course Phy 201
6/7 12I went ahead and finished the orientation for the Phy 231, I hope that is ok.
005. Calculus
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Question: `q001. The graph of a certain function is a smooth curve passing through the points (3, 5), (7, 17) and (10, 29).
Between which two points do you think the graph is steeper, on the average?
Why do we say 'on the average'?
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Your solution:
Slope 1: 17-5 / 7-3 = 3
Slope 2: 29-17 / 10-7 = 4
On the average, the graph is steeper between (7, 17) and (10, 29). You say on the average because the graph is a curve, which means the slope is not constant, but changing the whole time.
confidence rating #$&* 3
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Given Solution:
`aSlope = rise / run.
Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4.
The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3.
The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints.
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Question: 2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)?
1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)?
2. Will the value ever exceed a billion? Will it ever exceed one trillion billions?
3. Will it ever exceed the number of particles in the known universe?
4. Is there any number it will never exceed?
5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2?
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Your solution:
The values are 10, 100, 1000 and 10000.
As the values continue to get closer to 2, the numbers from the expression get larger and larger and approach infinity.
Yes, the value will exceed both a billion and 1 trillion billions and the number of particles in the universe because it is going to infinity. It will exceed all numbers.
When the graph approaches x=2 from the right side, it turns up sharply at x=2 because it is a vertical asymptote. When you approach it from the left, the graph turns down toward negative infinity.
confidence rating #$&* 3
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Given Solution:
`aFor x = 2.1, 2.01, 2.001, 2.0001 we see that x -2 = .1, .01, .001, .0001. Thus 1/(x -2) takes respective values 10, 100, 1000, 10,000.
It is important to note that x is changing by smaller and smaller increments as it approaches 2, while the value of the function is changing by greater and greater amounts.
As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001, etc.. Since we can put as many zeros as we want in .000...001 the reciprocal 100...000 can be as large as we desire. Given any number, we can exceed it.
Note that the function is simply not defined for x = 2. We cannot divide 1 by 0 (try counting to 1 by 0's..You never get anywhere. It can't be done. You can count to 1 by .1's--.1, .2, .3, ..., .9, 1. You get 10. You can do similar thing for .01, .001, etc., but you just can't do it for 0).
As x approaches 2 the graph approaches the vertical line x = 2; the graph itself is never vertical. That is, the graph will have a vertical asymptote at the line x = 2. As x approaches 2, therefore, 1 / (x-2) will exceed all bounds.
Note that if x approaches 2 through the values 1.9, 1.99, ..., the function gives us -10, -100, etc.. So we can see that on one side of x = 2 the graph will approach +infinity, on the other it will be negative and approach -infinity.
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Self-critique (if necessary): OK
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Self-critique rating #$&* OK
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Question: `q003. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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Your solution:
Using the coordinates given, you can determine the dimensions of the trapezoids and find the area.
Trapezoid A: Base 1 = 5; base 2 = 5; length = 5
Area = 1/2(5+9)*4 = 28 square units
Trapezoid B: Base 1 = 2; base 2 = 4; length = 40
Area = 1/2(2+4)*40 = 120 square units
The trapezoid with points (10,2) and (50,4) has the larger area.
confidence rating #$&* 3
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Given Solution:
`aYour sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area.
To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area.
This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.
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Self-critique (if necessary): OK
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Self-critique rating #$&* OK
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Question: `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which is steeper, the line segment connecting the x = 2 and x = 5 points on the graph of f(x), or the line segment connecting the x = -1 and x = 7 points on the same graph? Explain the basis of your reasoning.
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Your solution:
X y x y
2 4 -1 1
5 25 7 49
Slope 1: 25-4 / 5-2 = 21 / 3 = 7
Slope 2: 49-1 / 7- -1 = 48 / 8 = 6
The segment between x=2 and x=5 is steeper because it has a greater slope than the other segment.
confidence rating #$&* 3
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Given Solution:
`aThe line segment connecting x = 2 and the x = 5 points is steeper: Since f(x) = x^2, x = 2 gives y = 4 and x = 5 gives y = 25. The slope between the points is rise / run = (25 - 4) / (5 - 2) = 21 / 3 = 7.
The line segment connecting the x = -1 point (-1,1) and the x = 7 point (7,49) has a slope of (49 - 1) / (7 - -1) = 48 / 8 = 6.
The slope of the first segment is greater.
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Self-critique (if necessary): OK
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Self-critique rating #$&* OK
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Question: `q005. Suppose that every week of the current millennium you go to the jeweler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time (this is so), that the gold remains undisturbed (maybe, maybe not so), that no other source adds gold to your backyard (probably so), and that there was no gold in your yard before.
1. If you construct a graph of y = the number of grams of gold in your backyard vs. t = the number of weeks since Jan. 1, 2000, with the y axis pointing up and the t axis pointing to the right, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly?
