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.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises. 

11.3.

Let z = f(x, y) = 3 x^2 / y^3 + x * y.  It is easy to evaluate this function for any point (x, y) in the x-y plane, obtaining a value of z = f(x, y), and we can sketch the resulting point on a set of x-y-z coordinate axes.

If we replace y with a constant value we obtain a function of x only. 

For example, if y = 2 we get the function f(x, 2) = 3 x^2 / 8 + 2 x.

This function corresponds to the plane y = 2.  In this plane we can create an auxiliary x-z coordinate system and graph z = 3 x^2 / 8 + 2 x.  The function is quadratic so its graph will be a parabola.  ( We can analyze this parabola.  At any x value its slope is given by the derivative of 3 x^2 / 8 + 2 x, which you should verify is 6 x / 8 + 2.  The function f(x, 2) has a critical point, the point at which its derivative is zero.  A first- or second-derivative test reveals the critical point is a minimum.)

We could similarly replace x with a constant value.  For example if x = 1 we get the function f(1, y) = 3 / y^3 + y, the graph of which would occur in an auxiliary x-z plane x = 1.  (This function would be an odd function with a vertical asymptote as y = 0, and for large | y | would be asymptotic to the function z = y).

If we take the derivative of 3 x^2 / y^3 + x * y with respect to x, treating y as a constant number, we get 6 x / y^3 + y.  We call this the 'partial derivative with respect to x', denoted f_x (x, y), so that

f_x (x, y) = 6 x / y^3 + y

Note that if y = 2, our f_x function will be f_x ( x, 2 ) = 6 x / 8 + 2.  This agrees with the derivative of the function f(x, 2), see previously (6/8 reduces to 3/4, and should be so reduced, but to make the numbers easy to trace that reduction has not been done at this point)

Similarly f_y (x, y) is the derivative of f(x, y) with respect to y, where x is considered constant.  We get

f_y (x, y) = -9 x^2 / y^4 + x.

More generally we can use a related strategy to understand the behavior of f(x, y) in the neighborhood of any point (x0, y0).  The strategy is fairly straightforward:

Look at the function f(x0, y) in the plane x = x0.  Sketch a graph of this function on an auxiliary y-z coordinate system.  Take the y derivative of this function and use it to determine the slope at the point y0.  A tangent vector at that point is `j + slope * `k, where 'slope' is the value of the derivative.

Look at the function f(x, y0) in the plane y = y0.  Sketch the graph of this function on an auxiliary x-z coordinate system.  Take the y derivative of this function and use it to determine the slope at the point x0.  A tangent vector at that point is `i + slope * `k, where 'slope' is the value of the derivative.

The two planes x = x0 and y = y0 intersect in a vertical line above the point (x0, y0) in the x-y plane.  Sketch the two intersecting planes on a set of xyz coordinate axes, sketch the curve, and sketch the two tangent vectors.

The two tangent vectors, and the coordinates of the point (x0, y0, f(x0, y0) ), define the plane tangent to the surface z = f(x, y) at that point.

Until specified differently, the following questions address the function f(x, y) = x^2 * sqrt(y). 

Question:  `q001.  Let (x0, y0) be the point (1, 4).  What are the values of the following?

• f(x0, y0)
• (x0, y0, f(x0, y0))
• f(x0, y)
• f(x, y0)   x^2 * sqrt(y0) = x^2 * sqrt(4) = 2 x^2
• f(x0, y)    1^2 sqrt(y) = sqrt(y)
• f_y(x0, y)   f(x0, y) = sqrt(y) so f_y(x0, y) = 1 / (2 sqrt(y))
• f_x (x, y0)   f(x, y0) = 2 x^2 so f_x(x, y0) = 4 x
• f_x (x, y)   f(x,y) = x^2 * sqrt(y) so f_x(x, y) = sqrt(y) * (x^2) '  = 2 x sqrt(y)
• f_y (x, y)  f(x, y)=  x^2 sqrt(y) so f_y(x, y) = x^2 * (sqrt(y)) ' = x^2 * 1 / (2 sqrt(y))
• f_x (x0, y0)  f
• f_y (x0, y0)
• the derivative with respect to y of f(x0, y)
• the derivative with respect to x of f(x, y0)

Your solution: 

Confidence rating:
 

Given Solution: 

