average rate of change worksheet homework #5

• Derivatives

How to Find Average Rates of Change: Practice Problems

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Determine the average rate of change for $$\displaystyle f(x) = \frac{x+1}{x+2}$$ from $$x = 0$$ to $$x = 4$$ .

Calculate the average rate of change .

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

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

Determine the average rate of change for $$f(x) = \sin x$$ from $$x = \pi$$ to $$x = 2\pi$$ (where $$x$$ is measured in radians).

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

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

Determine the average rate of change for the function below, from $$t = -2$$ to $$t = 8$$ .

$$ f(x) = 60e^{0.5t} $$

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

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

Determine the average rate of change for the function below, from $$x = -6$$ to $$x = -3$$ .

$$ f(x) = 2 - 8x - 5x^3 $$

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

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

Suppose the average size of a particular population of cute, fluffy bunny rabbits can be described by the function

$$ P(t) = \frac{250}{1+4e^{-0.75t}}, $$

where $$t$$ is measured in years and $$P(t)$$ is measured in numbers of bunnies.

As time increases from $$t = 5$$ to $$t = 10$$ , what is the average rate of change in the bunny population?

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

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

At a particular company, the cost of producing $$x$$ pallets of goods can be described by the function

$$ C(x) = 25x + 4500, $$

where $$C(x)$$ is measured in dollars. Determine the average rate of change in the cost as production decreases from 150 pallets to 120 pallets.

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

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.

Suppose you invest $2000 in an account that earns 8% interest each year, but interest is compounded each month. Then the amount you have in the account is described by the function

$$ A(t) = 2000\left(1 + \frac{0.08}{12}\right)^{12t}. $$

If you make no deposits or withdrawals, what is the average rate of change in the amount of money in the account ...

• over the first 5 years?
• over the second 5 years?

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

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

• 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.

Suppose a particular electrical circuit is designed to keep the current, $$I$$, at a constant $$0.02$$ amps. However, both the voltage, $$V$$, and the resistance, $$R$$, can vary. Then according to Ohm's Law,

$$R = \frac{0.02} V,$$

where $$R$$ is measured in Ohms and $$V$$ is measured in volts.

What is the average rate of change in the resistance on the circuit as the voltage increases from 1.5 volts to 9 volts?

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

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.

Suppose $$P(t)$$ represents the proficiency achieved at a particular task after receiving $$t$$ hours training. Suppose the following equation applies when $$t$$ increases from 3 to 12. Interpret the equation in a complete sentence.

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

Rewrite the average rate of change as a fraction with a denominator of 1.

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

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

Suppose $$R(x)$$ represents the revenue (in thousands of dollars) earned by a particular company from the sale of $$x$$ tons of goods. Suppose the following equation applies when sales increase from 0.8 tons to 1.4 tons. Interpret the equation in a complete sentence.

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

Rewrite the average rate of change so it has a 1 in the denominator.

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

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.

Suppose the current in an electrical circuit increases at an average rate of 0.03 amps per second. Write an equation expressing this idea.

Define variables.

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

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

Suppose someone drives with an average velocity of 85 kilometers per hour. Write an equation expressing this idea.

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

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

Suppose someone has been driving for 45 minutes at a steady 50 kilometers per hour. Then they increased their speed and drove for the another 1.5 hours. When they arrived at their destination, their average speed for the entire trip was 80 kilometers per hour. How fast did they drive during the last 1.5 hours?

Find the total distance driven if the person had been driving at 80 kph for the entire 2.25 hours.

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

Determine the remaining distance that had to be driven during the last 1.5 hours.

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.

Determine the speed needed to cover the remaining distance in the remaining time.

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.

In electrical circuits, energy is measured in joules (pronounced jools ) and power is measured in watts. The relationship between the two is

$$ 1\mbox{ watt} = \frac {1\mbox{ joule}}{\mbox{second}} $$

So, watts are the rate of change of energy relative to time (just like speed is the rate of change of distance relative to time).

Suppose a variable wattage lightbulb (like a lightbulb on a dimmer switch) has been pulling 30 watts for the past 15 minutes. The wattage is then increased so that after another 5 minutes the average rate of change for the entire 20 minutes is 50 watts.

What was the higher wattage the bulb was set to in order to achieve this?

Determine the total amount of energy used during the 20 minutes.

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

Determine the amount of energy that needed to be used during the last 5 minutes.

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.

Determine the rate (in joules/second) that would be needed to use the remaining energy during the last 5 minutes.

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.

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Rate of Change Worksheets

The concept of rate of change is when we are track at least two quantities and see how their values get modified over time. The independent variable is variable that stands alone and the cause for this distortion. We normally define the independent variable as x. The variable that it has a direct effect on is referred to as the dependent variable. The dependent variable is normally stated as y. The rate of change is quantifiable by determining the quotient of these two variables in context of the equation: rate of change = difference in y (dependent variable) ÷ difference in x (independent variable). This is a wonderful selection of worksheets and lessons that show you how to predict the possible rate of change of various functions.

