course Mth 173 ??????????M?assignment #018
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21:06:05 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 figure used to derive the product rule is a box representing multiples of f(x) and g(x). The largest portion of the box is simply f(x) * g(x). In looking for the derivation of f(x)g(x), we add a little bit ('dx) to both f(x) and g(x), creating a slightly larger box than f(x)g(x). We then look to find the difference between this new box ( f(x+'dx)g(x+'dx) ) and the original box, so we simply subtract the original from the new. This leaves us with a little strip around the outside which can be broken into three little boxes: f(x)*dg, the change in f(x) g(x)*df, the change in g(x) df*dg, the two changes multiplied to form a corner. The two main areas, f(x)*dg and g(x)*df, are added together to find the new area. Since we are looking for the derivation of the original function, we need to divide the entire area found by the change in x, or 'dx, giving us this equation: ( f(x) * 'dg + g(x) * 'df ) / 'dx Now if 'df / 'dx is the same thing as the derivative of f(x) (and the same idea for 'dg), then we can simplify even further. ( f(x) * g' + g(x) * f' ) / 'dx Taking the limit of this leaves us with the product rule: f(x) * g' + g(x) * f' = y'
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21:07:08 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 --> Hmm. You have a lot more stuff there, but I think I got most of it. I missed that 'df * 'dg shrinks to 0 quickly.
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21:08:57 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 --> Multiply two functions together creates a new function which may not necessarily act like the old function. Constants or simply adding two functions together behave in a much more predictable manner.
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21:09:19 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 --> Hmm. Okay.
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21:09:50 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 --> I find the derivative to be: 8 / (t+4)^2
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21:10:05 *&*& 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 --> Ok.
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21:10:39 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 --> I believe: g(v) = 1 / f(v)
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21:11:53 ** 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 --> Ok.
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21:14:40 What are the meanings of f ' (80) and f(80)?
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RESPONSE --> f(80) means that at a velocity of 80 km/hour, the vehicle consumes fuel at a rate of .05 liters per kilometer. f'(80) is the rate at which the fuel consumption changes per km/hr of velocity. At this point, the fuel consumption is changing at a rate of .0005 liters per km/hr.
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21:14:54 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 --> Ok.
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21:24:27 What are g(80) and g'(80) and how do we interpret g ' (80)?
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RESPONSE --> g(80) is a graph representing the distance traveled per liter of fuel for the given velocity. g'(80) is the rate at which distance per liter of fuel is changing. I am not quite sure how to interpret it.
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21:26:43 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 --> Ah. Silly me. That would have helped the interpretation!
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21:49:29 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 --> Since h(v) is in L/hr, then we can multiply together f(v) (in units of L/km) and velocity (in units of km/hr) to find the value. This means: h(80) = f(x)*v = .05 * 80 = 4 L/hr. This means that at velocity of 80 km/hr, the vehicle is consuming 4 liters of fuel per hour. h'(v) = f'v + v'f = (.0005)(80) + (0)(.05) = 0.04 L/hr This means that the rate of fuel consumption is changing by .04 liters per km/hr of speed increase.
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21:57:31 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 --> Oh. I see why v =1. The derivative of v^1 is 1.
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21:57:40 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 -->
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21:57:54 ** 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 --> Ok.
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21:58:15 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Ah, relationships.
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