Query 5

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course MTH 173

5/27/2013 at 4:20AM

005. `query 5

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Question: `q 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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Your solution:

When multiplied by a value, the growth rate states how much the value will change over a single period. When multiplied by the value, the growth factor states the new value at the end of the next period.

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Given Solution:

** Specific statements:

When multiplied by a quantity the growth rate tells us how much the quantity will change over a single period.

When multiplied by the quantity the growth factor gives us the new quantity at the end of the next period. **

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Question: `q 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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Your solution:

A depth vs. time trapezoid’s slope is composed of a rise/run rate, which is exactly what the average rate of change represents.

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Given Solution:

** 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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Question: `q 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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Your solution:

The average rate of change can be measured by the average altitude(s) of rate vs. time trapezoid(s), which then multiplied by the width of the trapezoid represents the time interval. This gives the change in quantity.

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Given Solution:

**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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Question: `q ےےے #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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Your solution:

Growth rate: -.11%

Growth factor: 1 + r = 1 + (-.11) = .89

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

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

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

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Given Solution:

** 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)^tے **

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

** 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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Question: `q Describe your graph and explain how it was used to estimate half-life.

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Your solution:

The y-axis represents Q(t), so I found the point on the Q(t) axis will be half as high as the starting point. A horizontal line is drawn directly over to the graph, projecting this point down. The distance from the first point to the second will be the distance on the horizontal axis (t axis) between the two vertical lines drawn.

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Given Solution:

** 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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Question: `q What is the equation to find the half-life?ے What is its most simplified form?

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Your solution:

Q(DoublingTime) = ½ Q(0)

550 mg * .89^DoublingTime / 550 = .5 * 550 / 550

.89^DoublingTime = .5

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Given Solution:

** 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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Question: `q #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.

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

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Your solution:

Q(t) = .05 Q0

Q0 * 1.1^t = .05 Q0

1.1^t = .05

t = -31.4

Q(t) = .1 Q0

1.1^t = .1 Q0

1.1^t = .1

t = -24.2

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Given Solution:

** 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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Question: `q explain why the negative t axis is a horizontal asymptote for this function.

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Your solution:

The value of 1.1^t increases while t increases; as t approaches infinity, 1.1^t approaches infinity as well. Where 1.1^-t = 1 / 1.1^t, increasingly large negative values of t and the value of 1.1^t will become smaller. This would approach zero.

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Given Solution:

** 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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Question: `q #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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Your solution:

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

This function appears to be of the form y = A b^x when b = .61

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Given Solution:

** 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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Question: `q what is b for the function y = .007 ( e^(.71 x) )?

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Your solution:

.007 * 2.04^x

This function appears to be in the form y = A b^x when b = 2.041

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Given Solution:

** .007 e^(.71 x) = .007 (e^.71)^x = .007 * 2.04^x, approx.

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

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

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Your solution:

-13 e^(3.9 x) =

-13 (e^3.9)^x =

-13 * 49.4^x

This function is in the form y = A b^x when b = 49.4

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Given Solution:

** -13 e^(3.9 x) = -13 (e^3.9)^x = -13 * 49.4^x, approx.

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

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Question: `q List these functions, each in the form y = A b^x.

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Your solution:

y=12(.6065^x)

y=.007(2.03399^x)

y=-13(49.40244^x)

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Given Solution:

** The functions are

y=12(.6065^x)

y=.007(2.03399^x) and

y=-13(49.40244^x) **

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Question: `q query text problem 1.1.31 5th; 1.1.23 4th dolphin energy prop cube of vel

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Your solution:

The proportionality is E = k v^3

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Given Solution:

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

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Question: `q query text problem 1.1.37 5th; 1.1.32 4th temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts

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Your solution:

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.

H(30) = 10 means that when the clock time is 30, the temperature is 10.

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Given Solution:

** 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. **

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

** This means that when clock time t is 30, the temperature H is 10. **

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Question: `q What is the meaning of the vertical intercept?

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Your solution:

The vertical intercept is the temperature at clock time zero.

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Given Solution:

** This is the value of H when t = 0--i.e., the temperature at clock time 0. **

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Question: `q What is the meaning of the horizontal intercept?

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Your solution:

The horizontal intercept is the clock time when temperature reaches 0

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Given Solution:

** This is the t value when H = 0--the clock time when temperature reaches 0 **

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Question: `q query text problem 1.1.40 5th; 1.1.31 4th. Water freezes 0 C, 32 F; boils 100 C, 212 F. Give your solution.

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Your solution:

M = (212 - 32) / (100 - 0) = 1.8

F = 1.8 + 32

1.8 * 20 + 32 = 68

F = 1.8 C + 32

-.8 C = 32

C = 32 / (-.8) = -40

C = -40

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Given Solution:

** The graph contains points (0, 32) and (100, 212). The slope is therefore (212-32) / (100-0) = 1.8.

The y-intercept is 32 so the equation of the line is

y = 1.8 x + 32, or using F and C

F = 1.8 C + 32.

To find the Fahrenheit temp corresponding to 20 C we substitute C = 20 into F = 1.8 C + 32 to get

F = 1.8 * 20 + 32 = 36 + 32 = 68

The two temperatures will be equal when F = C. Substituting C for F in F = 1.8 C + 32 we get

C = 1.8 C + 32. Subtracting 1.8 C from both sides we have

-.8 C = 32 or

C = 32 / (-.8) = -40.

The scales read the same at -40 degrees. **

STUDENT QUESTION

I understand all work shown except the answer for d? I understand how to substaute and see how you worked it, but don’t understand the logic for why this works ?

