course mth 272
......!!!!!!!!...................................Applied Calculus II
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Asst # 26
07-30-2006
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11:42:52
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**** Query problem 7.3.14 f(x+`dx,y) and [ f(x, y+`dy) - f(x,y) ] / `dy for f(x,y) = 3 x y + y^2..
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12:36:38
(a). 3(x+dx)(y) +(y^2) = 3xy +3dxy +y^2
(b). 1, dy can't be 0
** 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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12:36:39
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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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12:41:37
3xy +3dxy+y*^2 and 1, dy cant be 0
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12:41:38
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**** If your expression for [ f(x, y+`dy) - f(x,y) ] / `dy is not simplified, give the simplified expression.
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12:41:59
was simplified.
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12:41:59
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**** What is your interpretation of the expression [ f(x, y+`dy) - f(x,y) ] / `dy, and what is its significance?
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12:42:24
because of the rules of division. dy cannot be 0
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12:42:25
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**** Query problem 7.3.18 domain of ln(x+y)
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12:49:22
the half plane below line y = -x + 0
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12:49:23
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**** Give the domain of the given function and describe this region in the xy plane.
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12:49:26
the half plane below line y = -x + 0
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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12:49:27
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**** Query Add comments on any surprises or insights you experienced as a result of this assignment.
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12:49:34
none.
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Applied Calculus II
Asst # 27
07-30-2006
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13:25:07
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**** Query problem 7.3.38 level curves of e^(xy) for c = 1, 2, 3, 4, 1/2, 1/3, 1 / 4
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14:02:53
a circle, not sure exactly how to determine
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14:02:54
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**** What is the level curve for c = 1? Describe the curve and give the value of e^(xy) at any point on the curve.
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14:03:26
the inner circle has the radius of one
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14:03:26
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**** Give the level curves for the other indicated values of c.
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14:07:00
radius of the others also 2, 3, 4, .5, 1/3, 1.4
** 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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14:07:01
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**** Describe the 'topographic map' you would get from these level curves.
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14:11:10
circles layering around each other.
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14:11:10
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**** If you walked along this 'map' at an angle or 45 degrees to the positive x axis, how would your indicated elevation change? Would your path indicate a hill of increasing, constant, or decreasing steepness, and how can you tell?
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14:19:09
increasing steepness. on the x axis the number are increasing positively
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14:19:09
"
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