Assignment 26

course Mth 272

7/30 2:30 pm

025.

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: To find f(x+`dx,y) and [ f(x, y+`dy) - f(x,y) ] / `dy, we need to use our original equation f(x,y) = 3xy + y^2. So f(x+dx, y) = 3(x+dx)(y) + (y)^2 = (3x+3dx)(y) +y^2 = 3xy + 3ydx + y^2. And [ f(x, y+`dy) - f(x,y) ] / `dy = ((3x(y+dy) + (y+dy)(y+dy)) – (3xy + y^2)) / dy = (3xy +3xdy + y^2 + 2ydy + dy^2 – 3xy – y^2)/dy = (3xdy + 2ydy + dy^2)/dy = 3x + 2y + dy.

confidence rating: 3

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

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

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: Because we know that [ f(x, y+`dy) - f(x,y) ] / `dy = 3x + 2y + dy from our previous solution, we can now explain its significance. We have changed the y component of f(x,y), making the point (x,y) change because of dy. Since our whole equation is also divided by dy, also makes the point (x,y) change as y and dy change.

confidence rating: 2

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

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

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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): I was not sure what interpret meant exactly. I understood the given answer and was trying to express that in my own. I should have taken the limit in my own answer however.

Your statement were very good, as is your self-critique. It seems clear that you knew this, and understand the need to also take the limit.

------------------------------------------------

Self-critique Rating: 3

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: Since f(x,y) = ln(x + y), we know that a log function cannot have a 0 or any negative numbers inside. So we need to understand that x + y must be positive and greater than 0. So x+y > 0. By rearranging this expression, we find that y = -x, which would make our equation ln(0). Because this doesn’t work, we can infer that our domain R is y> -x. Assuming that this is our domain, we can also conclude that the range is all reals, because that makes the inside of ln positive.

confidence rating: 3

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

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

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.

&#Your work looks very good. Let me know if you have any questions. &#

Assignment 26

course Mth 272

7/30 2:30 pm

025.

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: To find f(x+`dx,y) and [ f(x, y+`dy) - f(x,y) ] / `dy, we need to use our original equation f(x,y) = 3xy + y^2. So f(x+dx, y) = 3(x+dx)(y) + (y)^2 = (3x+3dx)(y) +y^2 = 3xy + 3ydx + y^2. And [ f(x, y+`dy) - f(x,y) ] / `dy = ((3x(y+dy) + (y+dy)(y+dy)) – (3xy + y^2)) / dy = (3xy +3xdy + y^2 + 2ydy + dy^2 – 3xy – y^2)/dy = (3xdy + 2ydy + dy^2)/dy = 3x + 2y + dy.

confidence rating: 3

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

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

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: Because we know that [ f(x, y+`dy) - f(x,y) ] / `dy = 3x + 2y + dy from our previous solution, we can now explain its significance. We have changed the y component of f(x,y), making the point (x,y) change because of dy. Since our whole equation is also divided by dy, also makes the point (x,y) change as y and dy change.

confidence rating: 2

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

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

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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): I was not sure what interpret meant exactly. I understood the given answer and was trying to express that in my own. I should have taken the limit in my own answer however.

Your statement were very good, as is your self-critique. It seems clear that you knew this, and understand the need to also take the limit.

------------------------------------------------

Self-critique Rating: 3

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: Since f(x,y) = ln(x + y), we know that a log function cannot have a 0 or any negative numbers inside. So we need to understand that x + y must be positive and greater than 0. So x+y > 0. By rearranging this expression, we find that y = -x, which would make our equation ln(0). Because this doesn’t work, we can infer that our domain R is y> -x. Assuming that this is our domain, we can also conclude that the range is all reals, because that makes the inside of ln positive.

confidence rating: 3

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

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

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.

&#Your work looks very good. Let me know if you have any questions. &#