INSTRUCTOR RESPONSE

You have a linear function which gives you F when you substitute C.

That means that the graph of F vs. C is a straight line.

According to the given information the straight line passes through the points (0, 32) and (100, 212).

Given two points, there are a variety of ways to get the equation of the corresponding straight line. The given solution finds the slope, then uses slope-intercept form of the equation of a straight line.

Alternatively you could use the point-point form of the equation, which would lead to the same result.

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&#Your work looks good. Let me know if you have any questions. &#

Query 5

#$&*

course MTH 173

5/27/2013 at 4:20AM

005. `query 5

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Question: `q 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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Your solution:

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Given Solution:

** Specific statements:

When multiplied by a quantity the growth rate tells us how much the quantity will change over a single period.

When multiplied by the quantity the growth factor gives us the new quantity at the end of the next period. **

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Question: `q 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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Your solution:

A depth vs. time trapezoid’s slope is composed of a rise/run rate, which is exactly what the average rate of change represents.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

** 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.

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Question: `q 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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Your solution:

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Given Solution:

**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.

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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Your solution:

Growth rate: -.11%

Growth factor: 1 + r = 1 + (-.11) = .89

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

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

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

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Given Solution:

** 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)^tے **

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

** 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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Question: `q Describe your graph and explain how it was used to estimate half-life.

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Your solution:

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Given Solution:

** 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 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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Question: `q What is the equation to find the half-life?ے What is its most simplified form?

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Your solution:

Q(DoublingTime) = ½ Q(0)

550 mg * .89^DoublingTime / 550 = .5 * 550 / 550

.89^DoublingTime = .5

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Given Solution:

** 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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Question: `q #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.

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

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Your solution:

Q(t) = .05 Q0

Q0 * 1.1^t = .05 Q0

1.1^t = .05

t = -31.4

Q(t) = .1 Q0

1.1^t = .1 Q0

1.1^t = .1

t = -24.2

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Given Solution:

** 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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Question: `q explain why the negative t axis is a horizontal asymptote for this function.

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Your solution:

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Given Solution:

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Question: `q #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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Your solution:

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

This function appears to be of the form y = A b^x when b = .61

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Given Solution:

** 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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Question: `q what is b for the function y = .007 ( e^(.71 x) )?

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Your solution:

.007 * 2.04^x

This function appears to be in the form y = A b^x when b = 2.041

confidence rating #$&*:

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Given Solution:

** .007 e^(.71 x) = .007 (e^.71)^x = .007 * 2.04^x, approx.

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

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

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Your solution:

-13 e^(3.9 x) =

-13 (e^3.9)^x =

-13 * 49.4^x

This function is in the form y = A b^x when b = 49.4

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Given Solution:

** -13 e^(3.9 x) = -13 (e^3.9)^x = -13 * 49.4^x, approx.

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

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Question: `q List these functions, each in the form y = A b^x.

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Your solution:

y=12(.6065^x)

y=.007(2.03399^x)

y=-13(49.40244^x)

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Given Solution:

** The functions are

y=12(.6065^x)

y=.007(2.03399^x) and

y=-13(49.40244^x) **

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Question: `q query text problem 1.1.31 5th; 1.1.23 4th dolphin energy prop cube of vel

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Your solution:

The proportionality is E = k v^3

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Given Solution:

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

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Question: `q query text problem 1.1.37 5th; 1.1.32 4th temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts

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Your solution:

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.

H(30) = 10 means that when the clock time is 30, the temperature is 10.

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Given Solution:

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

** This means that when clock time t is 30, the temperature H is 10. **

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Question: `q What is the meaning of the vertical intercept?

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Your solution:

The vertical intercept is the temperature at clock time zero.

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Given Solution:

** This is the value of H when t = 0--i.e., the temperature at clock time 0. **

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Question: `q What is the meaning of the horizontal intercept?

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Your solution:

The horizontal intercept is the clock time when temperature reaches 0

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Given Solution:

** This is the t value when H = 0--the clock time when temperature reaches 0 **

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Question: `q query text problem 1.1.40 5th; 1.1.31 4th. Water freezes 0 C, 32 F; boils 100 C, 212 F. Give your solution.

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Your solution:

M = (212 - 32) / (100 - 0) = 1.8

F = 1.8 + 32

1.8 * 20 + 32 = 68

F = 1.8 C + 32

-.8 C = 32

C = 32 / (-.8) = -40

C = -40

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Given Solution:

** The graph contains points (0, 32) and (100, 212). The slope is therefore (212-32) / (100-0) = 1.8.

The y-intercept is 32 so the equation of the line is

y = 1.8 x + 32, or using F and C

F = 1.8 C + 32.

To find the Fahrenheit temp corresponding to 20 C we substitute C = 20 into F = 1.8 C + 32 to get

F = 1.8 * 20 + 32 = 36 + 32 = 68

The two temperatures will be equal when F = C. Substituting C for F in F = 1.8 C + 32 we get

C = 1.8 C + 32. Subtracting 1.8 C from both sides we have

-.8 C = 32 or

C = 32 / (-.8) = -40.

The scales read the same at -40 degrees. **

STUDENT QUESTION

I understand all work shown except the answer for d? I understand how to substaute and see how you worked it, but don’t understand the logic for why this works ?

INSTRUCTOR RESPONSE

You have a linear function which gives you F when you substitute C.

That means that the graph of F vs. C is a straight line.

According to the given information the straight line passes through the points (0, 32) and (100, 212).

Alternatively you could use the point-point form of the equation, which would lead to the same result.

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