Assignment 4a

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course Mth 173

9/20/2010 @ 6:17pm

Set #34

Solve the following by describing a picture, and by constructing a graph:

1. If your total earnings up to week 8 are 104.6265 dollars, and your total earnings up to week 15 are 118.6265 dollars, then what are your average earnings per week? Why do we say average earnings per week rather than just earnings per week?

The average earnings per week would be 118.6265 divided by the total number of weeks, 15. 118.6265/15 = 7.90843 dollars per week average. The reason for the average per week is because the amount of income received on a weekly basis changes. In this case on a graph the amount of income from week 0 to week 8 would have a steeper slope than from week 8 to 15.

2. If an automobile is at milepost 510.4263 after having traveled for 10 hours and at milepost 562.4263 after having traveled for 15 hours, then what is its average speed? Why do we say average speed rather than just speed?

562.4263/15 = 37.4951 average miles per hour over the ten hour period. This is the average speed because it does not take into account the greater distance per hour covered in the first 10 hours of the trip. On a graph the slope of the first ten hours would be steeper than the final 5 hours of travel.

You have no information about where the automobile started. Your calculation is valid if it started at milepost 0. However the only information you have is where it was at the beginning and the end of the interval from the 10-hour time to the 15-hour time.

3. If the water in a uniform cylinder has depth 176.3919 cm at clock time 10 seconds and depth 195.3919 cm at clock time 15 seconds, then at what average rate is the depth changing? What we say average rate rather than just rate?

195.3919/15 = 13.0261 cm/s average depth increase. The amount is changing faster during the first ten seconds than the final 5 seconds, therefore the average is the change in the overall time.

You don't know where the water was at clock time 0. There is no reason to expect that it was at depth 0.

4. If you are earning money at an average rate of 7 dollars per week, and if at the end of week 9 your total earnings are 57.46883 dollars, then what will be your total earnings that the end of week 19?

Take the total amount of earnings at week 9 and add to that the product of 7 dollars per week * 10weeks. = 57.46883 + 7 * 10 = 57.46883 + 70 = $127.46883

Good.

Note, relevant to some of the preceding questions, that your total earnings at clock time 0 would not be 0. You are not given the time from which total earnings were calculated.

5. If an automobile is traveling at the rate of 70 mph, and is at milepost 684.2719 after having traveled for 10 hours, then what milepost would be after having traveled for 15 hours?

Take the mileage traveled up to milepost 684.2719 and add that to the product of 70mph * 5 hours. 684.2719 + 70 * 5 = 1034.27 miles.

Good.

6. If the depth of water in a uniform cylinder is changing at an average rate of 15 cm/second between clock times 8 s and 14 s, and if its depth at clock time 8 seconds is 104.3864 cm, then what is its depth at clock time 14 seconds?

104.3864 cm + 15cm/sec * 6seconds = 194.3864 cm

7. If your total earnings up to week 4 are 26.65315 dollars, and if you earn money at a constant rate of 7 dollars per week, then at what week will your total earnings be 33.65315 dollars?

33.65315 – 26.65315 = $7dollars

$7 / 7 dollars per week = 1 week

Week 4 + 1 week = week 5

8. If an automobile is traveling at the rate of 63 mph, and is at milepost 625.1494 after having traveled for 10 hours, then after how many hours of travel will it have reached milepost 16?

625.1494 – 16 = 609.149 miles to travel

609.149 / 63mph = 9.669 hours to travel distance

9.669 hours to go + 10 hours already traveled = 19.669 hours total

Good.

Observe once again that mile clearly 0 does not occur at clock time 0.

9. If the depth of water in a uniform cylinder is changing at an average rate of 26 cm/second as it depth changes from 100.4351 cm at clock time 4 seconds to 126.4351 cm at an unknown clock time, then what is the unknown clock time?

126.4351 – 100.4351 = 26cm change in water depth

26cm / 26cm.sec = 1 sec

4sec + 1 sec = 5 seconds

"

Good, but you made extraneous assumptions on some of these questions and obtained some results that were not justfied by the given information. Be sure to see my notes.