Since (x0, y0) = (1, 4), it is clear that x0 = 1 and y0 = 4.  It therefore follows that

f(x0, y0) = f(1, 4) = 1^2 * sqrt(4) = 2

(x0, y0, f(x0, y0)) = (1, 4, 2), which can be interpreted as a point in 3-dimensional coordinate space.

f_x (x, y) is the derivative with respect to x, obtained by treating y as a constant.  The derivative of the expression x^2 sqrt(y) is therefore the derivative of x^2, multiplied by the 'constant' factor sqrt(y).  We get

f_x (x, y) = 2 x * sqrt(y)

f_y (x, y) is found by treating x as constant, so that x^2 * sqrt(y) consists of the function sqrt(y) multiplied by the 'constant' factor x^2.  The derivative of sqrt(y) is 1 / (2 sqrt(y) ) so we get

f_y (x, y) = x^2 * 1 / (2 sqrt(y) ), which can be written more simply as x^2 / (2 sqrt(y) ).

x0 = 1 so f(x0, y) = 1^2 * sqrt(y) = sqrt(y).  This represents the function f(x, y), restricted to the plane x = 1. 

y0 = 4 so f(x, y0) = x^2 * sqrt(4) = 2 x^2.  This represents the function f(x, y), restricted to the plane y = 4. 

f_x (x, y) = 2 x sqrt(y), so f_x (x0, y0) = f_x (1, 4) = 2 * 1 sqrt(4) = 4.

f_y (x, y) = x^2 / (2 sqrt(y)) , so f_y (x0, y0) = f_y (1, 4) = 1^2 / (2 sqrt(4)) = 1/4.

f(x0, y) = f(1, y) = sqrt(y), so the derivative with respect to y of f(x0, y) is 1 / (2 sqrt(y))

f(x, y0) = f(x, 4) = 2 x^2, so the derivative with respect to x of f(x, y0) = 2 * (2x) = 4 x.

Question:   `q002.  This problem is a continuation of the preceding.

Sketch the graph of the function f(x, y0) on an auxiliary set of y-z axes, using the graphing techniques you learned in precalculus and calculus.

Determine the equation of the line tangent to this graph at the point x = x0.

Give a vector in the direction of the tangent line.

Sketch your graph and your tangent vector as they would appear within the plane x = x0 in an xyz coordinate system.

Your solution: 

Confidence rating:
 

Given Solution: 

f(x, y0) = f(x, 4) = x^2 * sqrt(4) = 2 x^2. 

The derivative of this function is 4 x.  A critical point occurs when 4 x = 0, so the x = 0 point is the critical point.  The second derivative of the function is 4, which is positive for all values of x.  So the critical point is a relative minimum.  The graph is concave upward, with a relative minimum at (0, 0).  The graph is in fact a parabola.

The function is z = 2 x^2 and the derivative function is z ' = 4 x. 

At x = x0 the value of the function is z = 2 x^2 = 2 * 1^2 = 2, and the value of the derivative function is z ' = 4 x0 = 4 * 1 = 4. 

So the tangent line passes through the point (1, 2) and has slope 4. The equation of the tangent line is (y - 2) = 4 * (x - 1), which can be rearranged to the form y = 4 x - 4.

A vector in the direction of the tangent line could be constructed with a 'run' of 1 and a 'rise' of 4.  The 'run' is parallel to the `i vector and the rise is parallel to the `k vector, so the corresponding vector is `i + 4 `k.

Question:  `q003.  Sketch the graph of the function f(x0, y) on an auxiliary set of y-z axes, using the graphing techniques you learned in precalculus and calculus.

Determine the equation of the line tangent to this graph at the point y = y0.

Give a vector in the direction of the tangent line.

Sketch your graph and your tangent vector as they would appear within the plane y = y0 in an xyz coordinate system.

x0 = 1, so f(x0, y) = f(1, y) = sqrt(y).

This function is undefined for y < 0, and takes value 0 at y = 0.  Its value is positive for y > 0. 

The derivative of this function is 1 / (2 sqrt(y)), which is undefined for y <= 0, but is positive for all y > 0.  The function is therefore increasing.  The limit of the derivative as y approaches zero is -infinity, so the graph is tangent to the y axis.

The second derivative is -1/4 * y(-3/2), and is negative for all y > 0, so the graph is concave down. 

