# Net Change Theorem

Let $$R\left( t \right)$$ 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 $$\int\limits_a^b {R\left( t \right)dt}$$ 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 $$f\left( t \right)$$ is the rate of change of some quantity, then the integral $$\int\limits_a^b {f\left( t \right)dt}$$ represents the net change in that quantity over the time interval $$\left[ {a,b} \right].$$

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:

$F\left( b \right) = F\left( a \right) + \int\limits_a^b {f\left( t \right)dt} .$

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$c\left( t \right) = 5 – {e^{ – t}}$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$f\left( t \right) = 10 + 5{t^{\frac{3}{2}}}$ 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 $$V = 1000\,\text{litres}.$$ Water is pumped into the tank at the rate of$R\left( t \right) = 60 – t$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$g\left( t \right) = 50 + 10\sqrt t$people per month. The initial population is 5000. What will be the population in $$3$$ years?

### Example 5

An initially empty swimming pool is being filled by water. The rate at which the height of water changes is given by the equation$\frac{{dH}}{{dt}} = 2 – \sqrt t ,$where $$H$$ is measured in meters, and $$t$$ is measured in hours. Calculate the maximum height of water.

### Example 6

The growth rate of a vegetable on a field linearly depends on temperature according to the law$g\left( T \right) = 2\left( {T – 10} \right),$where $${g\left( T \right)}$$ is measured in kg per hour and $$T \ge 15^{\circ}C.$$ The hourly temperature varies as follows:$T\left( t \right) = 20 + 5\sin \frac{{\pi t}}{{12}}.$Find the total growth of the vegetables (in kg) for the $$12-\text{hour}$$ period.

### Example 7

The relative output power of solar panels decreases with increasing temperature. Suppose that the relative power output is determined by the function$P\left( T \right) = 100 + \frac{1}{2}\left( {25 – T} \right),$where $$P\left( T \right)$$ is measured in percent and temperature $$T$$ is measured in $${^{\circ}C}.$$ The temperature varies during a month according to the law$T\left( t \right) = 15 + 10\sin \frac{{\pi t}}{{30}},$where $$t$$ is measured in days. What is the average power output of the solar panel?

### Example 8

A radioactive material decays according to the law$N\left( t \right) = {N_0}{e^{ – \lambda t}},$where $$N\left( t \right)$$ is the amount of the radioactive material at time $$t,$$ $${N_0}$$ is the initial amount at time $$t = 0,$$ $$\lambda$$ is the rate of decay. Calculate the cumulative exposure to radiation for all time and for the period of $$T = n{T_{1/2}},$$ where $${T_{1/2}}$$ is the half life time of the radioactive element and $$n$$ is an arbitrary positive number.

### Example 9

Oil leaks from a tank at a rate of$r\left( t \right) = {r_0}{e^{ – \lambda t}},$where $$\lambda = 1$$ cubic meters per hour. During the $$1\text{st}$$ hour, $${R_1} = 3\,{\text{m}^3}$$ of oil has leaked out. How much oil will leak out during the next hour?

### Example 10

Suppose a hormone production rate during a day is given by the cyclic function$p\left( t \right) = 1 + \sin \frac{{\pi t}}{{12}}$for $$0 \le t \le 24.$$ The removal rate of the hormone is described by the linear function$q\left( t \right) = \frac{t}{{12}}.$Find the highest hormone level over the period.

### Example 1.

The engine on a boat starts at $$t = 0$$ and consumes fuel at the rate of$c\left( t \right) = 5 – {e^{ – t}}$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:

${C = \int\limits_0^2 {c\left( t \right)dt} }={ \int\limits_0^2 {\left( {5 – {e^{ – t}}} \right)dt} }={ \left. {\left( {5t + {e^{ – t}}} \right)} \right|_0^2 }={ 10 + {e^{ – 2}} – 1 }={ 9 + \frac{1}{{{e^2}}} }\approx{ 9.14\,\text{}}$

### Example 2.

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

Solution.

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

${F = \int\limits_0^4 {f\left( t \right)dt} }={ \int\limits_0^4 {\left( {10 + 5{t^{\frac{3}{2}}}} \right)dt} }={ \left. {\left( {10t + 2{t^{\frac{5}{2}}}} \right)} \right|_0^4 }={ 40 + 2 \cdot 32 }{= 104.}$

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

### Example 3.

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

Solution.