Aligned Standard: HSF-IF.B.6

• Starting with Slope Step-by-step Lesson - This is more of a basic reminder of the skills that will be required for the other sections.
• Guided Lesson - I know that it is a pretty lame scenario with the potatoes. I was running low on creativity that day.
• Guided Lesson Explanation - I always like to encourage students to draw a graph even if all they needed was to calculate the slope.
• Practice Worksheet - Some of the word problems might stump kids. That is why it gets them into a set routine to adapt to the problem.
• Matching Worksheet - Many people will trip up between choices d and e.
• Answer Keys - These are for all the unlocked materials above.

Homework Sheets

Finding the slope on the first two leads us to finding an average trend on the third homework.

• Homework 1 - Slope is basically a measure of how fast a line ascends or descends.
• Homework 2 - The formula of slope is: y 2 - y 1 / x 2 - x 1
• Homework 3 - Mack had 4 kg Onions. He peeled 3 kg of them in 1 hour. Now he has 1 kg Onions left. The graph below shows Mack's situation.

Practice Worksheets

Now we have students start to interpret the meaning of the slope of the line.

• Practice 1 - Find the average rate of change of y with respect to x over the interval [11, 12].
• Practice 2 - To find the average rate of change put the interval values in equation and solve them.
• Practice 3 - Mary had 6 liters of milk in his jug. She was making milk shake at the rate of 3 liters of milk shake in 2 hours. He had 3 liters of milk left. The graph shows Mary's situation.

Math Skill Quizzes

We start with some simple word problems to target the concept of slope.

• Quiz 1 - A wall is 7 feet tall. After 2 years it was 9 feet tall. What does the slope tell us about the tree's growth situation is this negative or positive?
• Quiz 2 - A building was 100 feet tall. After 4 days construction was done and it was 150 feet tall. What does the slope tell us about the building growth situation is this negative or positive?
• Quiz 3 - Maria has 16 kg fresh fruits. After 2 hours she left with 8 kg fresh fruits. If an equation were written to represent this situation, what does the slope tell us about the growth of Maria's fresh flower is this negative or positive?

What Are the Degrees of Rates of Change?

Rates of change can be determined to be positive or negative. A positive rate of change is indicated by the upward trajectory of x and y on a graph. A negative rate of change is indicated by x and y values being in conflict, as the x value increase the y value decreases on a graph. These degrees vary by the elevation or decline in y-value. There are also situations where there is no or zero rate of change, this results in a flat horizontal line when it is graphed.

How to Predict the Rate of Change of Functions

Functions have extensive usage in advanced mathematics and calculus. Calculus is used extensively in other aspects of science. By definition, a function is a procedure in which every input gets associated with only one output. One of the implementations of the function is finding out its rate of change. What this does is describe the transformation in one quantity about the other. In other words, if y is the independent variable and x is the dependent one then: rate of change= (change in x )/(change in y).

There are numerous parameters in which the rate of change of functions can be used comprehensively. Such as finding out the acceleration of a vehicle. The acceleration deals with the change in velocity concerning time. In general, the change in the rate of a function can be denoted as: f(x) = (f(x + h) - f(a)) / (b - a ).

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Average Rate Of Change

Average Rate Of Change - Displaying top 8 worksheets found for this concept.

Some of the worksheets for this concept are Exercise set average rate of change, 03, Average rate of change, Average rates of change date period, Rate of change and slope, Activity average rate of change, 03, Rate of change 1.

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1. Exercise Set 2.5: Average Rate of Change

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Average Rate of Change

Related Topics: Lesson Plans and Worksheets for Grade 8 Lesson Plans and Worksheets for all Grades More Lessons for Grade 8 Common Core For Grade 8

Examples, solutions, and videos to help Grade 8 students learn how to compute the average rate of change in the height of water level when water is poured into a conical container at a constant rate.

New York State Common Core Math Grade 8, Module 7, Lesson 22

Worksheets for Grade 8

Lesson 22 Student Outcomes

• Students know how to compute the average rate of change in the height of water level when water is poured into a conical container at a constant rate.

Lesson 22 Summary

• We know intuitively that the narrower part of a cone will fill up faster than the wider part of a cone. • By comparing the time it takes for a cone to be filled to a certain water level, we can determine that the rate of filling the cone is not constant. Lesson 22 Classwork Watch this video of a cone being filled with water.

Exercise The height of a container in the shape of a circular cone is 7.5 ft. and the radius of its base is 3 ft., as shown. What is the total volume of the cone? • If we knew the rate at which the cone was being filled with water, how could we use that information to determine how long it would take to fill the cone? • Water flows into the container (in its inverted position) at a constant rate of ft3 when will the container be filled? • Now we want to show that even though the water filling the cone flows at a constant rate, the rate of change of the volume in the cone is not constant. For example, if we wanted to know how many minutes it would take for the level in the cone to reach 1 ft., then we would have to first determine the volume of the cone when the height is 1 ft. Do we have enough information to do that? • What equation can we use to determine the radius when the height is 1 ft.? Explain how your equation represents the situation. • Use your equation to determine the radius of the cone when the height is 1 ft. • Now determine the volume of the cone when the height is 1 ft. • Calculate the number of minutes it would take to fill the cone at 1 ft. intervals. Organize your data in the table below. • We know that the sand (rice, water, etc.) being poured into the cone is poured at a constant rate, but is the level of the substance in the cone rising at a constant rate? Provide evidence to support your answer.

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