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.
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
1 answer