estimate the amount of solid 1 second later

Question

Let f(t) be the weight (in grams) of a solid sitting in a beaker of water after t minutes have elapsed. Suppose that the solid dissolves in such a way that the rate of change (in grams/minute) of the weight of the solid at any time t can be determined from the weight using the forumula:

 

f′(t)=−4f(t)(1+f(t))

 

If there is 2 grams of solid at time t=2 estimate the amount of solid 1 second later.

Answer 1

f′(t)= −4 f(t) [1+f(t)]

 

df/dt = −4 f (1+f)

 

This is a separable equation.  Separating the variables gives:

 

df / f(f+1) = -4dt

 

1 / f(f+1) can be expanded by partial fractions:

 

1 / f(f+1) = A/f + B/(f+1)

 

1 = (f+1)A + fB = (A+B)f + A

 

Then A=1, A+B=0=1+B, so B = -1:

 

1 / f(f+1) = 1/f - 1/(f+1)

 

Then so far we have:

 

[1/f - 1/(f+1)] df = -4dt

 

Integrate both sides:

 

ln(f) - ln(f+1) = -4t + lnC, with lnC an unknown constant

 

Take e^() on both sides:

 

f / (f+1) = Ce^(-4t)

 

We still need to find C:

 

Given:  f(2) = 2 g  (I assume 2 is 2 minutes.)

 

2/(2+1) = Ce^(-4*2)

 

2/3 = Ce^(-8)

 

C = (2/3) / e^(-8) = 1987.31

 

f / (f+1) = 1987.31 e^(-4t)

 

Solve for f, simplify, and get:

 

f(t) = 1987.31 / [e^(4t) - 1987.31]

 

1 second later is 1/60 of a minute later = 0.0166667 min later

 

t = (2 + 0.0166667) min = 2.0166667 min

 

f(2.0166667) = 1987.31 / [e^(4*2.0166667) - 1987.31] = 1.65726 g