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idkfa
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Your approach is wrong because the left hand side represents the derivative of the composition of two time dependent functions. You completely ignored that fact and started cancelling terms. I have a different solutiontry to this problem in mind.give a simpler example:

$\frac{d 4x^3}{dx} = kx^2 \Leftrightarrow 12x^2=kx^2 \Leftrightarrow 12=k$

This is not the same as

$\frac{d 4x}{dx} = k \Leftrightarrow 4=k$

Therefore use the chain rule

$\frac{d(c \cdot V)}{dt} = V\frac{dc}{dt}+\frac{dV}{dt}c=-Ak\Delta c$

I would now assume a stationary diffusion, hence

$V\frac{dc}{dt} = 0$

$\frac{dV}{dt}c=-Ak\Delta c$

$A = \frac{dV}{dr}$

$\frac{dV}{dt}c=-\frac{dV}{dr}k\Delta c$

For this step I hope there are no mathematicians around, I cancel the differentials

$\frac{dr}{dt}c=-k\Delta c$

$\frac{dr}{dt}=-k\frac{\Delta c}{c}$

I have a different solution to this problem in mind.

$\frac{d(c \cdot V)}{dt} = V\frac{dc}{dt}+\frac{dV}{dt}c=-Ak\Delta c$

I would now assume a stationary diffusion, hence

$V\frac{dc}{dt} = 0$

$\frac{dV}{dt}c=-Ak\Delta c$

$A = \frac{dV}{dr}$

$\frac{dV}{dt}c=-\frac{dV}{dr}k\Delta c$

For this step I hope there are no mathematicians around, I cancel the differentials

$\frac{dr}{dt}c=-k\Delta c$

$\frac{dr}{dt}=-k\frac{\Delta c}{c}$

Your approach is wrong because the left hand side represents the derivative of the composition of two time dependent functions. You completely ignored that fact and started cancelling terms. I try to give a simpler example:

$\frac{d 4x^3}{dx} = kx^2 \Leftrightarrow 12x^2=kx^2 \Leftrightarrow 12=k$

This is not the same as

$\frac{d 4x}{dx} = k \Leftrightarrow 4=k$

Therefore use the chain rule

$\frac{d(c \cdot V)}{dt} = V\frac{dc}{dt}+\frac{dV}{dt}c=-Ak\Delta c$

I would now assume a stationary diffusion, hence

$V\frac{dc}{dt} = 0$

$\frac{dV}{dt}c=-Ak\Delta c$

$A = \frac{dV}{dr}$

$\frac{dV}{dt}c=-\frac{dV}{dr}k\Delta c$

For this step I hope there are no mathematicians around, I cancel the differentials

$\frac{dr}{dt}c=-k\Delta c$

$\frac{dr}{dt}=-k\frac{\Delta c}{c}$

Source Link
idkfa
  • 1.7k
  • 1
  • 8
  • 19

I have a different solution to this problem in mind.

$\frac{d(c \cdot V)}{dt} = V\frac{dc}{dt}+\frac{dV}{dt}c=-Ak\Delta c$

I would now assume a stationary diffusion, hence

$V\frac{dc}{dt} = 0$

$\frac{dV}{dt}c=-Ak\Delta c$

$A = \frac{dV}{dr}$

$\frac{dV}{dt}c=-\frac{dV}{dr}k\Delta c$

For this step I hope there are no mathematicians around, I cancel the differentials

$\frac{dr}{dt}c=-k\Delta c$

$\frac{dr}{dt}=-k\frac{\Delta c}{c}$