IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: zabady on April 28, 2010, 11:51:59 pm

Title: c4 fast differentiation doubt
Post by: zabady on April 28, 2010, 11:51:59 pm: attached down ;D
Title: Re: c4 fast differentiation doubt
Post by: T.Q 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 , $K=( \frac{k'}{\sqrt{A}})$

$\frac{dh}{dt}= -K\sqrt{h}$
Title: Re: c4 fast differentiation doubt
Post by: zabady on April 29, 2010, 02:06:35 am: but how u got k=(-k/squarroot A)
Title: Re: c4 fast differentiation doubt
Post by: T.Q 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