Question

A chemical in an inflatable
tank is being heated. The
volume of the tank depends on
time y in minutes. The rate at
which the volume is changing
at time y minutes is
1200 − 4y. The volume of the
tank after the first 20 minutes
is 24, 200 cubic centimeters.
Determine the volume of the
tank after the first 80 minutes

Answer 1

We can solve this using integration.

Given:

Rate of change of volume: dy/dt = 1200 - 4y

Initial volume: y(20) = 24200

To find:

Volume after 80 minutes: y(80)

Integration:

dy/dt = 1200 - 4y

Rearranging, we get:

dy/(1200-4y) = dt

Integrating both sides:

- 1/4 * ln|1200-4y| = t + C

where C is the constant of integration.

Applying the initial condition:

- 1/4 * ln|1200-4(24200)| = 20 + C

- 1/4 * ln|−94120| = 20 + C

C = - 1/4 * ln|−94120| - 20

C = -4.076

Substituting the value of C in the integration equation:

- 1/4 * ln|1200-4y| = t - 4.076

- ln|1200-4y| = -4t + 16.304

Taking antilogarithm:

|1200-4y| = e^-4t+16.304

Since the absolute value function is involved, we need to split this equation into two cases:

1200 - 4y = e^-4t+16.304 OR 1200 - 4y = -e^-4t+16.304

Solving for y in each case:

y = 300 - 75e^-4t OR y = 300 + 75e^-4t

Applying the initial condition:

When t = 20, y = 24200

Substituting these values in y = 300 - 75e^-4t:

24200 = 300 - 75e^-4*20

24200 = 300 - 75e^-80

e^-80 = (300-24200)/(-75)

e^-80 = 257.33

Taking natural logarithm:

ln(e^-80) = ln(257.33)

-80 = ln(257.33)

Using a calculator:

-80 = - 4.21

Therefore, y = 300 - 75e^-4t = 300 - 75e^-4*80 = 1254.42 cubic centimeters

Answer:

The volume of the tank after the first 80 minutes is 1254.42 cubic centimeters.