Question

One hundred grams of cane sugar in water are being converted into dextrose at a rate which is proportional to the amount unconverted. Find the differential equation expressing the rate of conversion after t minutes.

Answer 1

Answer to Question #69654 – Math – Differential Equations

Question

One hundred grams of cane sugar in water are being converted into dextrose at a rate which is proportional to the amount unconverted. Find the differential equation expressing the rate of conversion after T minutes.

Solution

$m_{s}$ – mass of sugar at the current moment of time

$m_{D}$ – mass of dextrose at the current moment $f$ time

$m_{s_{0}} = 100g$ – mass of sugar at the beginning

$T$ – the time that had passed

$C$ – proportion coefficient

The fact that the sugar is converted into dextrose at a rate which is proportional to the amount of sugar unconverted in terms of differentiates means

$$dm_{D} = C m_{s} dt$$

where

$$m_{s} = m_{s_{0}} - m_{D}$$

because sucrose is converted somehow to dextrose and we can in general neglect the mass of one molecule of water per a molecule of sucrose (thus finally two molecules of dextrose).

So

$$dm_{D} = C (m_{s_{0}} - m_{D}) dt$$

Solving it we get

$$\ln (m_{s_{0}} - m_{D}) = - C t + K$$ $$m_{D} = m_{s_{0}} - e^{K - C t} = m_{s_{0}} - \frac{e^{K}}{e^{C t}}$$

When $t=0$ there is no dextrose so

$$0 = m_{s_0} - \frac{e^K}{e^{-C*0}} \rightarrow e^K = m_{s_0}$$

and therefore

$$m_D = m_{s_0} \left(1 - \frac{1}{e^{Ct}}\right) = m_{s_0} (1 - e^{-Ct})$$

Thus, the rate of conversion after T minutes:

$$dm_D = C(m_{s_0} - m_D)dt = C m_{s_0} e^{-CT} dt$$

But we does not take the deferential equation! It should be treated in the terms of delta

$$\Delta m_D = C m_{s_0} e^{-CT} \Delta t$$

only when $T \gg \Delta t$.

The differential equation expressing the rate of conversion after T minutes will look like

$$dm_D = C(m_{s_0} - C m_{s_0} e^{-CT} - m_D)dt = C(m_{s_0} (1 - C e^{-CT}) - m_D) dt$$

In the last formula (after integrating) time should be counted starting from moment $T$.

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