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Accumulation Problems

- Calculus
- Absolute Maxima and Minima
- Absolute and Conditional Convergence
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Let's say you're the mayor of a small town. Although you know the rate at which your town grows, you want to know your town's population over the four years that you have served as mayor.

You could go knocking on the door of every house in your town asking how long each resident has lived there. Or, you could simply use a definite integral to interpret the **accumulation **of people over the time you've served as mayor.

The Fundamental Theorem of Calculus says that for a continuous function \(f\) on the interval \([a,b] \),

\[ \int_a^b f(x)\, dx = F(b) - F(a) \]

where \(F\) is the antiderivative of \(f\).

In terms of our population example, to find the town's population from your first year as mayor to your fourth, the Fundamental Theorem of Calculus says you should compute

\[ \int_0^4 f(x)\, dx = F(4) - F(0) \]

where \(f\) is a function of population of your town. In other words, all you have to do is subtract the population from your first year as mayor from the population from your fourth year as mayor.

Let's start by defining what an accumulation problem is.

An **accumulation problem** is a problem where the rate of change of a quantity is given, and you are tasked with finding the net change of the quantity over time.

Suppose there is a small leak below your sink such that it pours water at a constant rate of 1 milliliter per minute, and you place a bucket below the sink overnight, so it does not make a mess until a plumber comes to fix it. Here, the water will start to **accumulate **in the bucket, and since you know that the rate of change is constant, you can actually calculate how much water will be in the bucket!

For example, let's say that you place the bucket at 10:00 pm in the night and the plumber will come at 8:00 am the next morning. This means that the bucket will be filling for \(10\) hours, which in minutes is

\[ 10(60\text{ min})=600\text{ min.}\]

Next, you multiply this time by the rate at which the bucket fills, that is

\[600\text{ min} \left( \frac{1\text{ ml}}{\text{min}} \right) = 600\text{ ml}\]

so there will be 600ml (about 20 ounces) in the bucket before the sink gets fixed.

This calculation was simple because the rate of change was **constant,** but what happens if the rate of change is not constant? These problems are typically solved with definite integrals, so Calculus has the answer!

You can see an accumulation problem as a situation where you are stacking a quantity that changes over time. The **Net Change Theorem**, derived from the Fundamental Theorem of Calculus, states that the integral of a rate of change is the net change,

\[ \int_a^b F'(x) \, dx = F(b) - F(a).\]

Going back to the town population example, remember that you need to know how much the town's population has increased since you were chosen mayor. Let's say you were chosen \( 4\) years ago. Knowing a function \( F(x) \) that tells you the current population of the town would make this a trivial task, as you would only need to subtract the current population, \( F(4),\) from the town's population when you started as a mayor, \( F(0).\) That is

\[ F(4) - F(0).\]

As mentioned before, this would require a survey and would be very time-consuming. However, remember that you know the rate at which the town grows! Suppose this rate is given by the function

\[ f(x) = 1920e^{0.01x}.\]

Since the above function describes a rate of change, you can apply the Net Change Theorem to find the accumulation of population \(A\) in the four years of your government, so

\[ \begin{align} A &= \int_0^4 f(x) \, \mathrm{d}x \\ &= \int_0^4 1920 e^{0.01x}\,\mathrm{d}x. \end{align}\]

The above integral is the integral of an exponential function, which can be solved using basic integration techniques, so

\[ \begin{align} A &= \int_0^4 1920e^{0.01x}\,\mathrm{d}x \\ &= \left(\frac{1920}{0.01}e^{0.01(4)}\right)-\left(\frac{1920}{0.01}e^{0.01(0)}\right) \\ &= 192\,000e^{0.04}-192\,000 \\ &= 7835.6686, \end{align}\]

so the growth is of about \(7835\) people.

This means that over the four years that you have served as mayor, your town has grown by about \(7835\) people. This is the **net change** of the population, which can also be seen as the **accumulation** of the population over those four years.

Solving accumulation problems can be summarized as solving a definite integral. However, usually accumulation problems are word problems, so it is important to understand exactly what function you should be integrating, or if you should be integrating at all!

The Net Change Theorem states that the net change of a function is the definite integral of the **rate of change **of the function. So, if you are provided with a rate of change, you need to integrate it. Otherwise, you just need to find the difference of the function at two different values.

If the function represents a rate of change, and in this case you should integrate, the problem will use phrases like,

Rate

Derivative

Speed

Velocity

Here is an example.

The function

\[ g(t) = \frac{1}{5}t^2 \]

represents the speed ( in gallons per minute) at which an industrial tank is filled with a chemical, where \( t \) is given in minutes.

How much of the chemical is accumulated during the first \(10\) minutes of the filling process?

**Solution**

Begin by noting that the problem is telling you that the function represents a **speed.** This way you can identify the function

\[ g(t) = \frac{1}{5}t^2\]

as a rate of change, so you can use the Net Change Theorem to find the accumulation \( A\) of chemical. Since you are asked to find the accumulation during the first 10 minutes of the filling, the integration limits will be from 0 to 10, so

\[ \begin{align} A &= \int_0^{10} g(t) \, \mathrm{d}t \\ &= \int_0^{10} \frac{1}{5}t^2 \, \mathrm{d}t. \end{align} \]

You can solve the involved indefinite integral using the Power Rule, that is

\[ \begin{align} \int \frac{1}{5}t^2 \, \mathrm{d}t &= \frac{1}{5}\left( \frac{1}{3}t^3 \right) \\&= \frac{1}{15}t^3. \end{align}\]

Now that you know the indefinite integral, you can evaluate the definite integral, so

\[ \begin{align} A &= \int_0^{10} \frac{1}{5}t^2\,\mathrm{d}t \\ &= \left( \frac{1}{15}(10)^3 \right) - \left( \frac{1}{15} (0)^3 \right) \\ &= \frac{1000}{15} \\ &\approx 66.67.\end{align}\]

This means that about \( 66.67 \) gallons of chemical are accumulated during the first \(10\) minutes of the filling process.

