#$&* course mth173 018. `query 18
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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): I understand this way a lot better than previous ways it has been explained, the rectangle example made a lot of sense, the only part I had a hard time explaining through geometry is the division of 'dx for the average. ------------------------------------------------ Self-critique Rating: 2 ********************************************* 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 / g is (f ' g - g ' f) / g^2 f(x)= t-4 g(x) = t + 4 f'(x)=1 g'(x)=1 t+4 -(t-4) / (t+4)^2 t + 4 - t + 4 8/(t+4)^2 (t+4)(t+4) t^2 + 4t + 4t + 16 t^2 + 8t + 16 8/t^2+8t+16 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): DER, what does that stand for?
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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(v) is gas consumption in liters/km of a car at velocity v (km/hr); so f(80) represents the gas consumption in liters/km given a velocity of 80km/hr this means that f'(80) represents the current change at the instant f(80) and can be used to guess the values around it. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `af ' (80) is the rate at which the consumption rate per km is changing with respecting velocity. It would be the slope of the f(v) vs. v graph at v = 80. The rise on a graph of f(v) vs. v would represent the change in the number of liter / km -- in this case going faster increases the amount of fuel used per kilometer. The run would repesent the change in the velocity. So the slope represents change in liter/km / (change in speed in km / hr). f'(80) = .0005 means that for an increase of 1 km / hr in speed, fuel consumption per km goes up by .0005 liter/km. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat are g(80) and g'(80) and how do we interpret g ' (80)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: g(v) which represents the distance this car goes on one liter at velocity v g(80) at 80km/hr the car is going 20km/liter g'(80) means g(v) = 1/f(v) (g(x)f'(x)-f(x)g'(x))/g(x)^2 f(x)=1 g(x)= f(v) (f(v) * 0 -f'(v) )/ f(v) ^2 g ' (v) = - f ' (v) / (f(v))^2 .0005/.0025 g'(80) = .2 g'(80) is the rate of change at the instant 20km/hr, meaning ABOUT this much change will be around v=80. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 at .2 km for every km/hr of additional speed. STUDENT QUESTION I got f ' (80) and f(80) but I I don’t understand how we found the derivative. From before we know that f(80) = .05 and f ' (80) = -.0005. Since g(v) = 1 / f(v), it follows that g ' (v) = (1 / f(v) ) '. By the quotient rule (1 / f(v) ) ' = ( 1 ' * f(v) - 1 * f ' (v) ) / (f(v))^2 = (0 - f ' (v) ) / (f(v))^2 = f ' (v) / (f(v))^2. Thus g'(v) = - f ' (v) / (f(v))^2. Thus g ' (80) = -f ' (80) / (f(80))^2 = -.0005 / (.05)^2, etc.. STUDENT QUESTION Looking at the problem I could see the derivative of g(x) = -.2. I am not sure really why this made sense but it did. I’m still a little of unsure how you set this problem up using the quotient rule, I mean I see how it works and understand solution but do not understand how you know to set this equation up to find the derivative. INSTRUCTOR RESPONSE The distance you go per liter, in km/liter, is clearly the reciprocal of the number of liters you need per km. So g(v) = 1 / f(v), and of course f(v) = 1 / g(v). Using g(v) = 1 / f(v), we take the derivative to get g ' (v) = - f ' (v) / (f(v))^2. Since we know f ' (80) and f(80) we can therefore find g(80). It's actually fairly straightforward until we start asking what these quantities mean and why all this makes sense. However you seem to have done well with that part. (details of quotient rule: ((1') * f + 1 * f ') / f^2 = (0 * f + f ') / f^2. Might be a little confusing because the statement of the quotient rule is (f / g) ' = (f ' g - f g ') / g^2. In this problem the numerator is 1, taking the place of the f function in the rule, and the denominator is f, which takes the place of the g function. In other words the f function of the quotient rule is 1 and the g function of the quotient rule is our f function.) Let me know if this doesn't make sense. This problem is a great exercise in interpretation and it's worth understanding as well as possible. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): i got .2 instead of -.2 it states f ' (80) = -.0005 and using the quotient rule i come up with g ' (v) = - f ' (v) / (f(v))^2. giving g'(v) = -f'(80).... giving - (-.0005) or g'(v)= -(-.0005)/(.05)^2 .2 ------------------------------------------------ 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) is liters/hr f(v) gives liters/km v is km/hr so if f(80) = .05 that is .05liters/km traveling at 80km/hr so to know how many liters in that hour, multiply f(v) by v. .05 * 80. 4liters/hr h(v)= f(v) * v ????I had trouble discerning how h'(v) would be. I was thinking maybe v would be considered a constant, belocity being km/hr, but also recall velocity itself being a derivative when we worked with velocity/acceleration before. So im treating it as a constant but I wish for some further explaining on how this works. h'(v)=f(v) + vf'(v) ????I remember going over when you have a f(x) of a function, and taking the derivative of it, it became xf'(x), but It is still giving me some confusion, I wish for further detail with this as well h'(80)=.05 + 80 * -.0005 h'(80)=-05 - .04 h'(80)= .01 h(80)=4 h'(80) means at the instant where velocity is 80km/hr and it is 4liters/hr, the rate is changing at .01. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 during that hour. Thus we can get the number of liters used per hour by multiplying .05 liters / km by 80 km / hr, obtaining 4 liters / hour. Thus h(80) = 4, representing 4 liters / hour. We calculated our consumption rate by multiplying f(v), which is our number of liters used per kilometer, by our velocity. That is we multiplied v by f(v) to get the number of liters per hour. Thus if v is the speed in km / hr and f(v) the number of liters used per km, multiplying v * f(v) gives us the number of liters / km * km / hour = liters / hour, the number of liters used per hour. So h(v) = v * f(v). We need to calculate h ' (v) h(v) is the product of two functions, so we calculate our result using the product rule. h ' (v) = ( v * f(v) ) ' = v ' f(v) + v f ' (v) . The derivative is with respect to v so v ' = 1. We conclude that h ' (v) = ( v * f(v) ) ' = v ' f(v) + v f ' (v) = 1 * f(v) + v * f ' (v) = f(v) + v f ' (v). Thus 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. STUDENT COMMENT: INSTRUCTOR RESPONSE: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): it seems i am interpreting f'(v) differently, causing anything using f'(v) to vary from yours, when your values are .0005 mine are that *-1, and vice versa. ------------------------------------------------ Self-critique Rating: