course Mth 271
005. Calculus
*********************************************
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'?
*********************************************
Your solution:
(5-17)/(3-7) = 3
(17-29)/(7-10) = 4
The slope between points 2 and 3 is steeper because the slope (rise over run) will go up 4 and over 1.
Confidence Assessment:
OK
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?
*********************************************
Your solution:
1. X = 2.1
1/(x – 2)
1/(2.1 – 2)
= 1/(.1)
= 10
X = 2.01
1/(x – 2)
1/(2.01 – 2)
= 1/(.01)
= 100
X = 2.001
1/(x – 2)
1/(2.001 – 2)
= 1/(.001)
= 1000
X = 2.0001
1/(x-2)
1/(2.001 – 2)
= 1/(.001)
= 10000
2. There is no limit to x, as x is approaching zero, so the value of x can exceed any number.
3. Yes, because the equation is a limit meaning the equation is equal to infinity and can exceed any number.
4. No, see above.
5. If the value 2 were put into the equation, we would be dividing by zero, which cannot happen.
Confidence Assessment:
Great
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
OK
Self-critique Rating:
To find the answers to this problem, you put the values supplied into the equation and evaluation the found values.
*********************************************
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'.
*********************************************
Your solution:
The coordinates of the first line, (3.5) and (7, 9), indicate that the line is 4 units wide and the coordinates of the second line, (10, 2) and (50, 4), indicating that this line is 50 units wide. For a square, the width of the square does not change, but the height does. To find the height, you need to average the coordinates of y. So average 5 and 9 to find 7 for the first line and for the second line, average 2 and 4, to get 3. To get the area we multiply the width by the altitude.
Line 1: 7 * 4 = 28
Line 2: 3 * 40 = 120, meaning that the trapezoid made by line two has a greater area.
Confidence Assessment:
OK
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
OK
Self-critique Rating:
The steps to answering this question are outlined in the above solution.
*********************************************
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.
*********************************************
Your solution:
The points for the line x = 2 and x = 5 are (2, 4) and (5, 25). The points for line two are (-1,1) and (7, 49). To find the slope, we use (y1 – y2)/(x1 – x2):
(4 – 25)/(2 – 5) = (21/3) = 7
(1-49)/(-1-7) = -48/-8 = 6
The line segment for x = 2 and x = 5 is steeper.
Confidence Assessment:
Great
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
OK
Self-critique Rating:
To find the answer to this problem, we have to use the information provided. I was given the equation for the line and the values for x. To find y, I inserted the values of x into the line. I then used the values of x and y to find the slope and compared the slopes of the two lines.
*********************************************
Question: `q005. Suppose that every week of the current millenium you go to the jewler 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 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.
*********************************************
Your solution:
1. The slope of the points depend on the values of y and t. If I begin with 5 grams and never increase my amount, I will have a straight line. If I begin with 5 grams in week one and add 5 grams per week, the line will rise at a constant rate, but if I begin with 5 grams in week one and only add one gram per week, the slope will be very slight, but if I begin with 5 grams and increase the value by 1.2 times each week, the line will rise faster and faster.
2. As stated above, the line will increase faster and faster.
3. The line will increase by smaller and smaller amounts each week.
Confidence Assessment:
Good
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
OK
Self-critique Rating:
This was a problem of logic. I just had to think through what the problem was asking.
*********************************************
Question: `q006. Suppose that every week you go to the jewler 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 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.
*********************************************
Your solution:
1. The slope of the line depends on the rate of increase.
2. In this scenario, the rate is increasing, therefore, the line will be rising.
3. In this scenario, the rate is decreasing, therefore, the line will be decreasing.
Confidence Assessment:
Good
This questions is different from the preceding problem. In the first problem, we were asked about the amount of gold, but in this question we are asked about the rate at which the gold increases.
``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?
*********************************************
Your solution:
t = 30
d = 100 – 2t + .01t^2
= 100 – 2(30) + .01(30)^2
= 100 – 60 + .01(900)
= 100 – 60 + 9
= 49 cm
t = 40
d = 100 – 2t + .01t^2
= 100 -2(40) + .01(40)^2
= 100 – 80 + .01(1600)
= 100 -80 + 16
= 36 cm
t = 60
d = 100 – 2t + .01t^2
= 100 – 2(60) + .01(60)^2
= 100 – 120 + .01(3600)
= 100 – 120 + 36
= 16 cm
The depth is changing more rapidly during the second time interval.
Confidence Assessment:
Great
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
OK
Self-critique Rating:
To find the solution to this problem, you enter the value of t into the given equation to find the value of d.
*********************************************
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?
*********************************************
Your solution:
r = 10 - .1t
t = 10
= 10 - .1t
= 10 - .1(10)
= 9 cm/s
t = 20
= 10 - .1t
= 10 - .1(20)
= 8 cm/s
Because the water level is decreasing at a rate between 9 cm/s and 8 cm/s, I would expect the water level to change between 80 and 90 centimeters during this time interval.
Confidence Assessment:
Great
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
OK
Self-critique Rating:
To find the solution to this problem, enter the given value into the equation.
&#This looks very good. Let me know if you have any questions. &#