2.1 Average Rate Of Changeap Calculus



Distinguish between an average rate of change and an average value of a function. In this session, you will practice with 5 multiple choice problems and all or part of the following free response problems: 2007 Form B AB/BC3, 2011 Form B AB/BC 1, 2004 AB/BC 1, 2004 Form B AB 2. Calculus: Home List of Lessons Teacher Resources Previous Lesson. 2.1 Average and Instantaneous Rate of Change: Next Lesson. Calc2.1packet.pdf: File Size.

Answer

a) $frac{f(x)-f(3)}{x-3}$ b) $f'(3) = limlimits_{h to 0}frac{f(3+h)-f(3)}{h}$

Problems involving Limits and Average Rates. From the text on Page 66: 2, 8, 20, 40. For notes and practice problems, visit the Calculus course on (Version #1) is created for a 45-minute class period and fo.

Work Step by Step

For part a, we know that the secant line is the average rate of change on the interval between the two points from equation 6. To calculate the slope of this line, we need to calculate the change in y divided by the change in x: $frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$ We can then plug our two given points into the equation to get our answer: $frac{f(x)-f(3)}{x-3}$ For part b, we know from equation 4 that the equation to find the slope of the tangent line is: $f'(x) = limlimits_{h to 0}frac{f(a+h)-f(a)}{h}$ The question tells us that point P is located at (3, f(3)). Next we can plug in 3 for a to get the equation for the tangent line: $f'(3) = limlimits_{h to 0}frac{f(3+h)-f(3)}{h}$

Quick Overview

  • For the function, $$f(x)$$, the average rate of change is denoted $$frac{Delta f}{Delta x}$$.
  • In mathematics, the Greek letter $$Delta$$ (pronounced del-ta) means 'change'.
  • When interpreting the average rate of change, we usually scale the result so that the denominator is 1.
  • Average Rates of Change can be thought of as the slope of the line connecting two points on a function.

Familiar Example

Suppose you drive 120 miles in two hours. How fast were you going? The obvious answer is '60 miles per hour'.

Now, did you really drive at that speed the entire time? Maybe. Then again, you might have had to slow down because of heavy traffic for a while, but later you were able to drive a bit faster. So, over the two hours your speed averaged out to 60 mph.

This is called Average Velocity or Average Speed and it is a common example of using an average rate of change Playing with fire 2watermelon gaming. in our everyday lives.

Examples

Example 1

Find the average rate of change for $$f(x) = x^2 -3x$$ between $$x = 1$$ and $$x = 6$$.

Step 1

Calculate the change in function value.

$$ blue{f(6)}-red{f(1)} = blue{(6^2 -3cdot 6)} - red{(1^2 -3cdot 1)} = blue{18} - red{(-2)} = 20 $$

Step 2

Calculate the change in the variable value.

$$ 6 -1 = 5 $$

Step 3

Find the ratio of the changes.

2.1 Average Rate Of Changeap Calculus

$$ begin{align*} frac{Delta f}{Delta x} & = frac{f(6)-f(1)}{6 -1}[6pt] & = frac{20} 5[6pt] & = 4 end{align*} $$

Answer

The average rate of change in $$f$$ between $$x = 1$$ and $$x = 6$$ is 4.

Example 2

Suppose the monthly profit for a particular company can be described by the function

$$P(x) = -x^2 + 9x - 8,$$

where $$x$$ is the number of tons of goods sold by the company and $$P(x)$$ is the amount of profit earned, measured in thousands of dollars.

Determine and interpret the value of $$frac{Delta P}{Delta x}$$ when the amount of goods sold increases from 4 tons to 4.5 tons.

Step 1

Calculate $$frac{Delta P}{Delta x}$$

$$ begin{align*} frac{Delta P}{Delta x} & = frac{blue{P(4.5)} - red{P(4)}}{4.5 - 4}[6pt] & = frac{blue{(-(4.5)^2 + 9(4.5) - 8)} - red{(-(4)^2 + 9(4) - 8)}}{0.5}[6pt] & = frac{blue{12.25} - red{12}}{0.5}[6pt] & = frac{0.25}{0.5}[6pt] & = frac 1 2 end{align*} $$

Step 2

Interpret the results.

We found that $$frac{Delta P}{Delta x} = frac{1/2} 1$$. This tells us that as sales increase from 4 to 4.5 tons, the profit will increase by $500 per ton on average.

Note: Since $$P(x)$$ is measured in thousands of dollars, $$Delta P = frac 1 2$$ means '1/2 of a thousand dollars,' which is $500.

Answer

As sales increase from 4 tons of goods to 4.5 tons of goods, we expect to see profits increase at an average rate of $500 per ton.

Example 3

Suppose an electrical circuit contains a variable resistor. As the resistor is adjusted from 3 ohms to 3.15 ohms, the electrical current is decreasing by an average 0.03 amps per ohm.

Write an equation that expresses this situation in terms of an average rate of change.

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Step 1

Define variables.

Let $$I =$$ the amount of current flowing through the circuit (measured in amps).
Let $$R=$$ the amount of resistance provided by the variable resistor (measured in ohms).

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Answer

2.1 Average Rate Of Changeap Calculus Equation

$$frac{Delta I}{Delta R} = -0.03$$

(Note that we are told in the statement of the problem what the average rate of change is. We're not asked to calculate it.)

Example 4

Suppose $$P$$ is measured in hundreds of fish and represents the size of a particular population of Hawaiian trigger fish, and $$t$$ represents time, measured in years. Interpret the following equation in a complete sentence.

$$ frac{Delta P}{Delta t} = 0.05 $$

Step 1

Write the right-hand side of the equation as a fraction with a 1 in the denominator.

$$frac{Delta P}{Delta t} = frac{0.05} 1$$

Step 2

Identify the values of $$Delta P$$ and $$Delta t$$.

$$Delta P = 0.05$$ which is the same as an increase of 5 fish, since $$P$$ is measured in hundreds of fish.
$$Delta t = 1$$ represents 1 year.

2.1 Average Rate Of Changeap Calculus Formula

Answer

The population is increasing at an average rate of 5 fish per year.

Example 5

2.1 Average Rate Of Changeap Calculus Ab

Suppose someone has been driving at exactly 60 miles per hour for one hour. How fast would they need to drive for the next 30 minutes in order to increase their average speed to 70 miles per hour?

Step 1

Determine the total distanced traveled if the average speed were 70 mph for 1.5 hours.

2.1 Average Rate Of Changeap Calculus Calculator

$$ frac{70mbox{ miles}}{1 mbox{ hour}} cdot frac{1.5mbox{ hours}} 1 = 70(1.5)mbox{ miles} = 105mbox{ miles} $$

Step 2

After traveling at 60mph for an hour, determine the remaining distance left to travel.

Since the driver has been driving at 60mph for one hour, the distance traveled so far is 60 miles. This leaves 45 miles to travel.

Step 3

Determine the speed needed to cover 45 miles in 30 minutes.

$$ frac{45mbox{ miles}}{frac 1 2 mbox{ hour}} = frac{2cdot 45mbox{ miles}}{1mbox{ hour}} = frac{90mbox{ miles}}{1mbox{ hour}} $$

Answer

The person would need to drive 90 miles per hour for 30 minutes.

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