assignment 26  27

......!!!!!!!!...................................Applied Calculus II

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** 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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Good, but it would be the half plane above this line. Note the notation y > -x. ** 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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** The z = c level curve of e^(xy) occurs when e^(xy) = c.

We solve e^(xy) = c for y in terms of x. We first take the natural log of both sides:

ln(e^(xy)) = ln(c), or

xy = ln(c). We then divide both sides by x:

y = ln(c) / x.

For c = 1 we get y = ln(1) / x = 0 / x = 0. Thus the c = 1 level curve is the x axis y = 0.

For c = 2 we get y = ln(2) / x = .7 / x, approximately. This curve passes through the points (1,.7) and (-1, -.7), and is asymptotic to both the x and y axes.

For c = 3 we get y = ln(3) / x = 1.1 / x, approximately. This curve passes through the points (1,1.1) and (-1, -1.1), and is asymptotic to both the x and y axes.

For c = 4 we get y = ln(4) / x = 1.39 / x, approximately. This curve passes through the points (1,1.39) and (-1, -1.39), and is asymptotic to both the x and y axes.

For c = 1/2 we get y = ln(1/2) / x = -.7 / x, approximately. This curve passes through the points (-1,.7) and -1, -.7), and is asymptotic to both the x and y axes.

For c = 1/3 we get y = ln(1/3) / x = -1.1 / x, approximately. This curve passes through the points (-1,1.1) and -1, -1.1), and is asymptotic to both the x and y axes.

For c = 1/4 we get y = ln(1/4) / x = -1.39 / x, approximately. This curve passes through the points (-1,1.39) and -1, -1.39), and is asymptotic to both the x and y axes.

The c = 2, 3, 4 level curves form similar hyperbolas in the first and third quadrant which progressively 'bunch up' closer and closer together. Similar behavior is observed for the c = 1/2, 1/3, 1/4 hyperbolas, which occur in the second and fourth quadrants.

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Your work has not been reviewed.

Please notify your instructor of the error, using the Submit Work form, and be sure to include the date 07-31-2006.

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