course ??]????????assignment #005s?r??????????????Calculus I
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14:57:58 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 --> A growth rate is the specific rate at which the function grows. In compounding interest the growth rate might be 3.4%. The growth factor is the specific number by which the other parts of the function are multiplied. Going along with the growth rate above, the growth factor for the function would be 1.034.
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14:58:45 ** 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 --> That too.
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15:00:47 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 --> This slope is the average slope between the two chosen points. The rate of change is found for the two points, then the average of that is taken. The presents the average slope between the two points, allowing the average depth change to be projected.
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15:01:46 ** 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 got most of that.
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15:11:51 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 of a trapezoid is the average height multiplied by the width. In looking at a rate vs. time function, the average height is the average rate while the width is the time interval. A time multiplied by a rate gives a change of quantity.
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15:12:34 **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 --> I missed the integration bit. Ok.
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15:13:32 ??? #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 * (.89)^t
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15:13:44 ** 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 --> Ok.
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15:15:31 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> At 3pm, 4 hours have past, so t=4. Therefore: 550*(.89)^5 =307.12mg
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15:16:21 ** 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 --> Ok.
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15:17:56 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> After plotting the points, I drew lines between the points. After finding the half-life (275), I estimated where it would be on the y-axis and drew a line straight over until it crossed the plotted line. I then drew a line straight down to estimate the time at which the half-life was reached.
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15:18:05 ** 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 --> Yep.
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15:18:58 What is the equation to find the half-life?? What is its most simplified form?
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RESPONSE --> The equation to find half-life would be: 275 = 550*(.89)^t The reduced form: 1/2 = (.89)^t
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15:19:25 ** 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 --> Ok. I forgot to say t was equal to doublingTime.
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15:20:52 #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 --> Between .05 Q0 and .1 Q0: -31 < t < -24 Between .005 Q0 and .01 Q0: -55 < t < -48
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15:21:46 ** 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 --> Ok.
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15:23:07 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> Because as t grows more and more negative, the value of Q0 grows smaller and smaller. However, it does not completely disappear as it only gets very small, therefore creating an asymptote.
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15:23:24 ** 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 --> Yes, approaching infinity.
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15:27:19 #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 --> In the function y = A b^x, b = 2^k. In the function y = 12 (e^(-0.5x)), b = e^(-0.5) = 0.606.
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15:27:32 ** 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 --> Close enough.
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15:28:04 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> b = e^0.71 = 2.03
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15:28:11 ** 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 --> Ok.
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15:28:48 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> b = e^(3.9) = 49.4
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15:28:53 ** 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 --> Ok.
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15:29:54 List these functions, each in the form y = A b^x.
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RESPONSE --> 1. y = 12 (.606)^x 2. y = .007 (2.034)^x 3. y = -13(49.4)^x
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15:30:02 ** The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) **
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RESPONSE --> Ok.
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15:32:17 query text problem 1.1 #24 dolphin energy prop cube of vel
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RESPONSE --> Huh? Section 1.1 #24 in the text asks this question: ""The number of animal species, N, of a certain body length, l, is inversely proportional to the square of l."" Write a formula. N = 1 / (l^2)
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15:33:17 ** A proportionality to the cube would be E = k v^3. **
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RESPONSE --> This is question 23 in the 4th edition of the book. I take it k is the factor that makes it proportional, but not directly the same?
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15:40:18 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 --> Section 1.1 #27 is about graphing a flight from Dulles Airport to La Guardia Airport. Again, 4th edition text. However, Section 1.1 #32 sounds like the problem. a) In terms of temperature, in 30 minutes of being outside, the temperature of the object is 10 degrees C. b) At the vertical intercept, a, the object has just been put outside and so the temperature of the object has not changed. At the horizontal intercept, b, the time outside has progressed to the point where the object's temperature is equal to 0 degrees C.
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15:40:26 ** 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 --> Ok.
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15:41:24 what is the meaning of the equation H(30) = 10?
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RESPONSE --> This presents the relationship between time and temperature. What the exact formula is, we don't know, but we do know they are related.
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15:41:37 ** This means that when clock time t is 30, the temperature H is 10. **
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RESPONSE --> Eh, I forgot to include details.
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15:42:06 What is the meaning of the vertical intercept?
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RESPONSE --> The vertical intercept is the time at which the temperature has not dropped and the time is 0.
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15:42:11 ** This is the value of H when t = 0--i.e., the temperature at clock time 0. **
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RESPONSE --> Ok.
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15:42:28 What is the meaning of the horizontal intercept?
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RESPONSE --> The horizontal intercept is the time at which temperature is 0.
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15:42:32 ** This is the t value when H = 0--the clock time when temperature reaches 0 **
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RESPONSE --> Ok.
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15:44:34 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 --> This is Section 1.1 #31 in 4th edition text. a) Slope = 1.8. b) y = 1.8x + 32. c) At 20 degrees C, the temperature in F is 68. d) -40 degrees C.
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15:45:43 ** 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 --> Ok.
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