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

${V = \int\limits_0^T {R\left( t \right)dt} }={ \int\limits_0^T {\left( {60 – t} \right)dt} }={ \left. {\left( {60t – \frac{{{t^2}}}{2}} \right)} \right|_0^T }={ 60T – \frac{{{T^2}}}{2}.}$

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

${\frac{{{T^2}}}{2} – 60T + V = 0,\;\text{ or }\;}\kern0pt{{T^2} – 120T + 2000 = 0.}$

The roots of the equation are $${T_{1,2}} = 100, 20\,\text{min}.$$ The first root $${T_1} = 100\,\text{min}$$ does not make sense as the rate of the water flow becomes negative for $$T \gt 60\,\text{min}.$$

Thus, the answer is $$T = 20\,\text{min}.$$

### Example 4.

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

Solution.

By the net change theorem,

$G = {G_0} + \int\limits_0^T {g\left( t \right)dt} ,$

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:

${\int\limits_0^T {g\left( t \right)dt} }={ \int\limits_0^{36} {\left( {50 + 10\sqrt t } \right)dt} }={ \left. {\left( {50t + \frac{{20{t^{\frac{3}{2}}}}}{3}} \right)} \right|_0^{36} }={ 1800 + 1440 }={ 3240.}$

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

$G = 5000 + 3240 = 8240.$

### Example 5.

An initially empty swimming pool is being filled by water. The rate at which the height of water changes is given by the equation$\frac{{dH}}{{dt}} = 2 – \sqrt t ,$where $$H$$ is measured in meters, and $$t$$ is measured in hours. Calculate the maximum height of water.

Solution.

Determine the time $$T$$ it takes to fill the pool.

${\frac{{dH}}{{dt}} = 0,}\;\; \Rightarrow {2 – \sqrt T = 0,}\;\; \Rightarrow {T = 4\,\text{hours}}.$

Using the net change theorem we find the maximum height:

${{H_{\max }} = \int\limits_0^T {\frac{{dH}}{{dt}}dt} }={ \int\limits_0^4 {\left( {2 – \sqrt t } \right)dt} }={ \left. {\left( {2t – \frac{{2{t^{\frac{3}{2}}}}}{3}} \right)} \right|_0^4 }={ 8 – \frac{2}{3} \cdot 8 }\approx{ 2.67\,\text{meters}}$

### Example 6.

The growth rate of a vegetable on a field linearly depends on temperature according to the law$g\left( T \right) = 2\left( {T – 10} \right),$where $${g\left( T \right)}$$ is measured in kg per hour and $$T \ge 15^{\circ}C.$$ The hourly temperature varies as follows:$T\left( t \right) = 20 + 5\sin \frac{{\pi t}}{{12}}.$Find the total growth of the vegetables (in kg) for the $$12-\text{hour}$$ period.

Solution.

By substituting $$T\left( t \right)$$ into the equation for $$g\left( T \right)$$ we obtain the growth rate as an explicit function of time $$t:$$

${g\left( t \right) = 2\left( {T\left( t \right) – 10} \right) }={ 2\left( {20 + 5\sin \frac{{\pi t}}{{12}} – 10} \right) }={ 20 + 10\sin \frac{{\pi t}}{{12}}.}$

Integrating over the $$12-\text{hour}$$ period yields:

${G = \int\limits_0^{12} {g\left( t \right)dt} }={ \int\limits_0^{12} {\left( {20 + 10\sin \frac{{\pi t}}{{12}}} \right)dt} }={ \left. {\left( {20t – \frac{{120}}{\pi }\cos \frac{{\pi t}}{{12}}} \right)} \right|_0^{12} }={ 480 – \frac{{120}}{\pi }\left( {\cos \pi – \cos 0} \right) }={ 480 + \frac{{240}}{\pi } }\approx{ 556.4\,\text{kg}}$

### Example 7.

The relative output power of solar panels decreases with increasing temperature. Suppose that the relative power output is determined by the function$P\left( T \right) = 100 + \frac{1}{2}\left( {25 – T} \right),$where $$P\left( T \right)$$ is measured in percent and temperature $$T$$ is measured in $${^{\circ}C}.$$ The temperature varies during a month according to the law$T\left( t \right) = 15 + 10\sin \frac{{\pi t}}{{30}},$where $$t$$ is measured in days. What is the average power output of the solar panel?

Solution.

