course mth 173 assignment #018£l±Ý‘ñئ‰’íÀUxÍ´x‚óq ê‹T{©àŹ§
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18:12:16 Query class notes #22. Describe the figure used to derive the product rule and explain how the figure is used in that derivation.
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RESPONSE --> The product rule is f(x) * g(x) = f'g + g'f we use the chain rule to break down the equation into it's seperate functions and then plug these and there dirivatives into the product rule.
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18:20:34 The figure is a pair of nested rectangles, one whose dimensions are f(x) by g(x), the other with dimensions f(x + `dx) by g(x+`dx). The product of the two functions is initially f(x) * g(x), represented in the figure by the area of the smaller rectangle. The product after the change `dx is f(x + `dx) * g(x + `dx), represented by the area of the larger rectangle. The additional area is represented by three regions, one whose dimensions are f(x) * [ g(x + `dx) - g(x) ], another with dimensions g(x) * [ f(x + `dx) - f(x) ] and the third with dimensions [ f(x + `dx) - f(x) ] * [ g(x + `dx) - g(x) ]. We can approximate g(x + `dx) as g ' (x) * `dx, and we can approximate f(x + `dx) - f(x) by f ' (x) `dx, so the areas of the three regions are f(x) * g ' (x) `dx, g(x) * f ' (x) `dx and f ' (x) `dx * g ' (x) `dx, giving us total additional area f(x) g ' (x) `dx + g(x) f ' (x) `dx + f ' (x) g ' (x) `dx^2. This area, recall, represents the change in the product f(x) * g(x) as x changes by `dx. The average rate of change of the product is therefore ave rate = change in product / `dx = [ f(x) g ' (x) `dx + g(x) f ' (x) `dx + f ' (x) g ' (x) `dx^2 ] / `dx = f(x) g ' (x) + g(x) f ' (x) + `dx. As `dx -> 0 the average rate of change f(x) g ' (x) + g(x) f ' (x) + `dx approaches the instantaneous rate of change, which is f(x) g ' (x) + g(x) f ' (x).
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RESPONSE --> The product rule is found by the overlapping rectangles and simplifies to the (f*g)' = f'g + g'f
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18:22:38 Explain in your own words why the derivative of the product of two functions cannot be expected to be equal to the product of the derivatives.
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RESPONSE --> Because according to the overlapping rectangles the two derivatives would a have a rather large area with only a small square of said area (where the to rectangles of f('dx) and g('dx) overlap) which is what we are left with when we just multiply the drivatives.
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18:23:28 The product of the derivatives would involve just the limiting behavior of the small rectangle in the upper-right-hand corner and would not involve the behavior of the elongated rectangles along the sides of the figure.
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RESPONSE --> okay
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18:28:54 Query problem 3.3.16 (3d edition 3.3.14) (formerly 4.3.14) derivative of (t - 4) / (t + 4). What is the derivative of the given function?
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RESPONSE --> y= (t-4) / (t +4) y'= (f'g -g'f) / g^2 y'= ((t * t+4) - ( t * t-4)) / (t^2 + 8t + 16) y'= ((2t+4) - (2t-4)) / (t^2 + 8t + 16) y'= 8 / (t^2 + 8t +16)
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18:30:02 *&*& By the Quotient Rule the derivative of f / g is (f ' g - g ' f) / g^2. For this problem f(t) is the numerator t - 4 and g(t) is the denominator t + 4. We get f ' (t) = 1 and g ' (t) = 1 so that the derivative is [(1(t + 4)) - (1(t - 4))] / [(t + 4)^2] = 8 / ((t + 4)^2) = g'(t) ** DER
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RESPONSE --> we had the same answer except I expanded the denominator.
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18:48:10 Query problem 3.3.56 (3d edition 3.3.47) (was 4.3.40) f(v) is gas consumption in liters/km of a car at velocity v (km/hr); f(80) = .05 and f ' (80) = -.0005. What is the function g(v) which represents the distance this car goes on one liter at velocity v?
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RESPONSE --> It is given that f(80) = .05 and f'(80)= -.0005 meaning that the car gets uses .05 l/km at 80 km/hr and that this is decreasing at .0005 l at that point. So the km/liters is rising and the car is becoming more effcient. The function for the distance that the car goes on 1 liter is: using dimensional analysis we can say that at 80 km/hr you are getting 20km / liter 1/ f(v) = km / liter at a given velocity
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18:48:49 ** This is an interpretation problem. If you graph f(v) vs. v you are graphing the number of liters/km (just like gallon used per mile but measured in metric units). f(80) is the number of liters/km used at a speed of 80 km/hour--for this particular car .05 liters / km. If the car uses .05 liters every kilometer then it takes it 1/ (.05 liters / km) = 20 kilometers /liter: it takes 20 km to use a liter. Generalizing we see that liters / km and km / liter have a reciprocal relationship, so g(v) = 1 / f(v). g(80) represents the number of km traveled on a liter when traveling at 80 km/hour. f(80) = .05 means that at 80 km/hour the car uses .05 liters per kilometer. It therefore travels 20 km on a liter, so g(80) = 20. Using the reciprocal function relationship g(v) = 1 / f(v) we obtain the same result. **
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RESPONSE --> okay
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18:50:28 What are the meanings of f ' (80) and f(80)?
