5

course Mth 173

For some reason, I get confused on these type of quizes. I do better with paper form.

You should have paper handy when you work on these qa and query exercises, and you should be jotting things down.

When you do a query you should have completed the work for the assignment and should have it handy for reference when answering questions.

??O?n?“S????????assignment #005

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

??????[?H??????Calculus I

01-17-2007

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08:45:18

Growth rate and growth factor: Describe the difference between growth rate and growth factor and give a short example of how each might be used

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RESPONSE -->

The growth rate is the percentage(10% - GR of 10%)

The growth factor is where you multiply the principle in order to get the ending principle.(10% - GF of 1.10)

For example, a savings account of $5 at 1% interest for 5 years

$5 * .01 = .05 earned for the first year

$5 * (1.01 ^ 5) = 5.26 total at 5 years

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08:53:48

Class notes #05 trapezoidal representation.

Explain why the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval represented

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RESPONSE -->

We divide the rise by the run which is the change in velocity by the time. The area of the trapezoid gives us the change between the two times and the velocity is the rate the position changes and together that creates the slop.

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08:54:20

** GOOD ANSWER BY STUDENT WITH INSTRUCTOR COMMENTS:

The slope of the trapezoids will indicate rise over run

or the slope will represent a change in depth / time interval

thus an average rate of change of depth with respect to time

INSTRUCTOR COMMENTS:

More detail follows:

** To explain the meaning of the slope you have to reason the question out in terms of rise and run and slope.

For this example rise represents change in depth and run represent change in clock time; rise / run therefore represents change in depth divided by change in clock time, which is the average rate of change. **

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RESPONSE -->

I think I done what you asked for.

You didn't address the meaning of the rise, the meaning of the run, or the resulting meaning of the slope.

Your self-critique should address all these details.

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09:05:35

Explain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval.

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RESPONSE -->

The velocity is the rate of which the position changes.

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09:05:52

**STUDENT RESPONSE WITH INSTRUCTOR COMMENTS:

The area of a rate vs. time graph rep. the change in quantity.

Calculating the area under the graph is basically integration

The accumulated area of all the trapezoids for a range will give us thetotal change in quantity.

The more trapezoids used the more accurate the approx.

INSTRUCTOR COMMENTS: All very good but the other key point is that the average altitude represents the average rate, which when multiplied by the width which represents time interval gives the change in quantity

You have to reason this out in terms of altitudes, widths and areas.

For the rate of depth change example altitude represents rate of depth change so average altitude represents average rate of depth change, and width represents change in clock time.

average altitude * width therefore represents ave rate of depth change * duration of time interval = change in depth.

For the rate of change of a quantity other than depth, the reasoning is identical except you'll be talking about something besides depth. **

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RESPONSE -->

Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions.

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09:08:17

??? #17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour. What is the function Q(t)?

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RESPONSE -->

Q(t) = 550mg x (.89) ^ t

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09:08:25

** Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have

Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or

Q(t)=550(.89)^t? **

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09:11:09

How much antibiotic is present at 3:00 p.m.?

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RESPONSE -->

Q(t) = 550mg x (.89 ^ 5)

550 x .56 = 308.

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09:11:27

** 3:00 p.m. is 5 hours after the initial time so at that time there will be

Q(5) = 550 mg * .89^5 = 307.123mg

in the blood **

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RESPONSE -->

I rounded to the nearest whole number.

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09:12:59

Describe your graph and explain how it was used to estimate half-life.

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RESPONSE -->

You use the Y axis for 550 and more down half way which is 275 and use your finger to glide over to see which hour that your finger intercepts the slope.

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09:13:07

** Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point.

The vertical coordinate represents the quantity Q(t), so we find the point on the vertical axis which is half as high as the vertical coordinate of our starting point. We then draw a horizontal line directly over to the graph, and project this point down.

The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. **

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09:14:36

What is the equation to find the half-life?? What is its most simplified form?

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RESPONSE -->

550 mg (.89 ^ t) = 275

.89 ^ t = 1/2 (you get this equation by dividing both sides by 550)

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09:14:44

** Q(doublingTime) = 1/2 Q(0)or

550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have

.89^doublingTime = .5.

We can use trial and error to find an approximate value for doublingTIme (later we use logarithms to get the accurate solution). **

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09:15:36

#19. For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0.

For what values of t did Q(t) lie between .005 Q0 and .01 Q0?

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RESPONSE -->

I had trouble with these.

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09:16:12

** Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0.

Note that these values must be negative, since positive powers of 1.1 are all greater than 1, resulting in values of Q which are greater than Q0.

Solving Q(t) = .05 Q0 we rewrite this as

Q0 * 1.1^t = .05 Q0. Dividing both sides by Q0 we get

1.1^t = .05. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get

t = -31.4 approx.

Solving Q(t) = .1 Q0 we rewrite this as

Q0 * 1.1^t = .1 Q0. Dividing both sides by Q0 we get

1.1^t = .1. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get

t = -24.2 approx.

(The solution for .005 Q0 is about -55.6, for .01 is about -48.3

For this solution any value between about t = -48.3 and t = -55.6 will work). **

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RESPONSE -->

This helps explain.

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09:17:31

explain why the negative t axis is a horizontal asymptote for this function.

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RESPONSE -->

Every time we multiply by a number less than 1, our result decreases.

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09:17:39

** The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero. **

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09:18:54

#22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )?

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RESPONSE -->

The value of b is e or 2.718

y = 12(e ^ (-.5x))

y = 12 (.6065621044 ^ x)

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09:19:13

** 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx.

So this function is of the form y = A b^x for b = .61 approx.. **

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09:20:36

what is b for the function y = .007 ( e^(.71 x) )?

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RESPONSE -->

2.718

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09:20:56

** 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx.

So this function is of the form y = A b^x for b = 2.041 approx.. **

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RESPONSE -->

I thought you were asking a different question.

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09:21:10

what is b for the function y = -13 ( e^(3.9 x) )?

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RESPONSE -->

y = -13 (e ^ (3.9x))

y = -13 (49.40244911 ^ x)

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09:21:13

** 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx.

So this function is of the form y = A b^x for b = 49.4 approx.. **

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09:22:13

List these functions, each in the form y = A b^x.

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RESPONSE -->

y = 12(2 ^ (-.5x))

y = 12 (.03125 ^ x)

y = .007(2 ^ (.71x))

y = .007 (1.635804117 ^ x)

y = -13 (2 ^ (3.9x))

y = -13 (14.92852786x)

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09:22:37

** The functions are

y=12(.6065^x)

y=.007(2.03399^x) and

y=-13(49.40244^x) **

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RESPONSE -->

Wrong question.

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09:23:01

query text problem 1.1 #24 dolphin energy prop cube of vel

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RESPONSE -->

24. N = kl?.................................................

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09:23:31

** A proportionality to the cube would be E = k v^3. **

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RESPONSE -->

Okay, I am so lost with these questions.

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09:24:18

query text problem 1.1 #27 temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts

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RESPONSE -->

In 30 minutes, the temperature changed 10 degrees in C?

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09:24:24

** The interpretation would be that the vertical intercept represents the temperature at clock time t = 0, while the horizontal intercept represents the clock time at which the temperature reaches zero. **

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RESPONSE -->

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09:24:29

what is the meaning of the equation H(30) = 10?

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RESPONSE -->

In 30 minutes, the temperature changed 10 degrees in C?

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