First we write the relative power output as a function of time $$t:$$

${P\left( t \right) }={ 100 + \frac{1}{2}\left( {25 – T\left( t \right)} \right) }={ 100 + \frac{1}{2}\left( {25 – 15 – 10\sin \frac{{\pi t}}{{30}}} \right) }={ 105 – 5\sin \frac{{\pi t}}{{30}}.}$

Integrating the power over the period of $$1$$ month gives us the total energy produced for the period. The average power value is equal to the energy divided by the duration of the time interval. Hence,

${\bar P = \frac{1}{{30}}\int\limits_0^{30} {P\left( t \right)dt} }={ \frac{1}{{30}}\int\limits_0^{30} {\left( {105 – 5\sin \frac{{\pi t}}{{30}}} \right)dt} }={ \frac{1}{{30}}\left. {\left( {105t + \frac{{150}}{\pi }\cos \frac{{\pi t}}{{30}}} \right)} \right|_0^{30} }={ 105 + \frac{5}{\pi }\left( {\cos \pi – \cos 0} \right) }={ 105 – \frac{{10}}{\pi } }\approx{ 101.8\% }$

### Example 8.

A radioactive material decays according to the law$N\left( t \right) = {N_0}{e^{ – \lambda t}},$where $$N\left( t \right)$$ is the amount of the radioactive material at time $$t,$$ $${N_0}$$ is the initial amount at time $$t = 0,$$ $$\lambda$$ is the rate of decay. Calculate the cumulative exposure to radiation for all time and for the period of $$T = n{T_{1/2}},$$ where $${T_{1/2}}$$ is the half life time of the radioactive element and $$n$$ is an arbitrary positive number.

Solution.

We assume that the absorbed dose of radiation $$R\left( t \right)$$ is proportional to the amount of radioactive material:

${R\left( t \right) \sim N\left( t \right),}\;\; \Rightarrow {R\left( t \right) = {R_0}{e^{ – \lambda t}},}$

where $${R_0}$$ is the initial rate of radiation absorption.

The total amount of absorbed dose is given by the improper integral:

${{R_\infty } = \int\limits_0^\infty {R\left( t \right)dt} }={ {R_0}\int\limits_0^\infty {{e^{ – \lambda t}}dt} }={ {R_0}\lim \limits_{b \to \infty } \int\limits_0^b {{e^{ – \lambda t}}dt} }={ {R_0}\lim \limits_{b \to \infty } \left. {\left( { – \frac{1}{\lambda }{e^{ – \lambda t}}} \right)} \right|_0^b }={ \frac{{{R_0}}}{\lambda }\lim \limits_{b \to \infty } \left( {1 – {e^{ – \lambda b}}} \right) }={ \frac{{{R_0}}}{\lambda }.}$

Let’s now determine the amount of absorbed dose of radiation for the period of $${nT_{1/2}}.$$ The half life time $${T_{1/2}}$$ is defined from the condition:

${N\left( {{T_{1/2}}} \right) = \frac{{{N_0}}}{2} }={ {N_0}{e^{ – \lambda {T_{1/2}}}}.}$

It follows from here that

${T_{1/2}} = \frac{{\ln 2}}{\lambda }.$

Integrating from $$t = 0$$ to $$t = nT_{1/2}$$ yields

${{R_n} = \int\limits_0^{n{T_{1/2}}} {R\left( t \right)dt} }={ {R_0}\int\limits_0^{\frac{{n\ln 2}}{\lambda }} {{e^{ – \lambda t}}dt} }={ {R_0}\left. {\left( { – \frac{1}{\lambda }{e^{ – \lambda t}}} \right)} \right|_0^{\frac{{n\ln 2}}{\lambda }} }={ \frac{{{R_0}}}{\lambda }\left( {1 – {e^{ – n\ln 2}}} \right) }={ \frac{{{R_0}}}{\lambda }\left( {1 – \frac{1}{{{e^{\ln {2^n}}}}}} \right) }={ \frac{{{R_0}}}{\lambda }\left( {1 – \frac{1}{{{2^n}}}} \right).}$

Thus, at the moment equal to half life time, the cumulative absorbed dose is $$\large{\frac{1}{2}}\normalsize$$ of the total amount. At time $$2{T_{1/2}},$$ the absorbed dose is already $$\large{\frac{3}{4}}\normalsize$$ of the total amount, etc.

This pattern is typical for any processes described by the exponential decay function.

### Example 9.

Oil leaks from a tank at a rate of$r\left( t \right) = {r_0}{e^{ – \lambda t}},$where $$\lambda = 1$$ cubic meters per hour. During the $$1\text{st}$$ hour, $${R_1} = 3\,{\text{m}^3}$$ of oil has leaked out. How much oil will leak out during the next hour?

Solution.