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RESPONSE --> f(80) is how many kilometers you are getting per liter of fuel at 80 km/hr where as f'(80) is the rate of change that your fuel effciency is changing at 80 km/hr
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18:50:52 f ' (80) is the rate at which the consumption rate per km is changing with respecting velocity. It would be the slope of the f(v) vs. v graph at v = 80. The rise on a graph of f(v) vs. v would represent the change in the number of liter / km -- in this case going faster increases the amount of fuel used per kilometer. The run would repesent the change in the velocity. So the slope represents change in liter/km / (change in speed in km / hr). f'(80) = .0005 means that for an increase of 1 km / hr in speed, fuel consumption per km goes up by .0005 liter/km.
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RESPONSE --> okay
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18:52:51 What are g(80) and g'(80) and how do we interpret g ' (80)?
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RESPONSE --> g(80) is the number of kilometers you get per liter of fuel at 80 km/hr and g'(80) is the rate at which the fuel effciency is changing. (GO BACK AND CHANGE PREVOUS ANSWER)
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18:53:26 Since g(v) = 1 / f(v) the quotient rule tells us that g'(v) = - f ' (v) / (f(v))^2. So g ' (80) = -.0005 / (.05)^2 = -.2. Interpretation: At 80 km / hr the number of kilometers driven per liter is dropping by about -.2 for every additional km / hr. As your speed goes up the distance you can drive on a liter goes down.
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RESPONSE --> okay
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19:02:22 What is the function h(v) which gives gas consumption in liters per hour, and what are h(80) and h'(80)? What do these quantities mean?
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RESPONSE --> assuming that you stay at 80 km/hr at 20km / liter you will use about 4 liters of fuel or h(v) = v /g(v) and h'(80) would be the amount that the amount of fuel used/hr is changing.
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19:14:38 If you're using f(80) = .05 liters / km then if you travel 80 km in an hour you will use .05 liters / km * 80 km = 4 liters. Thus h(80) = 4, representing 4 liters / hour. Generalizing we see that if v is the speed in km / hr and f(v) the number of liters used per km, multiplying v * f(v) gives us liter / km * km / hour = liters / hour, the number of liters used per hour. So h(v) = v * f(v). Now h ' (v) = ( v * f(v) ) ' = v ' f(v) + v f ' (v) . The derivative is with respect to v so v ' = 1 and h ' (v) = f(v) + v f ' (v). At v = 80 we have h ' (80) = f(80) + 80 f ' (80) = .05 + 80 * .0005 = .09. Interpretation: At 80 km / hr, the number of liters used per hour is increasing by .09 for every km/hr of speed increase. This would tell any driver with an arithmetic background about how many additional liters will be used per hour for a given speed increase with speeds in the neigborhood of 80 km/hr. **
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RESPONSE --> since g(v)=1/ f(v) then v / g(v) = v * f(v) I see how you evaluated the h' I was having trouble with the v' = 1 but other than that it seems as though you just used the product rule.
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19:14:54 Student Comment: I still did not understand. The following explanation might help (click on Enter Answer to get the subsequent additional explanation)
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RESPONSE --> okay
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19:15:14 ** If you're using f(80) = .05 liters / km then if you travel 80 km in an hour you will use .05 liters / km * 80 km = 4 liters. Thus h(80) = 4, representing 4 liters / hours. Generalizing we see that if v is the speed in km / hr and f(v) the number of liters used per km, multiplying v * f(v) gives us liter / km * km / hour = liters / hour, the number of liters used per hour. So h(v) = v * f(v). Now h ' (v) = ( v * f(v) ) ' = v ' f(v) + v f ' (v) . The derivative is with respect to v so v ' = 1 and h ' (v) = f(v) + v f ' (v). At v = 80 we have h ' (80) = f(80) + 80 f ' (80) = .05 + 80 * .0005 = .09. Interpretation: At 80 km / hr, the number of liters used per hour is increasing by .09 for every km/hr of speed increase. **
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RESPONSE --> okay
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19:17:27 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I think that I understand the product rule: y'= f' g + g' and the quotent rule: y'= (f' g - g' f) / g^2 and how they were derived.
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