assignment 9 openquery

course Mth 163

11/6 5

009. `query 9

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Self-critique (if necessary):

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Question: `qSymbolic calculation of slope, preliminary exercise

What was the function, between which two points were you to calculate the average slope and how did you get this slope?

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Your solution:

The function y = .1x2 -3 was used with the given values of x to be x = -2 and 7.

Each value was inserted into the function for x and the values of y were determined to be

y = -2.6 and 1.9. The slope was then found by using the formula y2-y1 / x2-x1 for the points (-2, -2.6) and (7, 1.9)

1.9 – ( -2.6) = 4.5

7 – 2 = 5

4.5 / 5 = .9

Therefore the slope is .9

confidence rating: 3

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Given Solution:

** For the function y = .1 x^2 - 3 between x = -2 and x = 7 we get:

slope = (y2 - y1) / (x2 - x1).

For x1 = 2 and x2 = 7 we have y2 = .1 * 7^2 - 3 = 1.9 and y1 = .1 * 2^2 - 3 = -2.6, so

slope = (1.9 - (-2.6) ) / ( 7 - 2) = 4.5 / 5 = .9. **

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Question: `qproblem 2 symbolic expression for slope, fn depth(t).

What is the expression for the slope between the two specified t values?

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Your solution:

The t values for this problem were 10, 30

The first coordinate is (10, depth(10)

The second coordinate is (30, depth (30)

Therefore the slope is = depth (30) – depth (10) / 30-10

Or depth (30)-depth(10) / 20

confidence rating: 3

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Given Solution:

** The function is given a name: depth(t).

t values are 10 and 30.

So rise = depth(30) - depth(10) and run = 30 - 10 = 20.

Thus slope = [ depth(30) - depth(10) ] / 20 . **

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Question: `qWhat is the rise between the two specified t values?

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Your solution:

Rise is depth (30) – depth (10)

confidence rating: 3

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Given Solution:

** The rise is the change in depth. The two depths are depth(10) and depth(30).

The change in depth is final depth - initial depth, which gives us the expression

depth(30)-depth(10) **

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Question: `qWhat is the run between the two specified t values?

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Your solution:

Run = 30 -10 or 20

confidence rating: 3

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Given Solution:

** run = 30 - 10 = 20 **

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Question: `qWhat therefore is the slope and what does it mean?

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Your solution:

Depth (30) – depth (10) / 20

Represents the average change in depth compare to time

confidence rating: 3

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Given Solution:

** rise = depth(30)-(depth(10) indicates change in depth.

run = 30 - 10 = 20 = change in clock time.

Slope = rise / run = (depth(30) - depth(10) ) / 20, which is the average rate at which depth changes with respect to clock time between t = 10 and t = 30. **

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Question: `q problem 5 graph points corresponding to load1 and load2

What are the coordinates of the requested graph points?

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Your solution:

(load 1, spring length(load1)

(load 2, spring length (load2)

confidence rating: 3

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Given Solution:

** The horizontal axis is the 'load' asix, the vertical axis is the springLength axis.

The load axis coordinates are load1 and load2.

The corresponding spring lengths are springLength(load1) and springLength(load2).

The springLength axis coordinates are springLength(load1) and springLength(load2).

The graph points are thereofore (load1, springLength(load1) ) and (load2, springLength(load2) ). **

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Question: `qWhat is your expression for the average slope of the graph between load1 and load2?

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Your solution:

Spring length(load2) – spring length (load1) is the rise

Load 2 – load 1 is the run

Slope = spring length(load2) – spring length (load1) / load2 – load1

confidence rating: 3

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Given Solution:

** rise = springLength(load2) - springLength(load1)

run = load2 - load1 so

slope = [ springLength(load2) - springLength(load1)] / (load2 - load1). **

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Question: `q problem 6 symbolic expression for slope of depth function

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Your solution:

Run = t2 – t1

Rise = depth t2-depth t1

Slope = depth t2 – depth t1 / t2 – t1

confidence rating:3

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Given Solution:

** the name of the function is depth(t).

We need the slope between t = t1 and t = t2.

The depths are depth(t1) and depth(t2).

Thus rise is depth(t2) - depth(t1) and run is t2 - t1.

Slope is [ depth(t2) - depth(t1) ] / (t2 - t1). **

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Question: `q problem 8 average rate from formula f(t) = 40 (2^(-.3 t) ) + 25 intervals of partition (10,20,30,40)

What average rate do you get from the formula? Show your steps.

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Your solution:

Using the formula stated above and substituting the intervals of (10,20,30,40) the average rate was calculated for the following:

F(20) – f(10) = -4.375 / the run of 10, slope = -.4375

F(30) – f(20) = -.0546875 / the run of 10, slope = -.05469

F(40) – f(30) = -.06835937 / the run of 10, slope = -.00684

confidence rating: 3

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Given Solution:

** ave rate = change in depth / change in t. For the three intervals we get

(f(20)-f(10))/(20-10) = (25.625 - 30 )/(20 - 10) = -4.375 / 10 = -.4375

(f(30)-f(20))/(30-20) = (25.07813 - 25.625)/(30 - 20) = -.5469 / 10 = -.05469.

(f(40)-f(30))/(40-30) = (25.00977 - 25.07813)/(40 - 30) = -.0684 / 10 = -.00684. **

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