2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week.
{}3. Answer the same question assuming that every week you bury half the amount you did the previous week.
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Your solution:
For the first situation, the line would be a rising straight line because the rate at which the gold is added is constant: one gram per week.
If you bury one more gram than you did the week before, then the line will rise faster and faster.
If you bury half the amount than you did the previous week then the line will increase, but the rate will be increasing slower and slower.
confidence rating #$&* 3
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Given Solution:
`a1. If it's the same amount each week it would be a straight line.
2. Buying gold every week, the amount of gold will always increase. Since you buy more each week the rate of increase will keep increasing. So the graph will increase, and at an increasing rate.
3. Buying gold every week, the amount of gold won't ever decrease. Since you buy less each week the rate of increase will just keep falling. So the graph will increase, but at a decreasing rate. This graph will in fact approach a horizontal asymptote, since we have a geometric progression which implies an exponential function.
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Self-critique (if necessary): OK
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Self-critique rating #$&* OK
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Question: `q006. Suppose that every week you go to the jeweler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time, that the gold remains undisturbed, and that no other source adds gold to your backyard.
1. If you graph the rate at which gold is accumulating from week to week vs. the number of weeks since Jan 1, 2000, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly?
2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week.
3. Answer the same question assuming that every week you bury half the amount you did the previous week.
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Your solution:
If the rate is constant, then the line is going to be straight and level: 1 gram per week.
If you bury 1 gram more, then the graph will be a rising straight line; the rate of increase is the same.
If you bury half the amount, then the graph will be decreasing but slower and slower; also it’s not going to reach zero because you always add some gold..
confidence rating #$&*2
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Given Solution:
`aThis set of questions is different from the preceding set. This question now asks about a graph of rate vs. time, whereas the last was about the graph of quantity vs. time.
Question 1: This question concerns the graph of the rate at which gold accumulates, which in this case, since you buy the same amount each week, is constant. The graph would be a horizontal straight line.
Question 2: Each week you buy one more gram than the week before, so the rate goes up each week by 1 gram per week. You thus get a rising straight line because the increase in the rate is the same from one week to the next.
Question 3. Since half the previous amount will be half of a declining amount, the rate will decrease while remaining positive, so the graph remains positive as it decreases more and more slowly. The rate approaches but never reaches zero.
These two problems go to the heart of the Fundamental Theorem of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long run. The same is true of the last question in this document.
Self-critique (if necessary): Read the problem very carefully!!! I almost sais they were the same…
Lots of students do.
These two questions illustrate the difference between a quantity and its rate of change; i.e., the difference between a function and its derivative. Very important for applications in calculus-based physics.
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Self-critique rating #$&* 3
``q007. If the depth of water in a container is given, in centimeters, by 100 - 2 t + .01 t^2, where t is clock time in seconds, then what are the depths at clock times t = 30, t = 40 and t = 60? On the average is depth changing more rapidly during the first time interval or the second?
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Your solution:
Time Depth
30 49
40 36
60 16
Slope of interval 1: 36-49 / 40-30 = -13 / 10 = -1.3 units/second
Slope of interval 2: 16-36 / 60-40 = -20 / 20 = -1 unit/second
On the average, the water is decreasing faster during the first interval.
confidence rating #$&* 3
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Given Solution:
`aAt t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 = 49.
At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36.
At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16.
49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average.
36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average.
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Self-critique (if necessary): OK
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Self-critique rating #$&* OK
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Question: `q008. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate is it descending when t = 20? How much would you therefore expect the water level to change during this 10-second interval?
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Your solution:
T=10: 10-.1(10) = 10-1 = 9 cm/s
T=20: 10-.1(20) = 10-2 = 8 cm/s
I would expect the water level to have an average rate of change of 8.5 cm/s in the 10 second interval: take the average of the rates. So, the water level would change by 8.5 cm/s *10 s = 85 seconds.
confidence rating #$&* 3
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Given Solution:
`aAt t = 10 sec the rate function gives us 10 - .1 * 10 = 10 - 1 = 9, meaning a rate of 9 cm / sec.
At t = 20 sec the rate function gives us 10 - .1 * 20 = 10 - 2 = 8, meaning a rate of 8 cm / sec.
The rate never goes below 8 cm/s, so in 10 sec the change wouldn't be less than 80 cm.
The rate never goes above 9 cm/s, so in 10 sec the change wouldn't be greater than 90 cm.
Any answer that isn't between 80 cm and 90 cm doesn't fit the given conditions..
The rate change is a linear function of t. Therefore the average rate is the average of the two rates, or 9.5 cm/s.
The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm.
Self-critique (if necessary): OK
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Self-critique rating #$&* OK
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&#Very good work. Let me know if you have questions. &#
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