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Author Topic: c4 fast differentiation doubt  (Read 758 times)

Offline zabady

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c4 fast differentiation doubt
« on: April 28, 2010, 11:51:59 pm »
attached down  ;D
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Offline T.Q

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Re: c4 fast differentiation doubt
« Reply #1 on: April 29, 2010, 01:41:05 am »
\frac{dv}{dt}= -k'\sqrt{v}

V=Ah  

so, \frac{dv}{dh}=A

\frac{dh}{dt}=\frac{dh}{dv} X \frac{dv}{dt}

\frac{dh}{dt}=( \frac{1}{A}) X( -k'\sqrt{v})

\frac{dh}{dt}=( \frac{-k'\sqrt{A}\sqrt{h}}{A})

\frac{dh}{dt}=( \frac{-k'\sqrt{h}}{\sqrt{A}})

so , K=( \frac{k'}{\sqrt{A}})

\frac{dh}{dt}= -K\sqrt{h}
« Last Edit: April 29, 2010, 10:02:23 am by T.Q »
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Offline zabady

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Re: c4 fast differentiation doubt
« Reply #2 on: April 29, 2010, 02:06:35 am »
but how u got k=(-k/squarroot A)
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Offline T.Q

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Re: c4 fast differentiation doubt
« Reply #3 on: April 29, 2010, 10:11:31 am »
Quote from: zabady on April 29, 2010, 02:06:35 am
but how u got k=(-k'/squarroot A)

because K is a positive constant

and i used k prime (k') in solving ,

K is different from k'  ,

u can put (t) instead of (k') to make it more simpler   
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