If your solution to stated problem does not match the given solution, you should self-critique per instructions at

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

qa 11.1

`q001.  If f(x, y) = x^2 / 4 - y^2 / 25, then what curve do you get for each of the following?

• f(x, y) = 0?
• f(x, y) = 1?
• f(x, y) = 4?
• f(x, y) = 9?

Your solution: 

Confidence rating:
 

Given Solution:  For example, the curve f(x, y) = 16 would be

x^2 / 4 - y^2 / 25 = 16,

which is a hyperbola.  Its standard form is

x^2 / 64 - y^2 / 400 = 1.

It is asymptotic to the lines y = 5/2 x and y = -5/2 x.

If x = 0 then y^2 would have to be negative, so there is no vertex on the y axis.

If y = 0 then x = +- 8 so the vertices are (8, 0) and (-8, 0).

Question:  `q002.  Plot each of the curves obtained in the preceding.

Your solution: 

Confidence rating:
 

The hyperbola is asymptotic to the lines y = 5/2 x and y = -5/2 x.

If x = 0 then y^2 would have to be negative, so there is no vertex on the y axis.

If y = 0 then x = +- 8 so the vertices are (8, 0) and (-8, 0).

Question:  `q003.  If f(x, y, z) = x^2 / 4 + y^2 / 25 - z^2 / 16, then what equation corresponds to the condition f(x, y, z) = 9?  What would the plot of this equation look like in 3-dimensional space?

Your solution: 

Confidence rating:
 

f(x, y, z) = x^2 / 2 + y^2 / 25 - z^2 / 16, so the condition f(x,y,z) = 9 gives us

x^2 / 4 + y^2 / 25 - z^2 / 16 = 9.

The xy trace would be x^2 / 4 + y^2 / 25 = 9, an ellipse centered at the origin with vertices at (+-6, 0) and (+- 15, 0).

The x z trace would be x^2 / 4 - z^2 / 16 = 9, a hyperbola asymptotic to the lines z = +- x / 4 with vertices at (+- 6, 0).

The y z trace would be y^2 / 25 - z^2 / 16 = 9, a hyperbola asymptotic to the lines z = +- 5/4 * y with vertices at (+- 15, 0).

If z = c, a constant, we get

x^2 / 4 + y^2 / 25 = 9 + c^2 / 16.

an ellipse in the plane z = c with vertices at (+-2 sqrt(9 + c^2 / 16), 0) and (0, +-5 sqrt(9 + c^2 / 16)).

sqrt(9 + c^2 / 16) increases as c increases, and for large values of c is nearly equal to c / 4.  The ellipses grow accordingly as you move up or down from the y axis.

The parts of the ellipses above and near the x axis, for c = 0, 2, 4, 6, 8, 10, are depicted below:

Question:  `q004.  If f(x, y) = x * y then what curve do you get for each of the following?  Sketch each curve.

• f(x, y) = 0?
• f(x, y) = 1?
• f(x, y) = 4?
• f(x, y) = 9?

Your solution: 

Confidence rating:
 

f(x, y) = c corresponds to x * y = c, so that y = c / x.

For c = 0, all y values are 0 so the graph is just the x axis, excluding the point where x = 0 (at which the function is undefined).

For c = 1 the function is y = 1 / x, the graph of which should be familiar.

For c = 4 the function is y = 4 / x, the graph of which is just the c = 1 graph, stretched vertically by factor 4.

The graphs for c = 0, 1, 4, 9 are depicted below.

Question:  `q005.  If f(x, y, z) = x * y / z then what equation expresses the condition f(x, y, z) = 1?  What would the plot of this equation look like in 3-dimensional space?

Your solution: 

Confidence rating:
 

The equation would be x * y / z = 1.

The equation is undefined if z = 0.  So the graph does not intersect the x-y plane.

If x = 0 or y = 0 the equation cannot be satisfied, since in either case we would get 0 = 36.

If z = c then we have x * y / c = 1, so that

y = c / x.

For positive values of c the graphs in the z = c plane would be the same as in the preceding.  For negative values the graphs would be 'flipped' over the x axis.

