course MTH 173 7/4 5:20p If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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: Slope is `dy/`dt, which is the change in depth divided by the change in time, giving an average depth change per time interval. 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: Since in this example we are measuring a rate vs. time instead of a depth quantity vs. time, we're getting a rate change of a rate, or a rate of change of a velocity over a time. A rate multiplied by a time interval gives a total quantity of the rate change. Getting the average altitude of the trapezoid and multiplying it by the width gives the area, which in this example is the same as rate * time gives total quantity over a time. 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: Q(t) = 550(1+-.11)^t Q(t)=550(0.89)^2 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: Finding the half life of a material from any starting point is as easy as drawing a vertical line to the x axis at the chosen y (coordinate of the starting material size/amount), then going to the y coordinate of the number that is half of the starting point and draw another vertical line to the x axis. The run from the first x to the second (x2 - x1) will be the time for the material to degrade to 1/2 it's starting size or it's ""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: Q(doublinetime) = 550 * .89^doublingtime = 275, when divided by 550 gives .89^doublingtime = .5. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `q For what values of t did Q(t) lie between .005 Q0 and .01 Q0? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve for .005Q0, plug this into the equation. .005 Q0 = Q0 (1.1^ t), divide both sides by Q0 giving .005 = 1.1^t. Same for .01, giving .01 = 1.1^t. Solving for t in these two equations, you will have two amounts that anything in between will answer the problem. Confidence rating: 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I can't seem to find how to do negative exponents on my calculator so couldn't solve.
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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: y = 12( (e^-.5)^x) = 12 (1/e^.5)^x = 12(.61)^x This is in the form y = a b^x, so b = .61 approx. Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: y = .007(e^.71)^x = .007(2.03)^x This fits the form of y = a b^x, meaning b = 2.03 approx. Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: y = -13 (e^3.9)^x, y= -13 (49.40)^x, which is of the form y=a b^x, meaning b = 49.40 approx. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: y = 12 * .61^x y = .007 * 2.03^x y = -13 * 49.40^x Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #23 dolphin energy prop cube of vel YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: E = k v^3. 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 #32 temperature function H = f(t), meaning of H(30)=10, interpret vertical and horizontal intercepts YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f or H(30) = 10 means that at time t = 30 minutes temperature of the object is 10 degrees celsius. The vertical intercept represents the object temperature at clock time = 0 seconds, while the horizontal intercept represents the clock time when the temperature of the object reaches 0 degrees celsius. 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 initial temperature of the object at clock 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 clock time when the temp. of the object reaches 0 degrees celsius. 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 1.1.31. Water freezes 0 C, 32 F; boils 100 C, 212 F. Give your solution to problem 1.1.32. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope of the graph of F vs. C is 1.8. The equation of the line is y = 1.8x + 32. At 20 degrees C the temp is 68 degrees F. No temp. can be the same degrees in C and F. Confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn't think about this algebraically and try to solve it, I just assumed that there couldn't be a temperature that would be the same on both scales. Self-critique Rating: OK "