query 18

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course Mth 173

7/3 11:02 pm

018. `query 18

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Question: `qQuery 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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Your solution:

The figure is two rectangles, one with an area of f(x) * g(x) and the other f(x + dx) * g(x + dx)

The difference between the areas is a purple shaded region represented by three regions, f(x) * [ g(x + dx) - g(x) ], g(x) * [ f(x + dx) - f(x) ] and [ f(x + dx) - f(x) ] * [ g(x + dx) - g(x) ].

We approximate g(x + dx) as g ' (x) * dx, and we 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 a total area of f(x) g ' (x) dx + g(x) f ' (x) dx + f ' (x) g ' (x) dx^2.

The average rate of change of the product is:

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 goes to 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).

confidence rating #$&*:3

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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.

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Self-critique (if necessary):

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Self-critique Rating:

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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?

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Your solution:

F = (t-4)

G= (t+ 4)

(f/g)’ = f’g - fg’ / g^2

= (t-4)’ (t+4) - (t-4) (t+4)’ / (t+4)^2

= (1)(t+4) - (t-4)(1) / (t^2 + 8t + 16)

=(t+4) - (t-4) / (t^2 + 8t + 16)

= 8 / ((t+4)^2)

confidence rating #$&*:1

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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

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Self-critique (if necessary):

OK

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Self-critique Rating: OK

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Question: `qQuery problem 3.3.54 was 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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Your solution:

F(v) = gas consumption in liters/ km at a velocity of (km/ hr)

F(80) = .05 means at 80 km/hr the gas consumption is .05 liters/ km

F’(80) = -.0005 would be the rate of gas consumption with resepct to velocity

So at 80 km/ hr the consumption rate of gas is liters/ km is -.0005 liters/ km

confidence rating #$&*:1

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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. **

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Self-critique (if necessary):

I understand the first part

??????But what does f ' (80) = -.0005 mean?

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Self-critique Rating:3

@&

f ' (80) is the rate of change of gas consumption with respect to velocity.

f ' (80) = -.0005 means that when moving at 80 km/r, consumption is changing at an instantaneous rate of -.0005 liters / km per km/hr of velocity increase.

*@

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Question: `qWhat are the meanings of f ' (80) and f(80)?

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Your solution:

f ' (80) would the rate at which the consumption is changing

f(80) means that for every increase of 1 km / hr in velocity, consumption per km goes up by .0005 liter/km

confidence rating #$&*:1

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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 represent 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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Self-critique (if necessary):

Ok

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Self-critique Rating: OK

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Question: `qWhat are g(80) and g'(80) and how do we interpret g ' (80)?

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Your solution:

G(80) means at a velocity of 80 km/hr it takes 20 km to use a liter of gas

G’(80) means the rate at which the amount in km/ liter is changing at 80 km/ hr

We interpret g ‘ (80) as the slope of the g(x) function

confidence rating #$&*:1

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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.

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Self-critique (if necessary):

I did not find the numerical solution but understand the calculations and why you preformed them for this problem

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Self-critique Rating:

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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?

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Your solution:

H(v) would relate f(x) and g(x)

g(x) = 1/ f(x)

since f(80) = .05 liter/ km

.05 liter/ km * 80 km/ hour

= 4 liter/ hour

H(80) = 4 liter/ hour

H’(80) = the rate that this value is changing

confidence rating #$&*:

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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) = .01.

Interpretation: At 80 km / hr, the number of liters used per hour is increasing by .01 for every km/hr of speed increase.

STUDENT COMMENT:

INSTRUCTOR RESPONSE:

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Self-critique (if necessary):

The calculation for h’ (80) was very difficult for me to understand but I have reviewed it a few times and that has helped

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Self-critique Rating: 3

@&

That calculation is a straightforward application of the chain rule. Once you're comfortable with the chain rule the calculation won't be a problem.

Knowing to do the calculation, on the other hand, is somewhat challenging.

*@

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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.

The functions and derivatives of those functions were difficult. I understand the concepts but trying to apply them proved more difficult than I thought it would be.

"

Self-critique (if necessary):

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Self-critique rating:

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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.

The functions and derivatives of those functions were difficult. I understand the concepts but trying to apply them proved more difficult than I thought it would be.

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Good responses. See my notes and let me know if you have questions. &#