The value of the function at y = y0 = 4 is sqrt(4) = 2 and the derivative  is 1 / (2 sqrt(2)) = 1/4.  So the tangent line has equation (y - 2) = 1/4 ( x - 4), or y = 1/4 x + 1.

A vector in the direction of the tangent line would be `j + 1/4 `k.

Your solution: 

Confidence rating:
 

Question:  `q004.  In a single x-y-z coordinate system, sketch both of your graphs, one within the x = x0 plane and the other within the y = y0 plane.  Include your tangent vectors.

Your solution: 

Confidence rating:
 

Question:  `q005.  This is a summary of results you have already obtained, but you should be sure you can obtain them without looking back.  The same is true of the next few questions.

What is the function f(x, 4)?

Sketch the graph of this function.

What is the derivative of this function, and what is the value of this derivative when x = 1?

What is the equation of the line tangent to your graph at the x = 1 point?

In terms of the `i, `j and `k vectors, what is a vector in the direction of your tangent line?

Your solution: 

Confidence rating:
 

Question:  `q006.  What is the function f(1, y)?

Sketch the graph of this function.

What is the derivative of this function, and what is the value of this derivative when y = 4?

What is the equation of the line tangent to your graph at the y = 4 point?

In terms of the `i, `j and `k vectors, what is a vector in the direction of your tangent line?

Your solution: 

Confidence rating:
 

Question:  `q007.  Using a single set of coordinate axes, sketch the following:

The point (1, 4, 0)

The point (1, 4, f(1, 4)).

The plane y = 4, and in this plane also sketch the graph of the function f(x, 4).

A vector tangent to your graph at the point (1, 4, f(1, 4)).

In terms of `i, `j and `k, what is the vector you just sketched?

Your solution: 

Confidence rating:
 

Question:  `q008.  Using a single set of coordinate axes, sketch the following:

The point (1, 4, 0)

The point (1, 4, f(1, 4)).

The plane x = 1, and in this plane also sketch the graph of the function f(1, y).

A vector tangent to your graph at the point (1, 4, f(1, 4)).

In terms of `i, `j and `k, what is the vector you just sketched?

Your solution: 

Confidence rating:
 

Question:  `q009.  Sketch the 2 x 2 x 2 cube whose 'bottom' face rests on the x axis, whose 'back' face is in the plane x = 1, and whose 'left-side' fact is in the plane y = 4.

Sketch the part of the graph of f(1, y) that lies along the 'back' face of your cube.

Sketch the part of the graph of f(x, 4) that lies along the 'left' face of your cube.

Sketch the vectors tangent to your graph at the point (1, 4, 2).

Your solution: 

Confidence rating:
 

Question:  `q010.  Treating y as a constant, what is the derivative of f(x, y), with respect to x?

For y = 4, what is the value of this derivative?

Have you seen this result before?

For x = 1 and y = 4, what is the value of your derivative?

What does this value have to do with your picture?

Your solution: 

Confidence rating:
 

Question:  `q011.  Treating x as a constant, what is the derivative of f(x, y), with respect to y?

For x = 1, what is the value of this derivative?

Have you seen this result before?

For x = 1 and y = 4, what is the value of your derivative?

What does this value have to do with your picture?

Your solution: 

Confidence rating:
 

Question:  `q012.  You have just calculated the derivatives f_y (x, y) and f_x (x, y).  Specifically

f(x, y) = x^2 * sqrt(y).

f_y (x, y) treats x^2 as a constant.  The derivative with respect to y of sqrt(y) is 1 / (2 sqrt(y)).  So f_y (x, y) = x^2 * 1 / (2 sqrt(y), which simplifies to x^2 / (2 sqrt(y)).

f_x (x, y) treats sqrt(y) as a constant, and yields derivative (2 x) * sqrt(y).

Find f_x (x, y) and f_y ( x, y) for each of the following functions:

• f(x, y) = sqrt(x) y^3
• f(x, y) = e^(2 x) * y^2 + 1 / sqrt(x) * y

`q008.  For the function f(x, y) = x^2 sqrt(y):

Find the x derivative of f_x (x, y).

Find the y derivative of f_y (x, y).

Find the x derivative of f_y (x, y).

Find the y derivative of f_x (x, y).

We call these derivatives, respectively, f_xx, f_yy, f_yx and f_xy.

Your solution: 

Confidence rating: