If your solution to a 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.

022.

Question: `qQuery problem 7.1.24 picture of sphere, diam from (-1,-2,1) to (0, 3, 3).

What is the standard form of the equation of the pictured sphere?

Your solution:

Confidence Assessment:

Given Solution:

`a The midpoint between (-1, -2, 1) and (0, 3, 3) is (-1/2, 1/2, 2).

Thus the sphere will have the form (x - (-1/2) )^2 + (y - 1/2)^2 + (z-2)^2 = r^2.

r is half the diameter, which is half of `sqrt( (0 - -1)^2 + (3 - -2)^2 + (3 - 1)^2 ) = `sqrt(1+25+4) = `sqrt(30). The radius is therefore `sqrt(30) / 2 and r^2 = 30 / 4 = 15/2.

The equation is therefore (x - (-1/2) )^2 + (y - 1/2)^2 + (z-2)^2 = 15/2.

Self-critique (if necessary):

Self-critique Rating:

Question: `qQuery problem 7.1.39 (was 7.1.38) yz-trace of x^2 + y^2 + z^2 - 6x - 10 y + 6 z + 30 = 0

Your solution:

Confidence Assessment:

Given Solution:

`a The yz trace is characterized by x = 0.

The equation of the y-z trace is therefore y^2 - 10 y + z^2 + 6z + 30 = 0.

This is the equation of a circle in the y-z plane. Completing the square we get

(y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 or

(y-5)^2 + (z+3)^2 = 4 or

(y-5)^2 + (z+3)^2 = 2^2.

Self-critique (if necessary):

Self-critique Rating:

Question: `qWhat is the equation of the yz-trace of the given sphere and what shape does this equation define in the y-z plane?

Your solution:

Confidence Assessment:

Given Solution:

`a `a The yz trace is characterized by x = 0.

The equation of the y-z trace is therefore y^2 - 10 y + z^2 + 6z + 30 = 0.

This is the equation of a circle in the y-z plane. Completing the square we get

(y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 or

(y-5)^2 + (z+3)^2 = 4 or

(y-5)^2 + (z+3)^2 = 2^2.

Self-critique (if necessary):

Self-critique Rating:

Question: `qWhat is the center and what is the radius of the circle?

Your solution:

Confidence Assessment:

Given Solution:

`a `a The yz trace is characterized by x = 0.

The equation of the y-z trace is therefore y^2 - 10 y + z^2 + 6z + 30 = 0.

This is the equation of a circle in the y-z plane. Completing the square we get

(y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 or

(y-5)^2 + (z+3)^2 = 4 or

(y-5)^2 + (z+3)^2 = 2^2.

We thus have a circle of radius 2, in the y-z plane, centered at (0, 5, -3).

Self-critique (if necessary):

Self-critique Rating:

Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.