course Mth 173 8/11 10 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: `aThe 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). 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. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x) = (t-4) / (t + 4) f `(x) = (t-4)(1) – (t+4) (1) / (t+4)^2 = 8/ (t+4)^2 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a*&*& 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: g(v) = 1 / f(v). confidence rating: OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat are the meanings of f ' (80) and f(80)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(80) is just the rate of consumption f `(80) is the rate at which the consumption is changing with respect to velocity. confidence rating: OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `af ' (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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat are g(80) and g'(80) and how do we interpret g ' (80)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f ' (80) = -.0005 / (.05)^2 = -.2 As you increase your speed, the liter drops. confidence rating: OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qWhat 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: h(v) = v * f(v) .05 * 80 = 4 h ' (80) = f(80) + 80 f ' (80) = .05 + 80 * .0005 = .09 The amount of the liter used is increasing at a rate of .09 at the speed of 80. confidence rating: OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf 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. ** Student Comment: I still did not understand. The following explanation might help (click on Enter Answer to get the subsequent additional explanation) ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.