How to Find Average Rates of Change:
Practice Problems

$$ \begin{align*} \frac{\Delta f}{\Delta x} & = \frac{\blue{f(4)} - \red{f(0)}}{4 - 0}\\[6pt] & = \frac{\blue{\frac{4+1}{4+2}} - \red{\frac{0+1}{0+2}}} 4\\[6pt] & = \frac{\blue{\frac 5 6} - \red{\frac 1 2}} 4\\[6pt] & = \frac{1/3} 4\\[6pt] & = \frac 1 {12} \end{align*} $$

$$\displaystyle \frac{\Delta f}{\Delta x} = \frac 1 {12}$$

$$ \begin{align*} \frac{\Delta f}{\Delta x} & = \frac{\blue{f(2\pi)}-\red{f(\pi)}}{2\pi - \pi}\\[6pt] & = \frac{\blue{\sin 2\pi}-\red{\sin\pi}}{\pi}\\[6pt] & = \frac{\blue{0}-\red{0}}{\pi}\\[6pt] & = 0 \end{align*} $$

$$\displaystyle \frac{\Delta f}{\Delta x} = 0$$

$$ \begin{align*} \frac{\Delta f}{\Delta t} & = \frac{\blue{f(8)} - \red{f(-2)}}{8 - (-2)}\\[6pt] & = \frac{\blue{60e^{0.5(8)}} - \red{60e^{0.5(-2)}}}{8 - (-2)}\\[6pt] & = \frac{60\left(e^4 - e^{-1}\right)}{10}\\[6pt] & = 6\left(e^4 - e^{-1}\right)\\[6pt] & \approx 325.3816 \end{align*} $$

$$ \displaystyle \frac{\Delta f}{\Delta t} = 6\left(e^4 - e^{-1}\right)\approx 325.3816 $$

$$ \begin{align*} \frac{\Delta f}{\Delta x} & = \frac{\blue{f(-3)} - \red{f(-6)}}{-3 - (-6)}\\[6pt] & = \frac{\blue{(2 - 8(-3) - 5(-3)^3)} - \red{(2 - 8(-6) - 5(-6)^3)}}{-3 +6}\\[6pt] & = \frac{\blue{161} - \red{1130}} 3\\[6pt] & = -\frac{969} 3\\[6pt] & = - 323 \end{align*} $$

$$\frac{\Delta f}{\Delta x} = -323$$

$$ \begin{align*} \frac{\Delta P}{\Delta t} & = \frac{\blue{P(10)} - \red{P(5)}}{10 -5}\\[6pt] & = \frac{\blue{\frac{250}{1+4e^{-0.75(10)}}} - \red{\frac{250}{1+4e^{-0.75(5)}}}}{5}\\[6pt] & = \frac{250\left(\frac 1 {1+4e^{-7.5}} - \frac 1 {1+4e^{-3.75}}\right)}{5}\\[6pt] & = 50\left(\frac 1 {1+4e^{-7.5}} - \frac 1 {1+4e^{-3.75}}\right)\\[6pt] & \approx 4.2 \end{align*} $$

From year 5 to year 10 the population of cute, fluffy bunnies increases at an average rate of about 4.2 bunnies per year.

$$ \begin{align*} \frac{\Delta C}{\Delta x} & = \frac{\blue{C(120)} - \red{C(150)}}{120 - 150}\\[6pt] & = \frac{\blue{25(120)+4500} - \red{25(150)+4500}}{-30}\\[6pt] & = \frac{\blue{7500} - \red{8250}}{-30}\\[6pt] & = \frac{-750}{-30}\\ & = 25 \end{align*} $$

As the amount of goods produced drops from 150 pallets to 120 pallets, the cost of production decreases an average of $25 per pallet.

Note 1: We could have saved ourselves the effort of calculating $$\Delta C/\Delta x$$ by simply noticing $$C(x)$$ is a linear function. The average rate of change of any linear function is just its slope.

Note 2: When the average rate of change is positive, the function and the variable will change in the same direction. In this case, since the amount of goods being produced decreases, so does the cost.

