26-26 open query

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course Mth 272

11/16 7 pm EST

025.

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Question: `qQuery problem 7.3.14 f(x+`dx,y) and [ f(x, y+`dy) - f(x,y) ] / `dy.

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Your solution:

So, let us use f(x + ‘dx, y) = f (3x + x^2 + 4y)

= [pz][px] = 3 + 2x + 4y

F(x, y + ‘dy) – f(x,y) / ‘dy

= 3 + 2x + 4

Not sure exactly what is sought from the problem.

Does it correlate with the following?

f(x + ‘dx, y) – f(x,y) / ‘dx

f(x,y + ‘dy) – f(x,y) / ‘dy

Example:

z = 3x –x^2y^2 + 2x^3y

pz/px = 3 – 2xy^2 + 6x^2y

pz/py = -2x^2y + 2x^3

confidence rating #$&*: 2

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Give the expressions for f(x+`dx,y) and [ f(x, y+`dy) - f(x,y) ] / `dy.

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Your solution:

So, let us use f(x + ‘dx, y) = f (3x + x^2 + 4y)

= [pz][px] = 3 + 2x + 4y

F(x, y + ‘dy) – f(x,y) / ‘dy

= 3 + 2x + 4

Not sure exactly what is sought from the problem.

Does it correlate with the following?

f(x + ‘dx, y) – f(x,y) / ‘dx

f(x,y + ‘dy) – f(x,y) / ‘dy

Example:

z = 3x –x^2y^2 + 2x^3y

pz/px = 3 – 2xy^2 + 6x^2y

pz/py = -2x^2y + 2x^3

confidence rating #$&*: 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a If f(x,y) = 3 x y + y^2 then

[ f(x, y+`dy) - f(x,y) ] / `dy = [ 3 x (y + `dy) + (y + `dy)^2 - ( 3 x y + y^2) ] / `dy.

Simplifying the numerator we get

[ 3 x y + 3 x `dy + y^2 + 2 y `dy + `dy^2 - 3 x y - y^2 ] / `dy, or

[ 3 x `dy + 2 y `dy + `dy^2 ] / `dy.

Dividing the terms of the numerator by the denominator we have

3 x + 2 y + `dy.

Interpretation:

The expression 3 x + 2 y + `dy represents the change in f due to a small change `dy in the y value, divided by the change in the y value. This is the average rate of change of f with respect to y, over the y interval from y to `dy. This is therefore the average value of the partial derivative with respect to y.

As `dy -> 0 this expression gives us 3 x + 2 y.

The numerator of the expression [ f(x, y+`dy) - f(x,y) ] / `dy is the difference in f near the point (x, y) due to a change `dy in y, divided by that change `dy. That is, [ f(x, y+`dy) - f(x,y) ] / `dy is the average rate at which f(x, y) changes near (x,y) with respect to a change in y.

The limiting value 3 x + 2 y of this expression is the rate at which the function changes with respect to a change in y. This is the definition of the partial derivative of the function with respect to y.

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Self-critique (if necessary):

#### In continuation from main problem presented, I desire to learn the concept presented more clearly. I understand partial differentiation exemplified in the text, pg. 505, etc., and the problems relative to homework, but kind of unclear from the start on what are requested here in the original problem.

First you need to understand the notation:

f(x,y) = 3 x y + y^2

So, for example,

f(a, b) = 2 a b + b^2,

or as another example

f(x + a, y - b) = 2 (x + a) ( y - b) + (y - b)^2.

Continuing in the same way

f(x, y + `dy) = 3 x (y + `dy) + (y + `dy)^2

and, as in the original definition,

f(x, y) = ( 3 x y + y^2)

so that

f(x, y+`dy) - f(x,y) = 3 x (y + `dy) + (y + `dy)^2 - ( 3 x y + y^2) .

It follows that

[ f(x, y+`dy) - f(x,y) ] / `dy = [ 3 x (y + `dy) + (y + `dy)^2 - ( 3 x y + y^2) ] / `dy.