Using the net change theorem, we can write $${R_1}$$ in the form

${{R_1} = \int\limits_0^1 {r\left( t \right)dt} }={ \int\limits_0^1 {{r_0}{e^{ – \lambda t}}dt} }={ \left. { – \frac{{{r_0}}}{\lambda }{e^{ – \lambda t}}} \right|_0^1 }={ \frac{{{r_0}}}{\lambda }\left( {1 – {e^{ – \lambda }}} \right).}$

Similarly, we can express the volume of leaked oil for the $$2\text{nd}$$ hour $${R_2}:$$

${{R_2} = \int\limits_1^2 {r\left( t \right)dt} }={ \int\limits_1^2 {{r_0}{e^{ – \lambda t}}dt} }={ \left. { – \frac{{{r_0}}}{\lambda }{e^{ – \lambda t}}} \right|_1^2 }={ \frac{{{r_0}}}{\lambda }\left( {{e^{ – \lambda }} – {e^{ – 2\lambda }}} \right).}$

It then follows that

$\require{cancel}{\frac{{{R_2}}}{{{R_1}}} }={ \frac{{\cancel{\frac{{{r_0}}}{\lambda }}\left( {{e^{ – \lambda }} – {e^{ – 2\lambda }}} \right)}}{{\cancel{\frac{{{r_0}}}{\lambda }}\left( {1 – {e^{ – \lambda }}} \right)}} }={ \frac{{{e^{ – \lambda }} – {e^{ – 2\lambda }}}}{{1 – {e^{ – \lambda }}}} }={ \frac{{{e^{ – \lambda }}\cancel{\left( {1 – {e^{ – \lambda }}} \right)}}}{\cancel{1 – {e^{ – \lambda }}}} }={ {e^{ – \lambda }}.}$

Therefore the value of $${R_2}$$ is equal to

${{R_2} = {R_1}{e^{ – \lambda }} }={ 3 \cdot {e^{ – 1}} }={ \frac{3}{e} }\approx{ 1.10\,{\text{m}^3}}$

### Example 10.

Suppose a hormone production rate during a day is given by the cyclic function$p\left( t \right) = 1 + \sin \frac{{\pi t}}{{12}}$for $$0 \le t \le 24.$$ The removal rate of the hormone is described by the linear function$q\left( t \right) = \frac{t}{{12}}.$Find the highest hormone level over the period.

Solution.

We can make sure that the net change in hormone level over the $$24-$$hour interval is equal to zero:

$\require{cancel}{P – Q }={ \int\limits_0^{24} {\left[ {p\left( t \right) – q\left( t \right)} \right]dt} }={ \int\limits_0^{24} {\left[ {1 + \sin \frac{{\pi t}}{{12}} – \frac{t}{{12}}} \right]dt} }={ \left. {\left[ {t – \frac{{12}}{\pi }\cos \frac{{\pi t}}{{12}} – \frac{{{t^2}}}{{24}}} \right]} \right|_0^{24} }={ \cancel{24} – \frac{{12}}{\pi }\left( {\cancel{\cos 2\pi} – \cancel{\cos 0}} \right) – \cancel{24} }={ 0,}$

where $$P$$ and $$Q$$ denote the total amount of the hormone, respectively, produced and removed for the period.

Now let’s derive the expression for the hormone level $$H$$ at an arbitrary time $$T,$$ where $${0 \le T \le 24}.$$ Using the net change theorem, we have

${H\left( T \right)} = {P(T) – Q(T) }={ \int\limits_0^T {\left[ {p\left( t \right) – q\left( t \right)} \right]dt} }={ \int\limits_0^T {\left[ {1 + \sin \frac{{\pi t}}{{12}} – \frac{t}{{12}}} \right]dt} }={ \left. {\left[ {t – \frac{{12}}{\pi }\cos \frac{{\pi t}}{{12}} – \frac{{{t^2}}}{{24}}} \right]} \right|_0^T }={ T – \frac{{12}}{\pi }\left( {\cos \frac{{\pi T}}{{12}} – \cos 0} \right) – \frac{{{T^2}}}{{24}} }={ T – \frac{{{T^2}}}{{24}} – \frac{{12}}{\pi }\cos \frac{{\pi T}}{{12}} + \frac{{12}}{\pi }.}$

The hormone level $$H$$ is maximal when

${ p\left( t \right) – q\left( t \right) }={ 1 + \sin \frac{{\pi t}}{{12}} – \frac{t}{{12}} }={ 0.}$

Although this equation is nonlinear, we can notice that $$t = 12$$ is its solution. The total rate $$p\left( t \right) – q\left( t \right)$$ changes sign from positive to negative at this point. Therefore the hormone level $$H$$ reaches the maximum value at $$t = 12.$$

The maximum concentration of the hormone is equal to

${{H_{\max }} = P(12) – Q(12) }={ 12 – \frac{{{{12}^2}}}{{24}} – \frac{{12}}{\pi }\cos \pi + \frac{{12}}{\pi } }={ 6 + \frac{{24}}{\pi } }\approx{ 13.6\,\text{units}}$