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

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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. 

Question:  `q001.  Give the standard form equation for the tangent plane to the surface z(x,y) = ln(x^2 + y^2) at the point P_0 = (e,0,2).

Your solution: 

Confidence rating:
 

Given Solution: 

At (e, 0, 2) we have z_x = 2 * e / (e^2 + 0^2) = 2 / e (very approximately .75)

A unit vector tangent to the plane, in the x z plane, is therefore i + (2/e) k.

z_y = 2 y / (x^2 + y^2). 

At (e, 0, 2) we have z_y = 2 * 0 / (e^2 + 0^2) = 0.

The j vector is therefore tangent to the plane, in the yz plane.

The cross product of two tangent vectors is a normal vector. 

The cross product of the two tangent vectors obtained above is k + (2 / e) * (-i) = k - (2 / e) * i.

The point (x, y, z) lies on the tangent plane if (x - e) i + (y - 0) j + (z - 2) k is perpendicular to the normal vector. 

Setting the dot product of the two vectors equal to zero we get the equation

-(2/e) * (x - e) + (z - 2) = 0, which we simplify to

-(2/e) x + z = 0

Self-critique (if necessary):

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Question:  `q002.  Find the total differential of f(x,y,z) = 2xzy^3*cos(xy)*sin(z)

Your solution: 

Confidence rating:
 

Given Solution: 

The total differential is f_x ds + f_y dy + f_z dz.

f_x = ( 2 z y^3 * cos(xy) - 2 x z y^4 sin(x y) ) * sin(z)

f_y = (6 x z y^2 cos(x y) - 2 x^2 z y^3 sin(xy)) * sin(z)

f_z = (2xy^3*cos(xy)*sin(z) + 2 xzy^3*cos(xy)*cos(z)

The total differential is therefore

( 2 z y^3 * cos(xy) - 2 x z y^4 sin(x y) ) * sin(z) * f_x + (6 x z y^2 cos(x y) - 2 x^2 z y^3 sin(xy)) * sin(z) * f_y + ( 2xy^3*cos(xy)*sin(z) + 2 xzy^3*cos(xy)*cos(z) ) * dz

The expression would be simplified by factoring 2 z y^3 out of the coefficient of f_x , 2 x z y^2 out of the coefficient of f_y, and 2 x y^3 out of the coefficient of dz.

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Question:  `q003.  Use an incremental approximation to estimate f(sqrt(pi) + .01, sqrt(pi) - .01), where f(x,y) = cos(xy)

Your solution: 

Confidence rating:
 

Given Solution: 

We approximate by first finding f(x, y) at the point (sqrt(pi), sqrt(pi)).

Then we apply the differential with dx = .01 and dy = -.01.

f(sqrt(pi), sqrt(pi)) = cos(sqrt(pi) * sqrt(pi)) = cos(pi) = -1.

df = f_x dx + f_y dy indicates the change in f due to a given change dx in the value of x, and dy in the value of y.

df = -y sin(xy) dx - x sin(xy) * dy.

At the point (sqrt(pi), sqrt(pi) ) we have -y sin(x y) = - pi * sin(sqrt(pi) * sqrt(pi)) = -pi * sin(pi) = -pi * 0 = 0. 

In a similar manner we have -x sin(x y) = 0.

That is, both f_x and f_y are zero at this point.

Using our values of f_x and f_y at the original point, along with dx = .01 and dy = -.01 we get

df = 0 * .01 + 0 * (-.01) = 0.

The same procedure would apply to approximate the function at, say, the point (sqrt(pi) + .01, sqrt(pi) / 3 - .02):

At the point (x0, y0) = (sqrt(pi) , sqrt(pi) / 3 ) we have f(x, y) = cos(sqrt(pi) * sqrt(pi / 3) ) = cos ( pi / 3) = 1/2 or .5.

The differential of our function is still df = -y sin(xy) dx - x sin(xy) * dy.

At the point (x0, y0) = (sqrt(pi), sqrt(pi) / 3 ) we have

f_x = -y sin(x y) = - pi / 3 * sin(sqrt(pi) * sqrt(pi) / 3) = -pi * sin(pi / 3) = -pi / 3 sqrt(3) / 2, approximately -0.9.

f_y = In a similar manner we have -x sin(x y) = -pi sin(pi/3) = -pi sqrt(3) / 2 = -2.7.

Using our values of f_x and f_y at our (x0, y0) point, along with dx = .01 and dy = -.02 we get

df = -0.9 * .01 + (-2.7) * (-.02) = -.063.

Our approximation to the value of f at the given point is therefore .5 + (-.063) = .437.  That is

f(x, y) = f(sqrt(pi) + .01, sqrt(pi) / 3 -.02) = f(x0, y0) + df = .5 - .063 = .437.

You can assess the accuracy of this approximation by evaluating cos(x y) at the given point.

Self-critique (if necessary):

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Question:  `q004.  Find the equation of all horizontal tangent planes to the surface z = 4 - x^2 - y^2 + 6x.

Your solution: 

Confidence rating:
 

Given Solution: 

The tangent plane will be horizontal only if all tangent lines are horizontal.  That is, all tangent lines have to have slope zero.  Thus all the derivatives need to be zero.

This will be the case, for example, at a point where the x and y partial derivatives are both zero.

For this function z_x = -2 x + 6 and z_y = -2 y.  Thus our conditions z_x = 0 and z_y = 0 give us the two equations

-2 x + 6 = 0

-2 y = 0.

Each equation has only one solution.  We get x = 3 and y = 0.

Thus the point (3, 0) is a critical point.

We need to check to be sure that our critical point isn't a saddle point. 

Our second derivatives z_xx and z_yy are both negative, so (3, 0) is a candidate for a relative maximum.  So far so good.

We also need to test that the graph doesn't go off into a saddle point when we move at some nonzero angle to the x and y axes.  The test for this is that z_xx * z_yy - z_xy ^ 2 must be positive. 

In this case z_xy = 0.  We get z_xx * z_yy - z_xy ^ 2 = -2 * -2 + 0^2 = 4, which is > 0, so we don't have a saddle point

We conclude that our point (3, 0) does indeed give us a relative maximum.

The relative maximum therefore occurs at (3, 0, f(3, 0) ) = (3, 0, 13).

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