Then you will want to understand the meaning of this expression:

The expression f(x, y + `dy) represents the value of the function if start at point (x, y) in the plane, move a (presumably short) distance `dy in the y direction, and evaluate the function at that point. The value at that point will differ a bit from the value at (x, y).

If we then divide that difference by `dy, we get the approximate rate at which the value of the function is changing, with respect to the value of y, in the neighborhood of (x, y).

The change in the value of the function is f(x, y + `dy) - f(x, y). The change in y is `dy. So the average rate of change of f(x, y) with respect to y is [ f(x, y + `dy) - f(x, y) ] / `dy.

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self-critique rating #$&*: 1

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Question: `qWhat is your interpretation of the expression [ f(x, y+`dy) - f(x,y) ] / `dy, and what is its significance?

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Your solution:

So, let us use f(x + ‘dx, y) = f (3x + x^2 + 4y)

= [pz][px] = 3 + 2x + 4y

F(x, y + ‘dy) – f(x,y) / ‘dy

= 3 + 2x + 4

Not sure exactly what is sought from the problem.

Does it correlate with the following?

f(x + ‘dx, y) – f(x,y) / ‘dx

f(x,y + ‘dy) – f(x,y) / ‘dy

Example:

z = 3x –x^2y^2 + 2x^3y

pz/px = 3 – 2xy^2 + 6x^2y

pz/py = -2x^2y + 2x^3

confidence rating #$&*: 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a Interpretation:

** As `dy -> 0 this expression gives us 2 x + 2 y.

The numerator of the expression [ f(x, y+`dy) - f(x,y) ] / `dy is the difference in f near the point (x, y) due to a change `dy in y, divided by that change `dy. That is, [ f(x, y+`dy) - f(x,y) ] / `dy is the average rate at which f(x, y) changes near (x,y) with respect to a change in y.

The limiting value 2 x + 2 y of this expression is the rate at which the function changes with respect ot a change in y. This is the definition of the partial derivative of the function with respect to y. **

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Self-critique (if necessary):

#### #### In continuation from main problem presented, I desire to learn the concept presented more clearly. I understand partial differentiation exemplified in the text, pg. 505, etc., and the problems relative to homework, but kind of unclear from the start on what are requested here in the original problem.

My preceding note addresses the interpretation as well.

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self-critique rating #$&*:

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Question: `qQuery problem 7.3.18 domain of ln(x+y)

Give the domain of the given function and describe this region in the xy plane.

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Your solution:

The domain of ln x = x > 0

x + y > 0

y = -x

The domain is at all values above line of y = -x.

These values define the x-y plane.

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a The function is defined for all x, y such that x + y > 0, which is equivalent to y = -x. The domain is therefore expressed as the half-plane y > -x. It consists of all points above the line y = -x in the x-y plane.

Since the natural log function can take any value as x + y goes from 0 to infinity, the range of the function is all real numbers.

As we approach the line y = -x from points lying above the line, x + y approaches zero so ln(x+y) approached -infinity. So the surface defined by the function has a rapid dropoff whose depth exceeds all bounds as we approach y = -x.

Since ln(1) = 0, the graph intersects the xy plane where x + y = 1--i.e., on the line y = -x + 1, which lies 1 unit above the line y = -x. Between the line y = -x and y = -x + 1 the graph rises from unbounded negative values to 0.

The graph will reach altitude 1 when ln(x+y) = 1, i.e., when x + y = e. This will occur above the line y = -x + e, approximately y = -x + 2.718.

The graph will reach altitude 2 when ln(x+y) = 2, i.e., when x + y = e^2. This will occur above the line y = -x + e^2, approximately y = -x + 8.2.

The graph will reach altitude 3 when ln(x+y) = 3, i.e., when x + y = e^3. This will occur above the line y = -x + e^3, approximately y = -x + 22.

Note that the distances required to increase by 1 unit in altitude increase by greater and greater increments.

The graph will continue reaching greater and greater altitudes, but the spacing between integer altitudes will continue to spread out and the steepness of the graph will decrease fairly rapidly.

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Self-critique (if necessary):

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self-critique rating #$&*: ‘OK’

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

"

You have some good questions. See my notes and let me know if you have more.

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