#$&* course Mth 173
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Given Solution: ** Specific statements: When multiplied by a quantity the growth rate tells us how much the quantity will change over a single period. When multiplied by the quantity the growth factor gives us the new quantity at the end of the next period. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Class notes #05 trapezoidal representation. Explain why the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval represented YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If a graph is divided into a number of trapezoids, consider a single trapezoid. For a depth vs. time graph the slope of the trapezoid is average rate of change of depth with respect to time during the time interval represented. The slope of the trapezoid basically is rise / run. Rise for the trapezoid thus will be the change in depth during that interval. Run for that trapezoid will be time interval to which the trapezoid belongs. Thus slope will be the ratio of the change of depth to the time interval. And the ratio of the change in the depth and the time interval is average rate of change of depth with respect to time during the time interval. Thus the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval represented. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As explained in the previous question we know that the slope of a quantity vs. time trapezoid is the average rate of change of the quantity with respect to time during the time interval represented. Thus average rate = slope = change in quantity / time interval. Thus now if we have a rate vs. time trapezoid, the area of the trapezoid will be the product of average altitude of the trapezoid and the distance between the parallel lines. The average altitude of the trapezoid is thus the average rate of change of quantity for that trapezoid and the distance between the parallel lines is the time duration for which the rate is considered. Thus the area of the trapezoid = average rate of change of quantity * time interval, which previously explained is the change in the quantity for that time interval. Thus the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: **STUDENT RESPONSE WITH INSTRUCTOR COMMENTS: The area of a rate vs. time graph rep. the change in quantity. Calculating the area under the graph is basically integration The accumulated area of all the trapezoids for a range will give us thetotal change in quantity. The more trapezoids used the more accurate the approx. INSTRUCTOR COMMENTS: All very good but the other key point is that the average altitude represents the average rate, which when multiplied by the width which represents time interval gives the change in quantity You have to reason this out in terms of altitudes, widths and areas. For the rate of depth change example altitude represents rate of depth change so average altitude represents average rate of depth change, and width represents change in clock time. average altitude * width therefore represents ave rate of depth change * duration of time interval = change in depth. For the rate of change of a quantity other than depth, the reasoning is identical except you'll be talking about something besides depth. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q ÿÿÿ #17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour. What is the function Q(t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Initial penicillin P0 = 550 mg Rate at which penicillin changes = - 11% Thus the growth rate = -0.11 The growth factor = 0.89 Let t be the number of hours from 10:00 am Thus quantity of penicillin as a function of time be Q(t) = P0 * ( growth factor^t) = P0 *( 0.89 ^ t ) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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)^tÿ ** 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Describe your graph and explain how it was used to estimate half-life. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since the quantity of antibiotic is decreasing thus the graph must be a decreasing graph. By substituting t for 1 , 2 , 3 , 4 hours we obtain the quantity of antibiotic as 489.5 gm, 435.655 gm, 387.73295 gm and 345.0823255gm respectively. Thus we can see that the decrease in the quantity decreases as time passes. Thus the graph decreases with the slope increasing ( as the slope is negative and as the graph proceeds the slope approaches 0 ). Now in order to find the estimate half-life, the graph has the time on the x axis and the quantity on the y axis. Thus the half-life is the time at which the original quantity reduces to half of its value. Thus first locate the value of 550 / 2 = 275 gm on the y axis. Next draw an horizontal life from the point ( 0 , 275 ) to the graph. From the point of intersection construct a vertical line parallel to y - axis to meet the x - axis. The point of intersection will give the time after which the penicillin in the bloodstream will reduce to half its value(only if the initial quantity was located at t = 0 , if not then the difference between the initial time and the point of intersection time gives the half-life). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point. The vertical coordinate represents the quantity Q(t), so we find the point on the vertical axis which is half as high as the vertical coordinate of our starting point. We then draw a horizontal line directly over to the graph, and project this point down. The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q What is the equation to find the half-life?ÿ What is its most simplified form? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We know that half-life is the time a quantity takes to reduce to half its value. Thus Q(half life) = Q(0) / 2 = P0 / 2 For the penicillin problem Q(half life ) = 550 gm / 2 = 275 gm Thus in order to find the half life time we need to substitute the Q( half-life ) in the Q(t) equation Thus, Q (half-life) = 550 gm * (0.89) ^half life We know that Q (half-life) = 275 gm Thus 275 gm = 550 gm * (0.89) ^ half life 0.5 = (0.89)^ half life half-life = ln(0.5) / ln(0.89) = 5.948034 years thus the equation of half-life for any equation thus will be [Q(0) / 2] = Q(0) * [ ( 1 + r)^( half-life )] (1/2) = (1+r)^( half life ) Half life = ln(0.5) / ln( 1 + r ) ` For the penicillin problem half life is 5.948034 years confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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. Given function Q(t) = Q0(1.1 ^ t ) For Q(t) = 0.05 Q0 We calculate t as , 0.05*Q0 = Q0 ( 1.1 ^ t ) thus t = ln(0.05) / ln(1.1) = - 31.43139 units For Q(t) = 0.1Q0 We calculate t as, 0.1Q0 = Q0 ( 1.1 ^ t ) thus t = ln (0.1) / ln(1.1) = -24.15885 units Thus all values of t between -31.43139 and -24.15885, like -30 , -31 satisfies the function for values between 0.05Q0 and 0.1Q0 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q For what values of t did Q(t) lie between .005 Q0 and .01 Q0? