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This essential trains for: SAT-I, GMAT, AMC-8, Math Kangaroo 5-6, Math Kangaroo 7-8.

The average rate of a process is: Example:

In the following table, the growth of a tree is documented based on its height in feet. Compute the average rate of growth over the first 6 years of the sugar maple.

(data from the Firelands Electric Cooperative)

 SPECIES 1 Year 2 Years 3 Years 4 Years 5 Years Maple, silver 3.1 ft 5.6 ft 7.7 ft 9.1 ft 10.4 ft

The average rate of growth over the 5 year interval is: We notice that, if we graph the yearly growth over time, there are different rates of growth each year. We obtain graph 1 (red).

Graphing the average rate results in graph 2 (blue).

Notice how the average growth rate is a linear approximation of the growth process. More accurately, we should have in mind the duration of the process between an initial and a final state. For a quantity Q and time t: The rate may be positive in which case the quantity Q increases, or negative in which case the quantity Q decreases.

Example:

Measurements of the carbon dioxide concentration at the Mauna Loa volcano site in Hawaii, are summarized in the following table:

(data provided by NOAA)

 Month CO2 (ppm) January 378 February 379.8 March 381 April 382.2 May 382.5 June 382.2 July 380.5 August 378.3 September 376.5 October 376.6 November 378 December 380

Question: Between what months was the monthly rate of change of the carbon dioxide concentration negative? What was the average monthly rate over this period?

Solution: May to September. The rate is negative because the concentration has decreased. Question: Between what months was the monthly rate of change of the carbon dioxide concentration positive? What was the average monthly rate over the first such period?

Solution: January to May. The rate is positive because the concentration has increased. Question: What was the average monthly rate of change of the CO2 concentration from January to September?

Solution: A common error is to simply average the two rates from before. This results in a different answer. This answer is not the result of applying the definition of the average rate and is, therefore, wrong: Typical questions involving rates

Uniform motion problem:

In 1909, Louis Bleriot crossed the English Channel by plane for the first time in history, flying in 22 miles in 37 minutes. In 1927, Charles Lindbergh crossed the Atlantic Ocean by plane for the first time in history, flying 3,610 miles in 33 1/2 hours. What is the ratio of their average velocities in miles per hour?

First, convert both velocities to miles per hour (mph). Then, use the definition of the average rate (total distance over total time):    The ratio is approximately 33:100.

Uniform motion meeting problems:

To set up equations for meeting problems we have to use two main ideas:

• When two moving objects meet they are in the same place at the same time.

Example 1:

Monsieur Seguin's little goat makes jumps of 4 ft each. The wolf that pursues her makes strides of 2 ft each and is 20 ft away. The wolf makes 3 strides for each of the goat's jumps. How many feet will they have to run until the wolf catches up with the goat?

(characters from Alphonse Daudet's story "The Brave Little Goat of M. Seguin")

The ratio of the velocities of the wolf and the goat is: Distance is additive, therefore we add the distance that separates the goat from the wolf to the distance that the goat covers until the meeting - this is equal to the distance covered by the wolf in the same time:     Example 2:

In his "Travels with Charley: In Search of America," John Steinbeck traveled the distance between Bangor and Deer Isle (52.5 miles) in 2 hours. He drove at 30 mph for half the time. What was his average speed for the remaining time?

Half the time is 1 hour. Therefore, Steinbeck drove 30 miles in the first hour.

Since distance is additive, the total distance must be the sum of the distances covered in the first and in the second hour, respectively.

Denote the distance, in miles, traveled during the second hour: Solve for x to find the distance covered in the second hour: Since this distance was covered in 1 hour, then his average velocity for this segment of the trip was 22.5 mph.

Example 3:

Nasreddin Hodja was riding back home on his donkey when he decided to rest a bit and freshen up about 0.5 miles away from the house. The donkey, however, much more eager to be home, started to run on its own towards the house at a speed of 5 mph. Nasreddin rushed after him immediately but, being tired, only averaged 3 mph. The donkey did not stop home but continued to run past the home in the same direction towards a good grazing place. How many minutes after starting to run was the home midway between Hodja and the donkey?

(characters from Central Asian folk tales)

Denote the unknown time by t.

In the time t Hodja ran a distance equal to: The donkey ran a distance equal to: The distance between Nasreddin and the house is: The distance between the house and the donkey is:  Since the house must be exactly between Hodja and the donkey: Solve for t to obtain the desired time in hours (since the speed is in mph and the distance is in miles):  Multiply by 60 to convert the time from hours to minutes: 