A mixture contains two chemicals, A and B, with masses mA and mB. The total mass of the mixture, m, remains constant. B is converted into A at a rate that
is inversely proportional to the mass of A and directly proportional to the square of the mass of B in the mixture at any time t.
Form a differential equation and
find a general expression relating the mass of A to the time t.
ive got dma/dt = mb^2/ma
is this right and if so is the next bit
mb = m - ma
so dma/dt = (m-ma)^2/ma
then i get stuck
The differential equation you have formed is correct:
```
dma/dt = mb^2 / ma
```
To make progress, we need to solve for `mb` in terms of `ma` using the information given: `mb = m - ma`.
Now, substitute this expression for `mb` into the differential equation:
```
dma/dt = (m - ma)^2 / ma
```
To proceed further, we can simplify the equation by expanding the square:
```
dma/dt = (m^2 - 2mma + ma^2) / ma
```
Next, we can cancel out the common factor of `ma` in the numerator:
```
dma/dt = (m^2 - 2mma + ma^2) / ma = (m^2/ma) - 2m + ma
```
Now, let's rearrange the terms:
```
dma/dt = m^2 / ma - 2m + ma
```
This equation relates the rate of change of mass of A (`dma/dt`) to the mass of A (`ma`) at any given time `t`.