Worked Problems On Rates In Applied Models

Why Average Rates Matter in Applied Models

Many real situations involve quantities that change: the cost of producing more items, the temperature of a cooling object, or the fluid level in a filling tank. To describe how much one quantity changes in relation to another over an interval, we use an average rate of change.

The calculation is simple, but its interpretation depends on choosing the right endpoints, subtracting in the right order, and including units. A numerical answer without those details can be misleading. For example, a rate of $-2$ has little meaning until we know what is decreasing and how the input is measured.

Definition and Interpretation

Definition: If $f$ is defined at inputs $a$ and $b$, where $a<b$, the average rate of change of $f$ from $a$ to $b$ is the change in output divided by the change in input: $$ \frac{f(b)-f(a)}{b-a}. $$

The numerator $f(b)-f(a)$ is the output change across the interval. The denominator $b-a$ is the input change. The order must match: the output at $b$ is compared with the output at $a$, and $b$ is compared with $a$. If $f$ describes a cost as a function of the number of items produced, for example, this quotient measures the average cost change per additional item between the two production levels.

Units follow the same division. If the output is measured in dollars and the input in items, the rate has units of dollars per item. If the output is in degrees Celsius and the input is in hours, the rate has units of degrees Celsius per hour. The sign also carries information: a positive rate means the output is larger at the later endpoint, while a negative rate means it is smaller there.

A Reliable Endpoint Process

Before calculating, identify what the input represents, what the output represents, and which two input values define the interval. Then evaluate the model at those inputs. Keep the endpoint order consistent in both the numerator and denominator, and state the result with units and a contextual interpretation.

Identify the quantities.
Determine which variable is the input and which is the output, and note their units.

Choose the interval endpoints.
Use the two input values specified in the question. For an interval from $a$ to $b$, keep $a$ first and $b$ second.

Find the endpoint outputs.
Evaluate the model at each endpoint, or read the corresponding values from the given data.

Divide the changes.
Calculate $[f(b)-f(a)]/(b-a)$, then express its units and meaning in context.

Worked Example: Average Change in Production Cost

Worked Example: Average Cost Increase per Item

A manufacturer models total cost $C(q)$, in dollars, for producing $q$ items by $C(q)=120+8q+0.05q^2$. Find the average rate of change in total cost as production increases from 10 items to 30 items.

The input is production quantity $q$, measured in items. The output is total cost, measured in dollars. The endpoints are $q=10$ and $q=30$, so first evaluate the model at both values:

C(10)=120+8(10)+0.05(10)^2 =120+80+5 =205\text{ dollars}.

C(30)=120+8(30)+0.05(30)^2 =120+240+45 =405\text{ dollars}.

The cost change is $405-205=200$ dollars, and the production change is $30-10=20$ items. Therefore, the average rate is

\frac{C(30)-C(10)}{30-10} = \frac{405-205}{20} = \frac{200}{20} = 10\text{ dollars per item}.

Interpretation: Across the increase from 10 to 30 items, total cost rises by an average of 10 dollars for each additional item produced. This describes the interval as a whole; it does not claim that the cost rises by exactly 10 dollars for every individual item throughout the interval.

Check: The computed rate multiplied by the input change gives the output change: $10\text{ dollars per item}\times20\text{ items}=200\text{ dollars}$. Also, the model values differ by $405-205=200$ dollars, as required.

Worked Example: Temperature Change Over Time

Worked Example: Average Cooling Rate

The temperature $T(t)$, in degrees Celsius, of an object $t$ hours after observation begins is modeled by $T(t)=32-3t+0.25t^2$. Find the average rate of temperature change from $t=0$ to $t=4$ hours.

The input is time in hours, and the output is temperature in degrees Celsius. Evaluate the model at the two endpoints:

T(0)=32-3(0)+0.25(0)^2=32^\circ\text{C}.

T(4)=32-3(4)+0.25(4)^2 =32-12+4 =24^\circ\text{C}.

The temperature change from the first endpoint to the second is $24-32=-8$ degrees Celsius. The elapsed time is $4-0=4$ hours. Thus,

\frac{T(4)-T(0)}{4-0} = \frac{24-32}{4} = \frac{-8}{4} = -2^\circ\text{C per hour}.

Interpretation: Over these four hours, the temperature decreases by an average of 2 degrees Celsius per hour. The negative sign records that the later temperature is lower than the earlier one.

Check: An average change of $-2^\circ\text{C}$ per hour over 4 hours corresponds to a net change of $(-2)(4)=-8^\circ\text{C}$. Starting at $32^\circ\text{C}$, this gives $32-8=24^\circ\text{C}$, matching the model's endpoint value.

