Assignment 5 query

course Mth 271

005. `query 5

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Question: `qexplain why the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval represented

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Your solution:

The slope represents the average rate at which depth is changing (change in depth/change in time)

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Given Solution:

`a the specific idea is that ave rate of depth change is [change in depth / change in time] ; rise represents change in depth and run represents change in time so slope = rise/run represents ave rate of depth change. **

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Self-critique Rating:

I feel very confident about this problem.

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Question: `qexplain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval

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Your solution:

The area of a trapezoid is equal to the average altitude multiplied by the average width. The area gives us the change in position between two specific clock times.

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Given Solution:

`a The average altitude represents the avg. velocity. The area of a trapezoid involves the altitude, which represents the avg. velocity, and the width, which represents the change in clock time.

When you multiply ave altitude by width you are representing ave vel * change in clock time, which gives change in position.

This reasoning isn't confined to velocities. For any rate vs. clock time graph, average altitude represents approximate average rate, which multiplied by the change in time (not by the time itself) gives you the change in quantity **

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Self-critique Rating:

I feel fairly confident about this problem.

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Question: `qtext problem 0.5 #10 add x/(2-x) + 2/(x-2)

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Your solution:

x/(2-x) + 2/(x-2)

common denominator is (2-x)(x-2)

((2-x)(x-2))[(x/(2-x)) + (2/(x-2))]

(x-2)(x) + (2-x)(2)

x^2 – 2x + 4 – 2x

x^2 -4x + 4

(x – 2)^2

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Given Solution:

`a common denominator could be [ (2-x)(x-2) ]. In this case we have

x / (2-x) + 2 / (x-2)

= [ (x-2) / (x-2) ] * [ x / (2-x) ] + [ (2-x) / (2-x) ] * [ 2 / (x-2) ]

= x(x-2) / [ (2-x)(x-2) ] + 2 (2-x) / [ (2-x)(x-2) ]

= [x(x-2) + 2(2-x) ] / [ (2-x)(x-2) ]

= [ x^2 - 2x + 4 - 2x ] / [ (2-x)(x-2) ]

= (x^2-4x+4) / [ -x^2+4x-4 ]

= (x-2)^2 / [-(x-2)^2]

= -1.

NOTE however that there is a SIMPLER SOLUTION:

We can note that x-2 = -(2-x) so that the original problem is -x/(x-2) + 2 /(x-2) = (-x + 2) / (x-2) = -(x-2)/(x-2) = -1. **

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Self-critique Rating:

After looking over the given solution, it doesn’t make sense to me at all. You stated that the common denominator is (2 – x)(x-2) but then multiplied the two sides by (x-2)/(x-2) and (2-x)/(2-x) respectively. Why did you not multiply by the common denominator? I feel confident about my solution, but I’m completely confused about your solution.

If we had an equation, then we could multiply both sides by the common denominator. As long as the common denominator isn't zero the equality will be preserved.

However this isn't an equation.

For example if you had 1/6 + 4/15 the least common denominator would be 30. 1/6 = 5 * 1 / (5 * 6) = 5 / 30, and 4/15 = 2 * 4 / (2 * 15) = 8 / 30, and the sum is 5/30 + 8/30 = 13/30.

If you multiply 1/6 + 4/15 by 30, you get 5 + 8 = 13, which is not equal to 1/6 + 4/15.

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Question: `qtext problem 0.5 #50 cost = 6 x + 900,000 / x, write as single fraction and determine cost to store 240 units

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Your solution:

cost = 6x + 900000/x

multiply by x/x to obtain common denominator x

(6x + 900000/x) x/x

(6x^2 + 90000)/x

x = 240

cost = (6(240)^2 + 900000)/(240) = 5190

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Given Solution:

`a express with common denominator x:

[x / x] * 6x + 900,000 / x

= 6x^2 / x + 900,000 / x

= (6x^2 + 900,000) / x so

cost = (6x^2+900,000)/x

Evaluating at x = 240 we get cost = (6 * 240^2 + 900000) / 240 = 5190. **

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Self-critique Rating:

I feel confident about this problem.

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