Question

The rate of change of the mass, A, of salt at time t is proportional to the square of the mass of salt present at time t. Initially there is 10 grams of salt and 10 hour later there are 4 grams of salt. (a) Find the initial value problem (the differential equation and the initial condition) that fits this physical description. (b) Solve the I.V.P (initial value problem) from part (a) for the specific solution. (c) Based on your solution find the time when A(t) 1

Answer 1

Explanation:

The given question incomplete; here is the complete question given in the attachment.

The rate of change of the mass A of salt at the time 't' is proportional to the square of the mass of salt present at the time t.

`(dA)/(dt)` ∝ A²

`(dA)/(dt)` = kA² [k = proportionality constant]

And the given conditions are

`A_(0)=10` and
`A_(10)=4`

a).
`(dA)/(dt)=kA^(2)`

`A_(0)=10` and
`A_(10)=4`

b). From the differential equation
`(dA)/(dt)=kA^(2)`

`(dA)/(A^(2) )=kdt`

By integration on both the sides of the equation.

`\int {(1)/(A^(2) ) } \, dA=k\int dt`

`-(1)/(A)=kt+c` [c = integration constant]

From the given conditions in the question,

For
`A_(0)=10`,

`-(1)/(10)=k* 0+c`

c =
`-(1)/(10)`

For
`A_(10)=4`,

`-(1)/(4)=k* 10+c`

Since c =
`-(1)/(10)`

Therefore,
`-(1)/(4)=10k-(1)/(10)`

k =
`(1)/(10)((1)/(10)-(1)/(4))`

k =
`(1)/(10)* ((4-10))/(40)`

k = -
`(6)/(400)`

k = -
`(3)/(200)`

Now we place the values of constants in the solution of integration

`-(1)/(A)=-(3)/(200)t-(1)/(10)`

`(1)/(A)=(3t+20)/(200)`

A =
`(200)/(3t+20)`

c). Now we have to calculate the time when A(t) < 1

From part b,

`A_(t)=(200)/(3t+20)`

Therefore,
`(200)/(3t+20)<1`

By solving the inequality,

3t + 20 > 200

3t > 200 - 20

3t > 180

t > 60

Therefore, after 60 hours
`A_(t)<1`