course Mth173 I had an error last night with the querry program several times and one today, each time I had to reload the program, that is why I am late. Sorry for the inconvience. סdFņEpŻkgӂassignment #005
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10:34:56 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 by which the principal grows for example 10% or .1 the growth factor is what you multiply the principal by to get the new principal. If you have a principal that grows by 10% then your growth factor is 1.1
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10:35:06
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RESPONSE --> Blank
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10:35:21 ** 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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RESPONSE --> okay
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10:37:47 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 slope represents the average rate of change between the time intervals because you are getting the total depth change and dividing it by the elapsed time giving you an average quantity of cm/s etc.
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10:38:08 ** 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 --> okay
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10:41:22 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 --> Because when we see a line on a graph the area below is represntative to the actual area of the model that the graph represnts, therefore if the area changes from one trapazoid to the other it shows the actual change in quantity.
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10:42:17 **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 --> okay
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10:50:50 #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 --> Since the patient had 550mg in her bloodstream and this decreases by 11% every hour at the end of the first hour she would only have 89% of 550mg in her bloodstream. So we have the function of: 550 (.89^ t) = y
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10:51:11 ** 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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RESPONSE --> okay
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10:54:23 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> since she has 550mg in her bloodstream at 10:00am then at 3:00pm, a time of 5 hours has passed. we plug this into our equation: 550 (.89^ (5) = Q Q= 307.123mg at 3:00pm
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10:54:40 ** 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 --> okay
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11:00:07 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> The graph is a slowly curving line with points at (0,550), (5,307), (10, 171) To estimate the half life we use the equation: 550 (.89^ hl) = (550 / 2) (.89^hl) = .5 hl = (log.5) / (log.89) hl = 5.948 hrs
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11:02:01 ** 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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RESPONSE --> By using the graph I draw a line from half the orginol quantity then as the line intersects the model's curved lines draw a line straight down to the show the time.
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11:03:51 What is the equation to find the half-life? What is its most simplified form?
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RESPONSE --> the equation for half life in the simplest form would be: (R^hl) = .5 or Rate ^ (half life) = .5
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11:04:38 ** 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 --> okay
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11:13:15 #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 treated the lack of a quantity like having a half life, but more of a .075 life. we are looking for the a point in between .05 and .1, the average of these is .075 Q0 (1.1^t) = .075 (1.1 ^t)= .075 t= log(.075) / log(1.1) t= -27.177 and for a value between .005 and .01: Q0 (1.1^ t)= .0075 (1.1^ t)= .0075 t= log(.0075) / log(1.1) t= -51.336
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11:14:40 ** 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 --> I used the average of the two points in hindsight it probobly would have been better to make it an inequaltity type answer.
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13:14:36 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> because as t nears zero the times will continue to be smaller and smaller.
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13:15:25 ** 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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RESPONSE --> okay
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13:19:42 #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 the equation y=12 (e^(-.5x)) B=.607
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13:20:03 ** 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 --> okay
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13:21:35 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> y=Ab^x y= .007(e^ (.71x)) b= 2.034 13:24:17 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> B= 2.034
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13:24:37 ** 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 --> okay
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13:25:34 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> y=Ab^x y= -13 (e^(3.9 x)) B= 49.402
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13:25:43 ** 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 --> okay
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13:25:52 List these functions, each in the form y = A b^x.
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RESPONSE --> okay
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13:32:45 ** The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) **
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RESPONSE --> y=Ab^x (.6065) = B z = log(.6025)/ log(e) y= 12(e^zx) y= 12 (e^(-.5 x)) (2.03399) = B z= log(.2.03399)\ Log(e) y= .007 (e^zx) y= .007 (e^.71) (49.40244) = B z=log(49.40244) / Log(e) y= -13 (e^zx) y= .13(e^(3.9 x))
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13:35:02 query text problem 1.1 #24 dolphin energy prop cube of vel
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RESPONSE --> N= l / l^2
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13:37:46 ** A proportionality to the cube would be E = k v^3. **
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RESPONSE --> Number of animal species with a certain body Legnth is inversly proportional to L^2. If we cubed instead of squared the equation would be N = l/l^3
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13:42:56 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 --> Text problem 1.1 #27 is a graph of total distance traveled by a plane. But assuming if f was on the y axis and t was on the x this point would be(10,30).
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13:43:16 ** 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 --> okay
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13:45:40 what is the meaning of the equation H(30) = 10?
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RESPONSE --> H(f) = ft H(30) = (30) (t) =10 H(30) = (30) (.3333) = 10 H(30) = 10
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13:46:05 ** This means that when clock time t is 30, the temperature H is 10. **
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RESPONSE --> okay
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13:46:43 What is the meaning of the vertical intercept?
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RESPONSE --> The y axis portion of a point on a graph.
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13:47:24 ** This is the value of H when t = 0--i.e., the temperature at clock time 0. **
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RESPONSE --> when the function = 0 then the y intercept is given
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13:47:50 What is the meaning of the horizontal intercept?
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RESPONSE --> the x intercept of when the function is set equal to y.
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13:48:05 ** This is the t value when H = 0--the clock time when temperature reaches 0 **
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RESPONSE --> okay
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13:51:49 query text problem 1.1.32. Water freezes 0 C, 32 F; boils 100 C, 212 F. Give your solution to problem 1.1.32.
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RESPONSE --> A) f(30) = 10 means that when H=10 then T=30 B) The verticle intercept of is the heat or temperature of the object, where the horizontal intercept is the time of the object.
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13:53:33 ** 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. **
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RESPONSE --> for question 32 I had a different problem than is displayed, and therefore don't have the correct answer,.
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