Assignment 3

course Mth 158

ƾ򭔍olڶZassignment #003

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

003. Misc: Surface Area, Pythagorean Theorem, Density

qa areas volumes misc

06-06-2008

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assignment #002

002. Describing Graphs

qa initial problems

06-06-2008

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assignment #002

002. Volumes

qa areas volumes misc

06-06-2008

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assignment #003

003. PC1 questions

qa initial problems

06-06-2008

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10:14:42

`q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?

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RESPONSE -->

The line with the points [3,5] and [7,17] is steeper. It has a slope of 1/3 as cpmpared to 1/4 for the second line. It has the same verticle rise over a shorter distance.

confidence assessment: 3

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10:15:03

The point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction.

Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3.

Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.

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RESPONSE -->

ok

self critique assessment: 3

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Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions (to which I will respond).

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That said, the only thing you did wrong was reverse the rise and the run, and I expect you do understand that.

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10:19:54

`q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.

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RESPONSE -->

When x is 2, 2-2=0 and 0 *(2x+5) = 0.

when x = -2.5, 2x + 5=0 and (x-2) * 0=0

confidence assessment: 3

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10:20:13

If x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero.

If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero.

The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero.

Note that (x-2)(2x+5) can be expanded using the Distributive Law to get

x(2x+5) - 2(2x+5). Then again using the distributive law we get

2x^2 + 5x - 4x - 10 which simplifies to

2x^2 + x - 10.

However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form.

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RESPONSE -->

ok

self critique assessment: 2

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10:27:17

`q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?

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RESPONSE -->

x values will be [0,2,-4]

1. 3*2-6=0 and 0 times any number is 0.

2.-4+4=o and any number times 0 is 0.

3.-2^2=4 and 4-4=0; any number times 0 = 0.

Since all the quantities are multiplied, any quantity with a valule of zero will make the valule of the entire expression 0.

confidence assessment: 3

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10:27:43

In order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0.

3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.**

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RESPONSE -->

ok

self critique assessment: 3

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10:37:28

`q004. 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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RESPONSE -->

The trapezoid formed from the points [10,2] and [50,4] should have the greater area because its base is 40 units long as compared to the first trapezoid, which has a base of 4 units of length.

This when added to base 2 and multiplied by the height will gibe the second trapezoid a greater area.

confidence assessment: 2

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10:38:33

Your 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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RESPONSE -->

self critique assessment: ok

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11:01:46

* `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph:

As we move from left to right the graph increases as its slope increases.

As we move from left to right the graph decreases as its slope increases.

As we move from left to right the graph increases as its slope decreases.

As we move from left to right the graph decreases as its slope decreases.

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RESPONSE -->

As we move from left to right the graph increases as its slope increases applies to y=1/x

As we move from left to right the graph decreases as its slope increases applies to y= 1/x.

As we move from left to right the graph increases as its slope decreases applies to y=x^2

As we move from left to right the graph increases as its slope decreases applies to y=sqrt[x]

self critique assessment: 1

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11:51:28

* `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?

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RESPONSE -->

v=3[5]+9

v=24

E=800(24^2)

E=460,800

self critique assessment: 2

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11:52:49

For t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.

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RESPONSE -->

ok

self critique assessment: 3

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12:02:20

* `q009. Continuing the preceding problem, can you give an expression for E in terms of t?

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RESPONSE -->

E= 800(3t+9)^2

self critique assessment: 2

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12:03:21

end program

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RESPONSE -->

ok

self critique assessment: 3

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Your work looks good. See my notes. Let me know if you have any questions. &#