course Mth 173 For any question that i did not answer (any that I left blank) meanse that I could not find the questions and notes referrence in the question. Any of the question referring to the text 1.1, such as the dolphin question. ???????????assignment #005???????????
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16:42:08 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 --> Growth rate is the rate at which something grows over time. Growth factor is the factor by which we multiply the beginning amount to get the ending amount. If the beginning amount is $500 at a rate of 5%. The growth rate is 5%, or .05. The growth factor would be 1 + .05, or 1.05.
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16:42:11
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16:42:25 ** 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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16:49:00 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 --> The distance between the two times (x) shows the change in time. The difference between the two depths (y) shows the difference in depths. The sloping line shows the change in depth and the change in time; therefore it represents the average rate of depth change.
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16:49:07 ** 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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16:54:57 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 area is equal to the trapezoid's average altitude multiplied by its width. In this case it is the average altitude (y) multiplied by the width (x or change in time).
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16:54:59 **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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16:57:20 ??? #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) = 550 * (1 - .11) ^ t
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16:57:31 ** 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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16:58:03 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> Q(5) = 550 * (1 - .11) ^ 5 = 307.1232697mg
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16:58:06 ** 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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17:03:51 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> The graph showed that between 5 and 6 hours after 10:00am the penicillan = 275mg
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17:03:56 ** 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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17:04:58 What is the equation to find the half-life?? What is its most simplified form?
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RESPONSE --> 275 = 550 * (.89) ^ t .5 = (.89) ^ t
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17:05:15 ** 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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RESPONSE --> I did not use the doubleTime notation
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17:23:59 #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 --> Q(t) lies between .005 Q0 and .01 when -55.5 < t < -48.3
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17:24:15 ** 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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17:25:19 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> Because both .005 and .01 are less than 1.1; therefore we must raise 1.1 to a negative power to decrease its value.
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17:25:26 ** 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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17:55:35 #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 --> For y = 12 ( e ^ (-.5x) ), b = .7071067812
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17:55:48 ** 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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RESPONSE --> I have not yet grasped this concept.
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17:57:20 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> b = 1.63580
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17:57:39 ** 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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17:58:34 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> b = 49.3824
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17:59:09 ** 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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RESPONSE --> For the first two I did not realize to use 2.718 for the value of e. I now know that for the future
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18:00:11 List these functions, each in the form y = A b^x.
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18:00:25 List these functions, each in the form y = A b^x.
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RESPONSE --> I did not know what functions to list or how to find them.
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18:00:30 ** The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) **
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18:03:45 query text problem 1.1 #24 dolphin energy prop cube of vel
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18:03:58 query text problem 1.1 #24 dolphin energy prop cube of vel
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RESPONSE --> I do not know where to locate the dolphin query question
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18:04:02 ** A proportionality to the cube would be E = k v^3. **
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