Worked Example: Change in Fluid Level

Worked Example: Average Rise in a Tank

The fluid level $h(t)$, in metres, in a tank being filled is modeled by $h(t)=1.2+0.4t+0.1t^2$, where $t$ is time in minutes. Find the average rate of change in fluid level from $t=2$ minutes to $t=6$ minutes.

The input is time in minutes, and the output is fluid level in metres. Evaluate the level at each specified time:

h(2)=1.2+0.4(2)+0.1(2)^2 =1.2+0.8+0.4 =2.4\text{ metres}.

h(6)=1.2+0.4(6)+0.1(6)^2 =1.2+2.4+3.6 =7.2\text{ metres}.

The level changes by $7.2-2.4=4.8$ metres over $6-2=4$ minutes. The average rate is

\frac{h(6)-h(2)}{6-2} = \frac{7.2-2.4}{4} = \frac{4.8}{4} = 1.2\text{ metres per minute}.

Interpretation: Between 2 and 6 minutes, the fluid level rises by an average of 1.2 metres per minute. This is a rate of change in the level, not a statement about the tank's volume per minute.

Check: Multiplying $1.2\text{ metres per minute}$ by 4 minutes gives $4.8$ metres. Adding that change to the initial level, $2.4+4.8=7.2$ metres, recovers the endpoint value.

Why the Quotient Is a Secant Slope

The average-rate formula is the slope formula applied to the two points on a graph that correspond to the endpoints. For a function $f$, the endpoint inputs $a$ and $b$ give the points $(a,f(a))$ and $(b,f(b))$. The horizontal change between those points is $b-a$, and the vertical change is $f(b)-f(a)$. The slope of the line through them is therefore exactly the average rate.

Justification: By the definition of the slope between two points $(x_1,y_1)$ and $(x_2,y_2)$, the slope is $(y_2-y_1)/(x_2-x_1)$, provided $x_2\ne x_1$. Taking $(x_1,y_1)=(a,f(a))$ and $(x_2,y_2)=(b,f(b))$ gives $$ \frac{y_2-y_1}{x_2-x_1} = \frac{f(b)-f(a)}{b-a}. $$ Since $a<b$, the denominator is nonzero. Thus the average rate of change is the slope of the secant line through the two graph points.

This connection explains why the quotient is useful: it summarizes the overall change between two inputs using one number. But that summary does not describe every change within the interval. In the temperature example, $-2^\circ\text{C}$ per hour is the average across four hours. It does not, by itself, tell us the temperature's rate at a particular time.

Common Pitfall: Reversing the Sign

A frequent error is to subtract the endpoint outputs in one order while keeping the input change in the other order. In the temperature example, the incorrect calculation

\frac{32-24}{4-0} = \frac{8}{4} = 2^\circ\text{C per hour}

uses the initial temperature minus the final temperature in the numerator, but uses final time minus initial time in the denominator. The positive result falsely suggests that the temperature increased. For the change from $t=0$ to $t=4$, both differences must run from the first endpoint to the second:

\frac{T(4)-T(0)}{4-0} = \frac{24-32}{4} = -2^\circ\text{C per hour}.

A useful check is to ask whether the sign agrees with the endpoint outputs. If the output is lower at the second endpoint, the average rate must be negative when the inputs are listed in increasing order.

What to Include in an Applied Answer

A complete applied answer should make clear which interval was used, show the endpoint output values, and report the rate with units. It should also translate the sign and size into a sentence about the situation. For example, “$-2^\circ\text{C per hour}$” is the calculation, while “the temperature decreases by an average of 2 degrees Celsius per hour from 0 to 4 hours” is its interpretation.

Always distinguish the average over an interval from a value at a single input. The average-rate calculation uses two endpoint outputs and one endpoint interval. Its conclusion applies to that interval; it does not establish that the output changes steadily or at the same rate at every point within it.

Check Your Understanding

For each question, identify the endpoint changes before interpreting the result.

A cost model gives $C(5)=70$ dollars and $C(15)=190$ dollars. What is the average rate of change in cost from 5 to 15 items, including units?
A temperature changes from $18^\circ\text{C}$ at hour 2 to $10^\circ\text{C}$ at hour 6. Calculate and interpret the average rate of temperature change.
A fluid level is 1.5 metres at minute 3 and 4.5 metres at minute 9. Find its average rate of change in metres per minute.
Why must the output difference and input difference use the same endpoint order?
Does an average rate of $1.2$ metres per minute over an interval prove that the fluid level rises at exactly that rate at every moment in the interval? Explain.