course MTH 271 9/18,1800 assignment #004vvvv
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17:03:23 `questionNumber 40000 Query class notes #04 explain how we can prove that the rate-of-depth-change function for depth function y = a t^2 + b t + c is y' = 2 a t + b
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RESPONSE --> The rate of depth change is equal to the change of depth divided by the change in time. Given two points, this rate of depth change is equal to the slope of the line between them. Slope is equal the change in y-values divided by the change in x-values which is the same equation as our rate of depth change equation. Our two points are (t, y(t)) and ( t + `dt, y(t + `dt)). The rate of depth change (and the slope) is equal to y(t +`dt) - y(t) / (t + `dt) - t. We know that y(t)=at^2 + bt + c and substituting (t + `dt) into that equation, we can find y(t +`dt). We find (2at`dt + a`dt + b`dt)/(t + `dt -t). After factoring, our equation becomes 2at + a`dt + b. Since we want to know the value of our function as `dt nears 0, we can substitute 0 for `dt. Thus, our equation for y`=2at + b. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:04:07 `questionNumber 40000 explain how we know that the depth function for rate-of-depth-change function y' = m t + b must be y = 1/2 m t^2 + b t + c, for some constant c, and explain the significance of the constant c.
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RESPONSE --> Since y'=2at + b and y'=mt + b, we can see from these equations that m=2a m/2=a. So, substituting m into the quadratic for a, we must use m/2 or 1/2 m. Significance of c? The constant c is of no significance and is largely ignored when solving these equations. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:07:31 `questionNumber 40000 Explain why, given only the rate-of-depth-change function y' and a time interval, we can determine only the change in depth and not the actual depth at any time, whereas if we know the depth function y we can determine the rate-of-depth-change function y' for any time.
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RESPONSE --> To determine an actual depth, we must have an actual time and then, plug that into the depth function y(t). Having y' and a time interval, we do not have an exact time, only the change of time. Therefore, it makes sense that we can only find the change of depth and not the exact depth. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:07:56 `questionNumber 40000 In terms of the depth model explain the processes of differentiation and integration.
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RESPONSE --> Differentiation involves taking the derivative of a function. y=at^2 + bt + c is the model for depth function. By taking the derivative we obtain y'=2at + b which gives the rate of depth change. Integration is the process of turning the derivative of a function back into the original function. By taking the anti-derivative of y'=2at + b (the rate of depth change), we obtain the original function, y=at^2 + bt + c. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:10:09 `questionNumber 40000 query modeling project #2 #5. $200 init investment at 10%. What are the growth rate and growth factor for this function? How long does it take the principle to double? At what time does the principle first reach $300?
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RESPONSE --> Growth Rate=10%=.10 Growth Factor=1.1 How long does it take the principle to double? P(t)=200(1.1)^t P(t)=400=200(1.1)^t 2=1.1^t t=(log2)/(log1.1) t=7.3 years At what time does the principle first reach $300? P(t)=200(1.1)^t P(t)=300=200(1.1)^t 3/2=1.1^t t=(log 1.5)/(log 1.1) t=4.3 years confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:13:22 `questionNumber 40000 At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?
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RESPONSE --> P(t)=200(1.1)^t P(20)=200(1.1)^20 P(20)=1345.50 Half of t=20 value=1345.50/2=672.75 The doubling time for the principle $200 is the same, 7.3 yrs. The year value does not matter. confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:14:38 `questionNumber 40000 query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40%
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RESPONSE --> (x values=t, y values=P) r=10%=.1 P=1(1.1)^t (0,1.0) (1,1.10) (2,1.21) (3,1.33) (4,1.46) r=20%=.2 P=1(1.2)^t (0,1.0) (1,1.2) (2,1.44) (3,1.73) (4,2.07) r=30%=.3 P=1(1.3)^t (0,1.0) (1,1.3) (2,1.69) (3,2.20) (4,2.86) r=40%=.4 P=1(1.4)^t (0,1) (1,1.4) (2,1.96) (3,2.74) (4,3.84) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:15:31 `questionNumber 40000 query #11. equation for doubling time
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RESPONSE --> P0(1+r)^t=2P0 (1+r)^t=2 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:16:45 `questionNumber 40000 Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%.
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RESPONSE --> P0(1+r)^t=2P0 5000(1+.08)^t=2(5000) 1.08^t=2 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:17:23 `questionNumber 40000 Desribe how on your graph how you obtained an estimate of the doubling time.
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RESPONSE --> On a graph, to determine the doubling time, move up the y-axis until you are at double your original y value. You then find this point on your graphed curve. From the point on the curve, you go directly down to the x-axis where time is represented. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:21:33 `questionNumber 40000 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> Q(t)=550mg(.89)^t At 3pm, five hours have passed so t=5. Q(t)=550(.89)^5 =307.12mg confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:27:22 `questionNumber 40000 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> The graph is a smooth curve that starts high on the y-axis. As the x-axis value increases, the y values decrease. The initial drop in y value is large and becomes smaller and smaller. To find the half-life, find the value corresponding to half the original amount (275mg). You then locate a y-value of 275 on the curve and drop down to the x-axis to estimate the x value (time). This is approximately 6 years. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:29:28 `questionNumber 40000 What is the equation to find the half-life? What is the most simplified form of this equation?
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RESPONSE --> Q0/2=Q0(1+r)t 1/2=(1+r)^t confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:34:17 `questionNumber 40000 #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 --> Q(t)=Q0(1.1)^t .05
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17:35:52 `questionNumber 40000 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> Since the t values are negative, the are graphed on the negative t axis (aka negative x axis). This is a horizontal asymptote for this function because the y values are constantly becoming nearer and nearer to 0 without actually becoming 0. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:37:48 `questionNumber 40000 #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 --> b=2^k y= 12 (e^(-.5x) ) b=e^-.5 b=.61 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:38:41 `questionNumber 40000 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> b=e^.71 b=2.03 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:39:09 `questionNumber 40000 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> b=e^3.9 b=49.40 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:42:37 `questionNumber 40000 0.4.44 (was 0.4.40 find all real zeros of x^2+5x+6
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RESPONSE --> x^2 + 5x + 6 (x+3)(x+2) x+3=0 x=-3 x+2=0 x=-2 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:44:57 `questionNumber 40000 Explain how these zeros would appear on the graph of this function.
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RESPONSE --> These zeros are the points where the function actually crosses the x-axis. In this instance, the graphed function crosses the x-axis at (-3,0) and (-2,0). The y-values are 0 because the function is crossing the x-axis. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:49:31 `questionNumber 40000 0.4.50 (was 0.4.46 x^4-625=0
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RESPONSE --> x^4-625 First common multiple: 5 (x-5)(x^3 + 5x^2 + 25x +125) Second common multiple:-5 (x-5)(x+5)(x^2+25) x-5=0 x=5 x+5=0 x=-5 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:56:43 `questionNumber 40000 0.4.70 (was 0.4.66 P = -200x^2 + 2000x -3800. Find the x interval for which profit is >1000
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RESPONSE --> P=-200x^2 + 2000x -3800 P>1000 1000=-200x^2 + 2000x -3800 0=-200x^2 + 2000x -4800 0=-200(x^2 -10x +24) 0=-200(x-6)(x-4) x-6=0 x=6 x-4=0 x=4 This provides three intervals: (-infinity,4), (4,6) ,(6,infinity) We must test a sample value for each interval. If the value is negative, it is an undefined interval. Values less than 4 are undefined. Values greater than 6 are also undefined. The proper interval is therefore 4
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