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assignment #009

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19:43:40

`q001. Note that this assignment has 2 questions

For the function y = 1.1 x + .8, what are the coordinates of the x = x1 point, in terms of the symbol x1? What are the coordinates of the x = x2 point, in terms of the symbol x2?

What therefore is the rise between these two points, and what is the run?

What is the average slope of the graph between these two points? Be sure to simplify your result.

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RESPONSE -->

x=x1; y=1.1x1+.8; (x1,(1.1x1+.8))

x=x2 ; y=1.1x2+.8; (x2,(1.1x2+.8))

rise; (1.1x2+.8)-(1.1x1+.8)

run; x2-x1

slope; [(1.1x2+.8)-(1.1x1+.8)]/(x2-x1)

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20:10:06

In terms of the symbol x1 the coordinates of the x = x1 point are ( x1, 1.1 x1 + .8) and the coordinates of the x = x2 point are ( x2, 1.1 x2 + .8).

The rise between the two points is therefore

rise = (1.1 x2 + .8) - (1.1 x1 + .8) = 1.1 x2 + .8 - 1.1 x1 - .8 = 1.1 x2 - 1.1 x1.

The run is

run = x2 - x1.

The slope is therefore (1.1 x2 - 1.1 x1) / (x2 - x1) = 1.1 (x2 - x1) / (x2 - x1) = 1.1.

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RESPONSE -->

I forgot to simplify by doing the reverse distributive move of taking 1.1 out, but the process makes perfect sense.

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20:16:13

`q002. For the function y = 3.4 x + 7, what are the coordinates of the x = x1 point, in terms of the symbol x1? What are the coordinates of the x = x2 point, in terms of the symbol x2?

What therefore is the rise between these two points, and what is the run?

What is the average slope of the graph between these two points? Be sure to simplify your result.

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RESPONSE -->

x=x1: y=3.4x1+7; (x1, 3.4x1+7)

x=x2; y=3.4x3+7; (x2, 3.4x2+7)

Rise; (3.4x2+7)-(3.4x1+7)=3.4x2+7-3.4x1-7=3.4x2-3.4x1

Run; x2-x1

Slope; (3.4x2-3.4x1)/(x2-x1) = [3.4(x2-x1)]/(x2-x1) = 3.4

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20:16:20

In terms of the symbol x1 the coordinates of the x = x1 point are ( x1, 3.4 x1 + 7) and the coordinates of the x = x2 point are ( x2, 3.4 x2 + 7).

The rise between the two points is therefore

rise = (3.4 x2 + 7) - (3.4 x1 + 7) = 3.4 x2 + 7 - 3.4 x1 - 7 = 3.4 x2 - 3.4 x1.

The run is

run = x2 - x1.

The slope is therefore (3.4 x2 - 3.4 x1) / (x2 - x1) = 3.4 (x2 - x1) / (x2 - x1) = 3.4.

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ġxm̃Ȁߏϝ~f

assignment #009

ҶQ̫

Precalculus I

10-04-2006

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20:55:54

Symbolic 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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RESPONSE -->

function was y=.1x^2-3

between points x= -2 and x=7

To find the slope, you have to find the y coordinates that pair with these x coordinates.

when x= -2; .1(-2)^2-3 = .1(4)-3= .4-3 = -2.6; so the point is (-2,-2.6)

When x=7; .1(7)^2-3 = .1(49)-3= 4.9 - 3 = 1.9; so the point is (7, 1.9)

Run = 7- - 2 = 9

Rise = 1.9- - 2.6 = 4.5

slope 4.5/9 = .5

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20:57:00

** 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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RESPONSE -->

The Run would be 7- (-2) = 9 wouldn't it? Since x= -2 instead of 2.

That's correct.

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20:59:57

problem 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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RESPONSE -->

y=depth(t)

slope between t=10 and t=30

t1=10; y=depth(10); (10,depth(10))

t2=30; y=depth(30); (30,depth(30))

run 30-10=20

rise depth(30)-depth(10)

slope [depth(30)-depth(10)]/20

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21:00:05

** 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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21:00:33

What is the rise between the two specified t values?

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rise = depth(30)-depth(10)

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21:00:51

** 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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21:01:28

What is the run between the two specified t values?

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The run is the change in clock time

30-10=20, there was a 20 second change in clock time

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21:01:31

** run = 30 - 10 = 20 **

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21:02:37

What therefore is the slope and what does it mean?

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RESPONSE -->

[depth(30)-depth(10)]/20

This is the change in water depth in respect to clock time. For example, how many cm/s the water was flowing out of the uniform cylinder.

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21:02:44

** 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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21:03:19

problem 5 graph points corresponding to load1 and load2

What are the coordinates of the requested graph points?

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RESPONSE -->

(load1, springLoad(load1))

(load2, springLoad(load2))

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21:03:30

** 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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21:04:43

What is your expression for the average slope of the graph between load1 and load2?

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RESPONSE -->

[springLoad(load2)-springload(load1)]/(load2-load1)

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21:04:48

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

run = load2 - load1 so

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

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21:05:34

problem 6 symbolic expression for slope of depth function

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slope = [depth(t2)-depth(t1)]/(t2-t1)

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21:05:40

** 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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21:12:43

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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y=f(t)=40(2)^(-.3t)+25

between t=10 and t=20

y=f(10)=40(2)^(-.3*10)+25 = 40(.125)+25 = 5+25 = 30; (10,30)

y=f(20)=40(2)^(-.3*20)+25=40(.015625)+25 = 25.625; (20,25.625)

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

between t=20 and t=30

t=20; (20,25.625)

y=f(30)=40(2)^(-.3*30)+25 = 40+.078125=40.078125; (30, 40.078125)

(40.078125-25.625)/(30-20)= 14.453125/10 = 1.4453125

between t=30 and t=40

t=30; (30, 40.078125)

y=f(40)=40(2)^(-.3*40)+25 = 25.00976562; (40, 25.00976562)

(25.00976562-40.078125)/(40-30) = -1.506835938

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21:17:36

** 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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RESPONSE -->

between t=20 and t=30, on the y=f(30) function i added 40 to .078125 instead of adding 25, thus resulting in (30, 40.078125) instead of (30, 25.078125). This also causes the average rate between t=30 and t= 40 to be incorrect, but upon reworking the problem, I arrived at the same answers.

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21:17:45

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

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21:18:04

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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Your work on this assignment is very good. Let me know if you have questions.