Query 9

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course Mth 163

10/12 930

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:

y = .1x^2 - 3 when x = -2 and x = 7

y = .1(-2)^2 - 3

= -2.6

y = .1(7)^2 - 3

= 1.9

(-2, -2.6) ( 7, 1.9)

Rise = 1.9- (-2.6) = 4.5

Run = 7- (-2) = 9

Slope = 4.5/9 = .5 or 1/2

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 / 9 = 1/2. **

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Self-critique (if necessary):

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Self-critique Rating: ok

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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:

y = depth(t) when t = 30 and t = 10

rise = depth(30) - depth(10)

run = 30 - 10 = 20

slope = (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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Self-critique (if necessary):

ok

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Self-critique Rating: ok

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

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Your solution:

The rise is y = 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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Self-critique Rating: ok

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

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Your solution:

Run is 30 - 10 = 20

confidence rating #$&*: 3

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Given Solution:

** run = 30 - 10 = 20 **

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Self-critique (if necessary):

ok

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Self-critique Rating: ok

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

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Your solution:

The slope is (depth(30) - depth(10)) / 20. This is the average rate between clock time and water depth of 10 and 30.

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:

y = springLength(Load)

load = load 1

load = load 2

(load 1, springLength(load 1))

(load 2, springLength(load 2))

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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Self-critique (if necessary):

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Self-critique Rating: ok

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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:

Slope = (springLength(Load2)) - (springLength(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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Self-critique (if necessary):

ok

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Self-critique Rating: ok

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

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Your solution:

y = depth(t) when t = t1 and t = t2

(t1, depth(t1)) and (t2, depth(t2))

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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Self-critique (if necessary):

ok

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Self-critique Rating: ok

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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:

t = 10 and t = 20

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

= 30

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

= 25.625

25.625 - 30 = -4.375

20-10 = 10

Average rate = -.4375

t = 20 and t = 30

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

= 25.625

y = 40 (2^(30) ) ) + 25

= 25.078

25.078 - 25.625 = -.547

30- 20 = 10

Average rate = -.0547

t = 30 and t = 40

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

= 25.078

y = 40 (2^(40) ) ) + 25

= 25.0098

25.0098 - 25.078 = -.0682

40 - 30 = 10

Average rate = -.00682

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. **

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

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Your solution:

This was a neat assignment. It was fairly easy for me and understood well. On the last problem, I needed to carry out my decimals more to get a more defined answer.

confidence rating #$&*: 3

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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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