Sometimes you will not need to integrate! You have to play close attention to the words used in the problem.

The function

\[ P(t) = 200 e^{0.1t}\]

represents the population of wolves in a wolf sanctuary \(t \) years since its foundation. By how much the population increased between the fifth and tenth year since the foundation of the sanctuary?

**Solution**

This time, you can note that there is no word suggesting that the given function is a rate of change. Instead, it tells how many wolves are in the sanctuary in a given year. This means that you only need to find

\[ A = P(10) - P(5)\]

to find the answer, so

\[ \begin{align} A &= 200e^{0.1(10)}-200e^{0.1(5)} \\ &= 200(e-e^{0.5})\\ &= 200(1.06956) \\ &= 213.91, \end{align} \]

so the population of wolves increased by about \(213\) wolves between those years.

When studying about accumulation problems you might come up with the term **accumulation function. **

**Accumulation functions** are functions used in mathematical finances, which are used to represent things such as simple interest and compound interest.

For example, the function

\[ s(t) = 1+0.05t\]

represents simple interest, while the function

\[ c(t) = (1+0.05)^t\]

represents compound interest.

These functions are used in a different context, so you should not worry about relating the above functions to accumulation problems as seen in the context of AP calculus!

Here are more examples of accumulation problems.

There is a leak in a water tank, where water is draining out from a small hole. As time passes, the hole becomes bigger. The function

\[ V(t) = \frac{1}{2}t\]

models the rate at which water escapes through the hole, in gallons per hour. How much water drained out during the first three hours of the leak?

**Solution**

The given function models the **rate **at which water escapes through the hole, so you will need to integrate to find how much water was drained out, so

\[ A = \int_0^3 \frac{1}{2}t\,\mathrm{d}t. \]

You can use the Power Rule and the Fundamental Theorem of Calculus to evaluate the above definite integral, that is

\[ \begin{align} A &= \int_0^3 \frac{1}{2}t\, \mathrm{d}t \\ &= \left(\frac{1}{4}(3)^2 \right) - \left(\frac{1}{4}(0)^2 \right) \\ &= \frac{9}{4} \\ &=2.25 \end{align} \]

This means that \(2.25\) gallons of water drained out from the tank during the first three hours of the leak.

There are accumulation problems everywhere!

The function\[ S(t) = -t^2+2t \quad \text{for} \quad 0\leq t \leq 2\]models the rate at which rain pours from a storm that lasts two hours, in millions of gallons per hour. How much water is accumulated during the second hour of the storm?

**Solution**

Once again, note that the function is giving the rate at which rain is pouring down from the skies. This means that you only need to find the definite integral of the function over the requested interval. Since you are asked to find the accumulation during the second hour of the storm, the integration limits should be from \( 1\) to \(2.\) This will give you the definite integral

\[ A = \int_1^2 (-t^2+2t) \, \mathrm{d}t,\]

which you can also solve using the Power Rule and the Fundamental Theorem of Calculus, so

\[ \begin{align} A &= \int_1^2 (-t^2+2t) \, \mathrm{d}t \\ &= \left( -\frac{1}{3}(2)^3+(2)^2 \right) - \left( -\frac{1}{3}(1)^3+(1)^2 \right) \\ &= \frac{4}{3}-\left(\frac{2}{3}\right) \\ &= \frac{2}{3} \\ &\approx 0.67 . \end{align}\]

So about \( 0.67 \) millions of gallons of water rained down from the skies during the last hour of the storm.

Always remember to check if the function represents a change of rate or not!

The function

\[ S(m) = 125 + 20m \]

represents how much money, in dollars, is in a savings account in function of how many months have passed since its opening. How much money was deposited between the 4th and 8th month of its opening?

**Solution**

This time the function is **not **representing a rate of change. Instead it tells you how much money is present in the savings account. This means that you can find how much money was deposited by finding the difference

\[ S(8) - S(4),\]

which will give you

\[\begin{align} S(8) - S(4) &= \left( 125+20(8) \right) - \left( 125+20(4) \right) \\ &= (125+160)-(125+80) \\ &= 80.\end{align}\]

This means that a total of \( \$ 80 \) were deposited during those four months. You did not need to integrate anything!

- An
**accumulation problem**is a problem where the rate of change is given and you are tasked with finding the net change of the quantity over time.- Accumulation problems are usually solved using definite integrals.

- To find the accumulation, you need to
**integrate**the rate of change of the function over a given interval. - The rate of change of a function can be identified using some key words, like
**rate**,**speed**,**velocity**, or**derivative**. - It is possible that rather than giving you the rate of change, the problem will give you the actual quantity as a function. In such a case you only need to find the difference of the function evaluated at the ends of the requested interval.

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