Query 9

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course Mth 163

2/20

009. `query 9

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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 was y=.1x^2-3 between x=-2 and x= 7.

The slope would be (y2-y1/x2-x1).

For x1, we have y1= .1(2^2)-3, which would make y1 equal to -2.6.

For x2, we have y2= .1(7^2)-3, which would make y2 equal to 1.9.

The coordinates would be (-2, -2.6) and (7, 1.9).

1.9- -2.6/7--2= 4.5/9= .5

The slope equals .5.

confidence rating #$&*: 3

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Given Solution: OK

** 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 / 9 = 1/2. **

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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 function is depth(t).

The rise would be depth(30) - depth(10).

The run would be 30-10= 20.

The slope would be depth(30) - depth (10)/ 20.

confidence rating #$&*: 3

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Given Solution: OK

** 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:

The rise will be depth(30) - depth (10).

confidence rating #$&*: 3

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Given Solution: OK

** 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:

The run between the two points will be 30-10= 20.

confidence rating #$&*: 3

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Given Solution: OK

** 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:

The slope will be depth(30) - depth (10)/20.

The slope will describe the average rate the depth will change given clock time between 10 and 30.

confidence rating #$&*: 3

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Given Solution: OK

** 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:

The horizontal axis is the load axis. The vertical axis is the springLength axis.

The coordinates for the load axis are load1 and load2.

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

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

The graph points are (load1, springLength(load1) ) and (load2, springLength(load2)

confidence rating #$&*: 3

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Given Solution: OK

** 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:

The slope = springLength(load2) - springLength(load1)/load2 - load1

The rise will be springLength(load2) - springLength(load1).

The run will be load2 - load1.

confidence rating #$&*: 3

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Given Solution: OK

** 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:

The slope will be between t1 and t2.

The depth is depth(t1) and depth(t2).

Therefore, the rise will be depth(t2) - depth(t1).

The run will be t2-t1.

The slope equals depth(t2) - depth(t1)/t2-t1.

confidence rating #$&*: 3

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Given Solution: OK

** 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:

For y1= 10 and y2=20 you get,

40(2^(-.3(10)) +25= 30

40(2^-.3(20)) + 25= 25.625

Rate equals 25.625-30/20-10= -.4375.

For y1=20 and y2= 30

40(2^(-.3(20))+25= 25.625

40(2^(-.3(30))+25= 20.07813

Rate equals 20.07813-25.625/30-20= -.05469.

For y1=30 and y2=40

40(2^(-.3(30))+25= 20.07813

40(2^(-.3(40))+25= 25.00977

The rate will be 25.00977-20.07813/40-30= -.00684.

confidence rating #$&*: 3

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Given Solution: OK

** 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. **

Add comments on any surprises or insights you experienced as a result of this assignment.

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Your solution:

We are still using depth, slope, and rate in equations, so I feel like right now I am pretty comfortable with this information.

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Given Solution:

STUDENT RESPONSE:

Ummm I know the slope formula is (y2-y1)/(x2-x1), but I always just put the number into the expression in the order I see them, but that is ok because I keep the order and get the correct answere because the y2,y1,x2,x1 or all relative. I am correct in doing this?

INSTRUCTOR COMMENT:

In other words you use (y1 - y2) / (x1 - x2) instead of (y2 - y1) / (x2 - x1).

It's more conventional to regard, say, 10 as x1 and 20 as x2, so f(20) is y2 and f(10) is y1. If you start from the lower x number and change to the higher the difference is higher - lower, and this is the way we usually think about changes. According to this convention we calculate change in y as y2 - y1 and change in x as x2 - x1.

You are doing (y1 - y2) / (x1 - x2) and you get a negative change in x, a negative denominator, and if you are thinking about change from the first quantity to the second this is backwards.

However as you say both numerator and denominator follow the same order so you still get the right answer, since (y1-y2)/(x1-x2)= (y2-y1) / (x2-x1). **

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&#Very good responses. Let me know if you have questions. &#