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the value of Q(t) = 0.005Q0 , the t required to reach this quantity will be 0.005Q0 = Q0*(1.1^t) , t = ln(0.005) / ln(1.1) = -55.590256 units. For the value of Q(t) = 0.01Q0, the t required to reach this quantity will be 0.01Q0 = Q0*(1.1^t) , t = ln(0.01) / ln(1.1) = -48.317715 units Thus for all t values between - 55.590256 and - 48.317715 Q(t) will lie between 0.005Q0 and 0.01Q0. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q explain why the negative t axis is a horizontal asymptote for this function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As we see that the equation has a growth factor of greater than 1. Thus for decreasing quantities we need a negative time. If we go plotting the points on the graph with time on the x axis and quantity on the y axis we see that the initial point is plotted on the y axis. Now the graph would be a decreasing one for negative times as time decreases. The quantity decreases but at a decreasing rate, thus the graph for negative time seems to decrease by approaching zero. More technically speaking we see that the quantity equation has 1 / ( 1.1^t) factor , which with increasing t approaches 0. It takes the value of 0 when t reaches - infinity. Thus the negative x - axis is the asymptote for this function confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given y = 12*(e^(-.5x)) = 12*(e^(-0.5)^(x)) = A*(b^x) Comparing the 2 equation we see that A = 12 and b = e^(-0.5) = 0.6065 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx. So this function is of the form y = A b^x for b = .61 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q what is b for the function y = .007 ( e^(.71 x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given y = 0.007( e^(0.71 x)) = 0.007(e^(0.71)^(x)) = A b^x Comparing the 2 equations we get 0.007 = A and b = e^(0.71) = 2.03399 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q what is b for the function y = -13 ( e^(3.9 x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given y = -13( e^(3.9x)) = -13( e^(3.9)^x ) = A b^x Comparing the 2 equations we get A = -13 and b = e^3.9 = 49.402449 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** -13 e^(3.9 x) = -13 (e^3.9)^x = -13 * 49.4^x, approx. So this function is of the form y = A b^x for b = 49.4 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q List these functions, each in the form y = A b^x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The functions obtained are y =12*(0.6065 ^ x) y = 0.007*(2.03399 ^ x) y = -13*(49.402449 ^ x) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q query text problem 1.1.31 5th; 1.1.23 4th dolphin energy prop cube of vel YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.1.31 and 1.1.23 submitted in the query text assignment. Given that the energy expended by a swimming dolphin = S The velocity of the dolphin = v Since energy is directly proportional to the energy thus we know S = k(v^3) , where k is the constant of proportionality. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** A proportionality to the cube would be E = k v^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q query text problem 1.1.37 5th; 1.1.32 4th temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Text problem 1.1.37 and 1.1.32 submitted using the text query assignment. The function given H = f(t) The meaning of H(30) = 10 means that after 30 mins the temperature is 10 C The vertical intercept that is the y axis intercept is the initial temperature of the body at time 0 The horizontal intercept that is the x axis intercept is the time in min that the body will take to reach temperature of 0 C. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q What is the meaning of the vertical intercept? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The vertical intercept that is the y axis intercept is the initial temperature of the body at time 0 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** This is the value of H when t = 0--i.e., the temperature at clock time 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q What is the meaning of the horizontal intercept? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The horizontal intercept that is the x axis intercept is the time in min that the body will take to reach temperature of 0 C. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** This is the t value when H = 0--the clock time when temperature reaches 0 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.1.40 and 1.1.31 submitted using text query entry. !!!!!! Note- for the last part of the question given, where is the question I could not find it in the text book , the below solution I have given is based upon my interpretation of the question by viewing the solution before handed, please try to specify where the problem is or what the problem is!!!!!! Given points ( 0 , 32 ) and ( 100 , 212 ) when plotted on a x - y graph with temperature in C on the x axis and temperature in F on the y axis. Rise of the line between the points = 212 - 32 = 180 Run for the line = 100 - 0 = 100 Thus slope for the line = rise / run = 180 / 100 = 1.8 Since it passes through ( 0 , 32 ) thus c ( y intercept ) = 32 Thus the equation of the line will be y = 1.8x + 32 where x is temperature in Celsius and y temperature in Fahrenheit. Fahrenheit corresponding to 20 C is calculated by substituting x for 20 in the equation obtained above. y = 1.8*20 + 32 = 68 F in order to find the temperature common to both , we substitute y = x thus , x = 1.8*x + 32 , -0.8*x = 32 , x = - 32 / 0.8 = - 40 thus - 40 F = - 40 C confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!