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17:19:35 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 percentage by which the function increases on the Y axis. growth factor is the actual number by which the function uses to calculate the increase. if the rate is 1% or .01, it cannot be the actual multiple of the function because that means 1/100th of the number we want to be increased. so the factor adds the actual quantity present to that number to include the most recent amount thats already there.so the factor is 1.01 in this case
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17:20:28 ** 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. **......!!!!!!!!...................................
RESPONSE --> i wish i was as concise as you..........
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18:27:35 ** 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. **......!!!!!!!!...................................
RESPONSE --> ok. basically, the trapezoid's top line IS the function. all the shape is, is an illustration of the steps performed to get the average slope, in relation to the axis that it is connecting to the function.
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19:18:06 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 quantity is is the average rate x the time it has to travel that rate, just like the trapezoid. only it is a visual representation of the travel, or distance x the average height, or 2-dimensional quantity.
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19:18:15 **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. **......!!!!!!!!...................................
RESPONSE --> ok.
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19:20:00 #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 --> y=550(1.11^x)
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19:21:27 ** 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 **......!!!!!!!!...................................
RESPONSE --> oooohhhh crap. yeah, i see. i rushed it and forgot it was decreasing/not increasing.
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19:26:10 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> 550(.89)^x 15:00-10:00=5 hrs. 550(.89)^5=y 307.1233=y
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19:26:21 ** 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 **......!!!!!!!!...................................
RESPONSE --> yup.
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19:34:03 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> the graph is a swooping curve that continiously gets closer to the axis, but never will touch it. the function of the half-life was decreasing over an interval, just like the amount of decomposition was decreasing over time.
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19:34:35 ** 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. **......!!!!!!!!...................................
RESPONSE --> ok.
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19:40:52 What is the equation to find the half-life? What is its most simplified form?
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RESPONSE --> it is y @next = b((y @last +r) ^x )
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19:41:38 ** 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). **......!!!!!!!!...................................
RESPONSE --> oh. ok.
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20:01:05 #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?......!!!!!!!!...................................
RESPONSE --> uuhh....05 or .005? ok, say .05 so. .05
(-.9,.4241) um, isnt there an infinate# of these decimals?.................................................
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20:02:21 ** 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). **......!!!!!!!!...................................
RESPONSE --> uuuhhhhhhh.......????????
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20:04:32 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> because a neg # is needed to fit the interval, hence the (-) axis
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20:07:11 ** 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 --> ok.
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20:14:41 #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 --> um, the value of b would be the the growth factor. b is the nat. log.
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20:16:28 ** 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.. **......!!!!!!!!...................................
RESPONSE --> so thats what u wanted me to do.
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20:24:12 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> y = .007 ( 2.034^ x)
b=2.034.................................................
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20:25:06 ** 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.. **......!!!!!!!!...................................
RESPONSE --> ok.
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20:26:15 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> 49.4024 says my ti92plus
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20:26:27 ** 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.. **......!!!!!!!!...................................
RESPONSE --> ok
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20:31:33 List these functions, each in the form y = A b^x.
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RESPONSE --> y=12 * 49.4^x y=12 * 2.04^x y=12 * .61^x
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20:32:41 ** The functions are
y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) **......!!!!!!!!...................................
RESPONSE --> ok
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20:35:01 query text problem 1.1 #24 dolphin energy prop cube of vel
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RESPONSE --> whhhhhaaaaatttttt???dolphin cube something.....?????
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20:36:04 ** A proportionality to the cube would be E = k v^3. **
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RESPONSE --> the cube of the dolphin, i presume???
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20:38:49 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 --> ok ill quit before the puter forgets what its doing...
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