$$ \begin{align*} \frac{\Delta A}{\Delta t} & = \frac{\blue{A(5)} - \red{A(0)}}{5-0}\\[6pt] & = \frac{\blue{2000\left(1 + \frac{0.08}{12}\right)^{12(5)}} - \red{2000\left(1 + \frac{0.08}{12}\right)^{12(0)}}}{5}\\[6pt] & = \blue{400\left(1 + \frac{0.08}{12}\right)^{60}} - \red{400\left(1 + \frac{0.08}{12}\right)^{0}}\\[6pt] & = \blue{400\left(1 + \frac{0.08}{12}\right)^{60}} - \red{400}\\[6pt] & \approx 195.94 \end{align*} $$

$$ \begin{align*} \frac{\Delta A}{\Delta t} & = \frac{\blue{A(10)} - \red{A(5)}}{10-5}\\[6pt] & = \frac{\blue{2000\left(1 + \frac{0.08}{12}\right)^{12(10)}} - \red{2000\left(1 + \frac{0.08}{12}\right)^{12(5)}}} 5\\[6pt] & = \blue{400\left(1 + \frac{0.08}{12}\right)^{120}} - \red{400\left(1 + \frac{0.08}{12}\right)^{60}}\\[6pt] & \approx 291.92 \end{align*} $$

• During the first five years, the account grows by an average of $195.94 per year.
• During the second five years, the account grows by an average of $291.92 per year.

$$ \begin{align*} \frac{\Delta R}{\Delta V} & = \frac{\blue{R(9)}-\red{R(1.5)}}{9-1.5}\\[6pt] & = \frac{\blue{\frac{0.02} 9}-\red{\frac{0.02}{1.5}}}{7.5}\\[6pt] & = \left(\blue{\frac{0.02} 9}-\red{\frac{0.02}{1.5}}\right)\cdot \frac 1 {7.5}\\[6pt] & = -\frac 1 {675}\\[6pt] & \approx -0.00148 \end{align*} $$

As the voltage increases from 1.5 volts to 9 volts the resistance will decrease at an average rate of $$\frac 1 {675}$$ ohms per volt, or approximately 0.00148 ohms per volt.

$$ \frac{\Delta P}{\Delta t} = 12\% = \frac{12\%} 1 $$

As $$t$$ increases from 3 hours to 12 hours of training, proficiency increases at an average rate of 12% per hour.

$$ \frac{\Delta R}{\Delta x} = -0.2 = -\frac{0.2} 1 $$

When sales increase from 0.8 to 1.4 tons, the company's revenue decreases at an average rate of $200 per ton of goods sold.

Note 1: Since the average rate of change is negative, the two quantities change in opposite directions. Since the amount of goods sold is increasing, revenue must be decreasing. Note 2: Even though the average rate of change in revenue is negative, this does not mean that the company is losing money. It only means they are earning less per ton than previously. This might happen if the company decreases the price of their goods. They sell more goods, but earn less per item.

• Let $$I = $$ the amount of electrical current flowing through the circuit, measured in amps.
• Let $$t$$ represent time, measured in seconds.

$$\displaystyle \frac{\Delta I}{\Delta t} = 0.03$$

• Let $$d$$ represent the persons distance from their starting point, in kilometers.
• Let $$t$$ represent time, in hours.

$$\displaystyle \frac{\Delta d}{\Delta t} = 85$$

$$ \frac{80\mbox{ kilometers}}{1\mbox{ hour}} \cdot \frac{2.25\mbox{ hours}} 1 = (80)(2.25) \mbox{ kilometers} = 180 kilometers $$

The driver has spent $$3/4$$ of an hour driving at 50 kph, and so had traveled $$50\cdot 0.75 = 37.5$$ kilometers. This left $$180-37.5 = 142.5$$ kilometers to travel.

The person needed to travel 142.5 kilometers in 1.5 hours. So the speed had to be

$$ \frac{142.5\mbox{ kilometers}}{1.5\mbox{ hours}} = 95\mbox{ kph.} $$

The person drove at a speed of 95 kilometers per hour for the last 1.5 hours.

$$ \frac{50\mbox{ joules}}{\mbox{second}} \cdot \frac{20\mbox{ minutes}} 1 = \frac{50\mbox{ joules}}{\mbox{second}} \cdot \frac{1200\mbox{ seconds}} 1 = 60{,}000\mbox{ joules}. $$

Since the bulb had been burning at 30 watts for 15 minutes, it had already used

$$ \frac{30\mbox{ joules}}{\mbox{second}} \cdot \frac{15\mbox{ minutes}} 1 = \frac{30\mbox{ joules}}{\mbox{second}} \cdot \frac{900 \mbox{ seconds}} 1 = 27{,}000\mbox{ joules}. $$

The remaining energy to be used would be 60,000-27,000=33,000 joules.

The remaining energy would have to be used in 5 minutes which is the same as 300 seconds. So, we have

$$ \frac{33{,}000\mbox{ joules}}{300\mbox{ seconds}} = \frac{110\mbox{ joules}}{1\mbox{ second}} = 110\mbox{ watts}. $$

The bulb would have burned at 110 watts during the last 5 minutes.