The surface in the first octant, and its intersection with the plane z = 1, are depicted below.

Question:  `q006.  Find the x and y derivatives of the following functions:

• f(x, y) = x^2 sqrt(y)
• f(x, y) = cos(x y)
• f(x, y) = e^-(x^2 + y^2)
• f(x, y) = sqrt(x) * e^y
• f(x, y) = cos(x y) + e^(x^2 y)
• f(x, y) = x y / (x^2 + y^2)
• f(x, y) = sqrt(x^2 / 4 - y^2 / 9)
• f(x, y) = y e^(-k x^2) + x y cos(x + y)

Your solution: 

Confidence rating:
 

f(x, y) = x^2 sqrt(y)
f_x(x, y) = 2 x sqrt(y)
f_y(x, y) = x^2 / (2 sqrt(y))

f(x, y) = cos(x y^2)
f_x(x, y) = (x y^2) ' * (-sin(x y^2)), where ' denotes derivative with respect to x, so f_x = - y^2 sin(x y^2)
f_y(x, y) = (x y^2) ' * (-sin(x y^2), where ' denotes derivative with respect to y, so f_y = - 2 x y sin(x y^2)

f(x, y) = e^-(x^2 + y^2)
f_x(x, y) = -(x^2 + y^2) ' e^(x^2 + y^2), where ' denotes derivative with respect to x, so f_x = -2 x e^(-x^2 + y^2)
f_y(x, y) = -(x^2 + y^2) ' e^(x^2 + y^2), where ' denotes derivative with respect to y, so f_y = -2 y e^(-x^2 + y^2)

f(x, y) = sqrt(x) * e^y
f_x(x, y) = 1 / (2 sqrt(x)) * e^y. e^y is a function of y, hence constant when y is held constant, so we treat e^y as a constant when taking the derivative with
respect to x.
f_y(x, y) = sqrt(x) * e^y. sqrt(x) is a function of x, hence constant when x is held constant, so we treat sqrt(x) as a constant when taking the derivative with
respect to y.

f(x, y) = cos(x y) + e^(x^2 y)
f_x(x, y) = y sin(x y) + 2 x y e^(x^2 y)
f_y(x, y) = x sin(x y) + x^2 e^(x^2 y)

f(x, y) = x y / (x^2 + y^2)
f_x(x, y) = ( (x y) ' ( x^2 + y^2) - x y ( x^2 + y^2) ' ) / (x^2 + y^2)^2, where ' denotes derivative with respect to x, so that f_x = ( y ( x^2 + y^2) - x y ( 2
x^) ) / (x^2 + y^2)^2. The numerator simplifies further.
f_y(x, y) = ( (x y) ' ( x^2 + y^2) - x y ( x^2 + y^2) ' ) / (x^2 + y^2)^2, where ' denotes derivative with respect to y, so that f_y = ( x ( x^2 + y^2) - x y ( 2
y) ) / (x^2 + y^2)^2. This simplifies further.

f(x, y) = sqrt(x^2 / 4 - y^2 / 9)
f_x(x, y) = (x^2 / 4 - y^2 / 9) ' * 1 / (2 sqrt( x^2 / 4 - y^2 / 9), where ' denotes derivative with respect to x, so that f_x = ( 2 x / 4 ) / (2 sqrt(x^2 / 4 -
y^2 / 9), which simplifies further.
f_y(x, y) = (x^2 / 4 - y^2 / 9) ' * 1 / (2 sqrt( x^2 / 4 - y^2 / 9), where ' denotes derivative with respect to y, so that f_y = ( 2 y / 9 ) / (2 sqrt(x^2 / 4 -
y^2 / 9), which simplifies further.

f(x, y) = y e^(-k x^2) + x y cos(3x + 7y)
f_x(x, y) = 2 k x y e^(- k x^2 ) + y cos(3 x + 7 y) - 3 x y sin(3 x + 7 y)
f_y(x, y) = e^(- k x^2 ) + x cos(3 x + 7 y) - 7 x y sin(3 x + 7 y)