query9

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course Mth 173

2/17 1:06 AM

009. Finding the average value of the rate using a predicted point

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Question: `qNote that there are 9 questions in this assignment.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q001. The process we used in the preceding qa to approximate the graph of y corresponding to the graph of y ' or the y ' function (the function is y ' = .1 t - 6 for t = 0 to t = 100) can be significantly improved. Recall that we used the initial slope to calculate the change in the y graph over each interval.

For example on the first interval we used the initial slope -6 for the entire interval, even though we already knew that the slope would be -5 by the time we reached the t = 10 point at the end of that interval. We would have been more accurate in our approximation if we had used the average -5.5 of these two slopes.

Using the average of the two slopes, what point would we end up at when t = 10?

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Your solution:

what is the orginal equation?

y=.05t^2-6t+c

what is c?

i will give it a value of 0

y=c at t=0

(0,0)

and at t=10

.05 * 100 - 60

is -55

(10,-55)

y'=.1t-6 so y'= -6

y'=.1(10-6)

1-6 is 5.

between that is -5.5

confidence rating #$&*: 2

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Given Solution:

`aIf the average value slope = -5.5 is used between t = 0 and t = 10, we find that the rise is -55 and we go from (0,100) to (10,45).

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Self-critique (if necessary):

was I suppose to know c=100?

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Self-critique Rating:

@&

This question is a continuation of the last question on the preceding document. I split the document, and failed to make that clear.

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Question: `q002. We find that using the average of the two slopes we reach the point (10, 45). At this point we have y ' = -5.

By the time we reach the t = 20 point, what will be the slope? What average slope should we therefore use on the interval from t = 10 to t = 20? Using this average slope what point will we end up with when t = 20?

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Your solution:

well it seems y=.05t^2-6t+100

and y'=.1t-6

at t=20

y'=2-6

y'=-4 which the average would be -4.5 and that is ten seconds lapsed, so -45. and from (10,45) that would be 0, so (20,0)

confidence rating #$&*: 3

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Given Solution:

`aThe slope at t = 20 is y ' = -4. Averaging this with the slope y ' = -5 at t = 10 we estimate an average slope of (-5 + -4) / 2 = -4.5 between t = 10 and t = 20. This would imply a rise of -45, taking the graph from (10,45) to (20, 0).

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q003. We found that we reach the point (20, 0), where the slope is -4. Following the process of the preceding two steps, what point will we reach at t = 30?

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Your solution:

there is a pattern, each ten is a rate of 1 more in the direction we are heading, it was -6, -5, -4, and now will be -3. and the average will be -3.5, and again on the ten second interval -35, so (30,-35)

confidence rating #$&*:3

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Given Solution:

`aThe slope at t = 20 is y ' = -3. Averaging this with the slope y ' = -4 at t = 20 we estimate an average slope of (-4 + -3) / 2 = -3.5 between t =20 and t = 30. This would imply a rise of -35, taking the graph from (20,0) to (30, -35).

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `q004. Continue this process up through t = 70. How does this graph differ from the preceding graph, which was made without averaging slopes?

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Your solution:

I don't know if I did this without averaging slopes.

if original point (0,100)

and we kept using the intial rate and we start at -6

(0,100) with rate of -6

(10,40) with rate of -5

(20,-10) with rate of -4

(30,-50) with rate of -3

(40,-80) with rate of -2

(50,-100) with rate of -1

(60,-110) which would give us a rate of 0? this would seem to give an odd result if you dont find the average

(70,-110)

(0,100) (10,40) (20,-10) (30,-50) (40,-80) (50,-100) (60, -110) (70,-110)

(0,100) (10,45) (20,0) (30, -35) (40,-60) (50, -75) (60,-80) (70,-75)

if i did that correctly in my head quickly. at any rate the first causes a horizontal line at the end point the averaged graph is a lot lower, and you can tell the values start to switch around.

confidence rating #$&*:

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Given Solution:

`aThe average slopes over the next four intervals would be -2.5, -1.5, -.5 and +.5. These slopes would imply rises of -25, -15, -5 and +5, taking the graph from (30,-35) to (40, -60) then to (50, -75) then to (60, -80) and finally to (70, -75). Note that the graph decreases at a decreasing rate up to the point (60, -80) before turning around and increasing to the ponit (70, -75).

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Self-critique (if necessary): OK

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Self-critique Rating:OK

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Question: `q005. Now suppose that y = -.2 t^2 + 5 t + 100 represents depth vs. clock time. What is the corresponding rate function y ' ?

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Your solution:

y'=2at+b

y'=-.4t+5

confidence rating #$&*: 3

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Given Solution:

`aThe rate function corresponding to y = a t^2 + b t + c is y ' = 2 a t + b. In this example a = -.2, b = 5 and c = 100 so y ' = 2(-.2) t + 5, or y ' = -.4 t + 5.

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Self-critique (if necessary):OK

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Self-critique Rating: OK

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Question: `q006. From the depth function and the rate function we can find the coordinates of the t = 30 point of the graph, as well as the rate of change of the function at that point. The rate of change of the function is identical to the slope of the graph at that point, so we can find the slope of the graph.

What are the coordinates of the t = 30 point and what is the corresponding slope of the graph? Sketch a graph of a short line segment through that point with that slope.

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Your solution:

guessing it is for y = -.2 t^2 + 5 t + 100

y=-.2 * 900 + 150 + 100

y=-180 + 150 = 100

y=70

the point would be (30,70)

????I seem to get slopes, instantaneous rate of change, average rate of change, and when to use which mixed up, so by asking slope here, is it asking for the instantaneous rate of change, which the derivative gives?

at the point (30,70) the rate of change is -7

trying to remember how you said to find an average rate of change, which i believe is what the slope is, if i recall dy/dt? and that the rate of change right?

