a gallon of milk is slowly leaking inside your refrigerator. the number of ounces in the container, t minutes after the milk began leaking is given by the equation V(t)= 128e^(-0.007t), how fast is the milk leaking from the gallon container at the end of 5minutes

Answer: Rate of leaking milk from the gallon container at the end of 5minutes = 6.987598074×10^-3 %


Solution:

The quantity decreases slowly after which the rate of change and the rate of growth decreases over a period of time rapidly. This decrease in growth is calculated by using the exponential decay formula
A = P e^-k t
A = remaining Amount after t time
P = Initial amount
t = time intervals (time can be in minutes, hours, years, days, (or) months, whatever you are using should be consistent throughout the problem).
k = constant of proportionality
e- Euler's constant

Comparing the equation V(t)= 128e^(-0.007t) by general equation of exponential decay equation
Initial amount of milk,
P = 128 ounce
K = 0.007
Time t = 5 minutes
Milk remained in container after 5 minutes A = V(t) = 128e^(-0.007×5)
= 128e^(-0.035)
= 123.59 ounces

Another exponential decay equation
A = P(1 - r)^t
Where
r = Rate of decay in % (for exponential decay)
By putting the values of A, P and t

123.59 = 128(1- r)^5
1 - r = (123.59/128)^(1/5)
r = 1 - (123.59/128)^(1/5)
r = 6.987598074 × 10^-3
Rate of leaking milk from the gallon container at the end of 5minutes = 6.987598074×10^-3 %