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Rates of Change

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Rates of Change

What if when you bought a koi fish it measured 4 inches long? Three months later you measure it again and it is 6 inches long? At what rate is the fish growing? After completing this Concept, you'll be able to find the rate of change taking place in problems like this one.

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CK-12 Foundation: 0407S Rate of Change (H264)

Guidance

The slope of a function that describes real, measurable quantities is often called a rate of change . In that case the slope refers to a change in one quantity (y) per unit change in another quantity (x) . (This is where the equation m= \frac{\Delta y}{\Delta x} comes in—remember that \Delta y and \Delta x represent the change in y and x respectively.)

Example A

A candle has a starting length of 10 inches. 30 minutes after lighting it, the length is 7 inches. Determine the rate of change in length of the candle as it burns. Determine how long the candle takes to completely burn to nothing.

Solution

First we’ll graph the function to visualize what is happening. We have 2 points to start with: we know that at the moment the candle is lit ( time = 0 ) the length of the candle is 10 inches, and after 30 minutes ( time = 30 ) the length is 7 inches. Since the candle length depends on the time, we’ll plot time on the horizontal axis, and candle length on the vertical axis.

The rate of change of the candle’s length is simply the slope of the line. Since we have our 2 points (x_1, y_1) = (0, 10) and (x_2, y_2) = (30, 7) , we can use the familiar version of the slope formula:

\text{Rate of change}& = \frac{y_2 - y_1}{x_2 -x_1}\\ &= \frac{(7 \ \text{inches})-(10 \ \text{inches})}{(30 \ \text{minutes})-(0 \ \text{minutes})}\\ &= \frac{-3 \ \text{inches}}{30 \ \text{minutes}}\\ &= -0.1 \ \text{inches per minute}

Note that the slope is negative. A negative rate of change means that the quantity is decreasing with time—just as we would expect the length of a burning candle to do.

To find the point when the candle reaches zero length, we can simply read the x- intercept off the graph (100 minutes). We can use the rate equation to verify this algebraically:

\text{Length burned} & = \text{rate} \times \text{time}\\10 & = 0.1 \times 100

Since the candle length was originally 10 inches, our equation confirms that 100 minutes is the time taken.

Example B

The population of fish in a certain lake increased from 370 to 420 over the months of March and April. At what rate is the population increasing?

Solution

Here we don’t have two points from which we can get x- and y- coordinates for the slope formula. Instead, we’ll need to use the alternate formula, m=\frac{\Delta y}{\Delta x} .

The change in y- values, or \Delta y , is the change in the number of fish, which is 420 - 370 = 50 . The change in x- values, \Delta x , is the amount of time over which this change took place: two months. So \frac{\Delta y}{\Delta x} = \frac{50 \ \text{fish}}{2 \ \text{months}} , or 25 fish per month.

Interpret a Graph to Compare Rates of Change

Example C

The graph below represents a trip made by a large delivery truck on a particular day. During the day the truck made two deliveries, one taking an hour and the other taking two hours. Identify what is happening in the first 3 stages of the trip (stages A through C).

Solution:

The first 3 stages of the trip are:

A. The truck sets off and travels 80 miles in 2 hours.

B. The truck covers no distance for 2 hours.

C. The truck covers (120 - 80) = 40 \ \text{miles} in 1 hour.

A. \text{Rate of change} = \frac{\Delta y}{\Delta x}=\frac{80 \ \text{miles}}{2 \ \text{hours}}=40\ \text{miles per hour}

Notice that the rate of change is a speed— or rather, a velocity. (The difference between the two is that velocity has a direction, and speed does not. In other words, velocity can be either positive or negative, with negative velocity representing travel in the opposite direction. You’ll see the difference more clearly in part E.)

Since velocity equals distance divided by time, the slope (or rate of change) of a distance-time graph is always a velocity.

So during the first part of the trip, the truck travels at a constant speed of 40 mph for 2 hours, covering a distance of 80 miles.

B. The slope here is 0, so the rate of change is 0 mph. The truck is stationary for one hour. This is the first delivery stop.

C. \text{Rate of change} = \frac{\Delta y}{\Delta x}=\frac{(120 - 80) \ \text{miles}}{(4-3) \ \text{hours}} = 40 \ \text{miles per hour}. The truck is traveling at 40 mph.

Watch this video for help with the Examples above.

CK-12 Foundation: Rate of Change

Vocabulary

  • Slope is a measure of change in the vertical direction for each step in the horizontal direction. Slope is often represented as “ m ”.
  • Slope can be expressed as \frac{\text{rise}}{\text{run}} , or \frac{\Delta y}{\Delta x} .
  • The slope between two points (x_1, y_1) and (x_2, y_2) is equal to \frac{y_2 - y_1}{x_2 -x_1} .
  • Horizontal lines (where y = a constant) all have a slope of 0.
  • Vertical lines (where x = a constant) all have an infinite (or undefined) slope.
  • The slope (or rate of change ) of a distance-time graph is a velocity.

Guided Practice

Continue where we left of in Example C, by identifying what is happening in the last two stages of the trip (stages D and E).

Solution:

The last two stages of the trip are:

D. The truck covers no distance for 1 hour.

E. The truck covers -120 miles in 2 hours.

Let's find the slopes:

D. Here the slope is 0, so the rate of change is 0 mph. The truck is stationary for two hours. This is the second delivery stop. At this point the truck is 120 miles from the start position.

E.

\text{Rate of change} &= \frac{\Delta y}{\Delta x}\\ &=\frac{(0-120) \ \text{miles}}{(8-6) \ \text{hours}}\\ &=\frac{-120 \ \text{miles}}{2 \ \text{hours}}\\ &=-60 \ \text{miles per hour.}

The truck is traveling at negative 60 mph.

Wait – a negative speed? Does that mean that the truck is reversing? Well, probably not. It’s actually the velocity and not the speed that is negative, and a negative velocity simply means that the distance from the starting position is decreasing with time. The truck is driving in the opposite direction – back to where it started from. Since it no longer has 2 heavy loads, it travels faster (60 mph instead of 40 mph), covering the 120 mile return trip in 2 hours. Its speed is 60 mph, and its velocity is -60 mph, because it is traveling in the opposite direction from when it started out.

Practice

For 1-6, the graph below is a distance-time graph for Mark’s three and a half mile cycle ride to school. During this ride, he rode on cycle paths but the terrain was hilly. He rode slower up hills and faster down them. He stopped once at a traffic light and at one point he stopped to mend a punctured tire. The graph shows his distance from home at any given time. Identify each section of the graph accordingly.

  1. Section A.
  2. Section B.
  3. Section C.
  4. Section D.
  5. Section E.
  6. Section F.

For 7-12, approximate the slope of each part of Mark's ride.

  1. Section A.
  2. Section B.
  3. Section C.
  4. Section D.
  5. Section E.
  6. Section F.

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