example at t = 12 -- I is decreasing still
    I is stationary when dI/dt=0.

    Stationary means the rate of change is zero.

    RonL
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  8. #8
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    Thanks your a legend
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  9. #9
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    Small correction, Cap'n . . .


    \cos(x)=1 only when x is a multiple of 2\pi,

    so the stationary points are solutions of: . \frac{\pi}{12}t \,=\, 2 \pi\cdot n

    or: . t \,= \,24n,\;\;n=0,\,1,\,2, ....

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  10. #10
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban
    Small correction, Cap'n . . .


    \cos(x)=1 only when x is a multiple of 2\pi,

    so the stationary points are solutions of: . \frac{\pi}{12}t \,=\, 2 \pi\cdot n

    or: . t \,= \,24n,\;\;n=0,\,1,\,2, ....

    Hello Soroban
    Does it make any sense that I is stationary at t=0(n=0).
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  11. #11
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    Hello, malaygoel!

    Does it make any sense that I is stationary at t=0\,(n=0).
    Yes, it does . . .

    An analogous example:
    A particle can be at rest at t = 0\:(v = 0).
    A nano-jiffy later it is moving (v > 0) due to some acceleration.
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  12. #12
    Grand Panjandrum
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    Quote Originally Posted by Soroban
    Small correction, Cap'n . . .


    \cos(x)=1 only when x is a multiple of 2\pi,

    so the stationary points are solutions of: . \frac{\pi}{12}t \,=\, 2 \pi\cdot n

    or: . t \,= \,24n,\;\;n=0,\,1,\,2, ....

    Oops

    Thanks for spotting that.

    Also perhaps a reminder to everyone to check what the helpers write,
    after all we all make mistakes at some time

    RonL
    Last edited by CaptainBlack; June 17th 2006 at 08:09 AM.
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  13. #13
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Soroban
    Hello, malaygoel!


    Yes, it does . . .

    An analogous example:
    A particle can be at rest at v = 0)" alt="t = 0\v = 0)" />.
    A nano-jiffy later it is moving (v > 0) due to some acceleration.
    Imagine the motion of the same particle.
    Let x=f(t) describes the motion of the particle. Consider the motion from t=0 to  t=10.Can we find  \frac{dx}{dt} at   t=0.. In other words, if a function is defined on the closed interval [a,b], can it be differentiable on one of its end-points e.g.'a'?
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