Net Change Theorem

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Let $R (t)$ represent the rate at which water flows into a tank. The rate can be measured in cubic meters per second or, for example, in gallons per minute.

Then the definite integral $\int_{a}^{b} R (t) d t$ expresses the total change in volume from time $a$ to time $b .$

Instead of a water flow, we can consider any other quantity. In general case, if $f (t)$ is the rate of change of some quantity, then the integral $\int_{a}^{b} f (t) d t$ represents the net change in that quantity over the time interval $[a, b] .$

This leads us to the Net Change Theorem, which states that if a quantity changes and is represented by a differentiable function, the final value equals the initial value plus the integral of the rate of change of that quantity:

F (b) = F (a) + \int_{a}^{b} f (t) d t .

The Net Change Theorem can be applied to various problems involving rate of change (such as finding volume, area, population, velocity, distance, cost, etc.)

Solved Problems

Click or tap a problem to see the solution.

Example 1

The engine on a boat starts at $t = 0$ and consumes fuel at the rate of $c (t) = 5 - e^{- t}$ litres per hour. How much fuel does it consume for the first 2 hours?

Example 2

Suppose a fish population in a lake is increasing with a rate of $f (t) = 10 + 5 t^{\frac{3}{2}}$ thousands of fish per year, where $t$ is the number of years from now. How much will the fish population increase in $4$ years?

Example 3

A tank has a capacity of $V = 1000 litres .$ Water is pumped into the tank at the rate of $R (t) = 60 - t$ litres per minute, where time $t$ is measured in minutes. How long will it take to fill the tank to its full capacity.

Example 4

Suppose that $t$ months from now the population of a town will be growing at the rate of $g (t) = 50 + 10 \sqrt{t}$ people per month. The initial population is 5000. What will be the population in $3$ years?

Example 1.

The engine on a boat starts at $t = 0$ and consumes fuel at the rate of $c (t) = 5 - e^{- t}$ litres per hour. How much fuel does it consume for the first 2 hours?

Solution.

To estimate the fuel consumption, we use the net change theorem. This yields:

C = \int_{0}^{2} c (t) d t = \int_{0}^{2} (5 - e^{- t}) d t = {(5 t + e^{- t}) |}_{0}^{2} = 10 + e^{- 2} - 1 = 9 + \frac{1}{e^{2}} \approx 9.14

Example 2.

Solution.

The integral of the rate of change from $t = 0$ to $t = 4$ is the net change of the fish population:

F = \int_{0}^{4} f (t) d t = \int_{0}^{4} (10 + 5 t^{\frac{3}{2}}) d t = {(10 t + 2 t^{\frac{5}{2}}) |}_{0}^{4} = 40 + 2 \cdot 32 = 104.

Hence, the fish population will increase by $F = 104$ thousands.

Example 3.

Solution.

Let $T$ be the time required to fill the tank. Using integration, we have

V = \int_{0}^{T} R (t) d t = \int_{0}^{T} (60 - t) d t = {(60 t - \frac{t^{2}}{2}) |}_{0}^{T} = 60 T - \frac{T^{2}}{2} .

To find the time $T,$ we must solve the quadratic equation

\frac{T^{2}}{2} - 60 T + V = 0, or T^{2} - 120 T + 2000 = 0.

The roots of the equation are $T_{1, 2} = 100, 20 min .$ The first root $T_{1} = 100 min$ does not make sense as the rate of the water flow becomes negative for $T > 60 min .$

Thus, the answer is $T = 20 min .$

Example 4.

Solution.

By the net change theorem,

G = G_{0} + \int_{0}^{T} g (t) d t,

where $G_{0}$ and $G$ are the initial and final populations, respectively.

By integrating from $t = 0$ to $T = 36$ months, we get the total increase in the number of people:

\int_{0}^{T} g (t) d t = \int_{0}^{36} (50 + 10 \sqrt{t}) d t = {(50 t + \frac{20 t^{\frac{3}{2}}}{3}) |}_{0}^{36} = 1800 + 1440 = 3240.

Thus, the final population in $3$ years is expected to be

G = 5000 + 3240 = 8240.