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Right

The slope at a point is identified with the instantaneous rate of change.

The average slope between two points is identified with an average rate of change over the corresponding interval.

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if i recall it is before we take the x to zero for the derivative so 2at+b+a'dt

-.4 * 0 + 5 +- .4 * 30

0 + 5 + -12

-7

that is the same as what I got for the derivative...

???I recall that confusing me and why I had a hard time determining what I was suppose to do, but this only happens if we have a linear derivative, and if it wasnt, then my answer would have been different, yes?

@&

You can calculate the value of the derivative at a point using a limiting process. If done correctly, this will give you the correct result, whether the function is linear or not.

A linear derivative implies a quadratic function, and also implies that the average rate of change on an interval is equal to the instantaneous rate of change at the midpoint of the interval.

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confidence rating #$&*:2

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Given Solution:

`aAt t = 30 we have y = -.2 * 30^2 + 5 * 30 + 100 = 70, and y ' = -.4 * 30 + 5 = -7. The graph will therefore consist of the point (30, 70) and a short line segment with slope -7.

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Self-critique (if necessary):

should we have used the derivative for a slope?

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That would be the easiest way to get the result.

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Self-critique Rating: OK

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Question: `q007. What is the equation of the straight line through the t = 30 point of the depth function of the preceding example, having a slope matching the t = 30 rate of change of that function?

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Your solution:

(30,70) is my point, and -7 was my slope. knowing that I go back to my algebra

y-70= -7(x-30)

y-70=-7x+210

y=-7x+280

????I know there are quite a number of linear equations that could go through that point, how would we find a good range of them?

@&

There is only one line through the given point having the required slope.

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confidence rating #$&*:3

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Given Solution:

`aA straight line through (30, 70) with slope -7 has equation

y - 70 = -7 ( x - 30),

found by the point-slope form of a straight line.

This equation is easily rearranged to the form

y = -7 x + 280.

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Self-critique (if necessary):OK

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Self-critique Rating: OK

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Question: `q008. Evaluate the depth function y at t = 30, 31, 32. Evaluate the y coordinate of the straight line of the preceding question at t = 30, 31, 32. What are your results?

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Your solution:

???just thinking, couldnt we find parallel lines for all of these points using a linear equation and just shifting it

y = -.2 t^2 + 5 t + 100

at (30,70) it is y = -7 x + 280. as found previously

-192+ 155 + 100

(31,63)

y'=-.4 * 31 + 5

y'=-7.4

y-63=-7.4(x-31) y-63=-7.4x+229.4 y=-7.4x + 292.4

y(32)=- 204.8 + 160 + 100

(32,55.2)

y'=-12.8 + 5

y'=-7.8

y-55.2=-7.8(x-32) y-55.2= -7.8x + 249.6 y=-7.8x+304.8

so

(30,70) is y=-7x=280

(31,63) is y=-7.4x+292.4

(32,55.2) is y=-7.8x+304.8

???noticing each slope is changing by .2 and c by 12.4, is this something to pay attention to?

confidence rating #$&*: 2

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Given Solution:

`aPlugging t = 30, 31, 32 into the original function y = -.2 t^2 + 5 t + 100 we get y = 70, 62.8, 55.2 respectively.

Plugging t = 30, 31, 32 into the straight-line equation y = -7 t + 280 we get y = 70, 63, 56 respectively.

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Self-critique (if necessary): OK

I read the question wrong, I thought it wanted me to find linear equations for each ordered pair of the depth function.

what is the purpose of seeing the values of y on the linear equation and comparing them to the depth function?

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You see how the linear function, which constitutes the tangent line at the t = 30 point, approximates the quadratic function better for points close to the t = 30 point. By comparing the deviations at the t = 31 and t = 32 points, you begin to get a feel for how the approximation gradually diverges.

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Self-critique Rating:OK

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Question: `q009. By how much does the straight-line function differ from the actual depth function at each of the three points? How would you describe the pattern of these differences?

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Your solution:

depth y = 70, 62.8, 55.2

linear y= 70, 63, 56

for a pattern, the linear decreases by the same amount. the depth I see no apparent pattern with the three points. and of course we found the linear equation when their y value was the same, so the points will be the same here. the difference has increased between them the farther off you get, which i would expect.

confidence rating #$&*:2

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Given Solution:

`aAt t = 30 the two functions are identical.

At t = 31 the values are 62.8 and 63; the function is .2 units below the straight line.

At t = 32 the values are 55.2 and 56; the function is now .8 units below the straight line.

The pattern of differences is therefore 0, -.2, -.8. The function falls further and further below the straight line as we move away from the point (30, 70).

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q009. By how much does the straight-line function differ from the actual depth function at each of the three points? How would you describe the pattern of these differences?

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Your solution:

depth y = 70, 62.8, 55.2

linear y= 70, 63, 56

for a pattern, the linear decreases by the same amount. the depth I see no apparent pattern with the three points. and of course we found the linear equation when their y value was the same, so the points will be the same here. the difference has increased between them the farther off you get, which i would expect.

confidence rating #$&*:2

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Given Solution:

`aAt t = 30 the two functions are identical.

At t = 31 the values are 62.8 and 63; the function is .2 units below the straight line.

At t = 32 the values are 55.2 and 56; the function is now .8 units below the straight line.

The pattern of differences is therefore 0, -.2, -.8. The function falls further and further below the straight line as we move away from the point (30, 70).

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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