Assigenment 22

course Mth 272

5/4/10 @ 9:03 pm

022.

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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?

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Your solution:

First apply the midpoint Formula:

(h,k,l) = ([X1 + X2] / 2 , [Y1 + Y2] / 2 , [Z1 + Z2] / 2)

= ([-1 + 0] / 2, [-2 + 3] / 2, [1 + 3] / 2)

= (-1 / 2 , 1 / 2, 4 / 2)

= ( -1/2, 1/2, 2)

Next use the distance Formula:

r = sgrt[(x2 – x1)^2 + (y2 – y1)^2 + (z2 – z1)^2]

= sqrt[(0 - -1)^2 + (3 - -2)^2 + (3 – 1)^2]

= sqrt[ 1 + 25 + 4]

=sqrt 30

Radius is equal to sqrt 30 / 2 = sqrt 15

Standard equation of the sphere:

(x – h)^2 + (y – k)^2 + (z – l)^2 = r^2

(x – (-1/2))^2 + (y – 1/2) + (z – 2) = 15

confidence rating #$&*2

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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.

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Self-critique (if necessary):

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Self-critique rating #$&*

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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

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Your solution:

Complete the square with out x since we want yz-trace

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

(y – 5)^2 + (z + 3)^2 + 30 - 34 = 0

(y – 5)^2 + (z + 3)^2 = 4

(y – 5)^2 + (z + 3)^2 = 2^2

confidence rating #$&*2

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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.

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Self-critique (if necessary):

I believe I just stumbled on the 4 not sure if my math was right

(y – 5)^2 + (z + 3)^2 = 4 follows pretty much immediately from the preceding line

(y – 5)^2 + (z + 3)^2 + 30 - 34 = 0

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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?

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Your solution:

(y – 5)^2 + (z + 3)^2 = 2^2

Which is a circle

confidence rating #$&*2

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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.

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Self-critique (if necessary):

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Self-critique rating #$&*

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

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Your solution:

The radius is 2 and the center is (0, -5, 3)

confidence rating #$&*2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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).

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Self-critique (if necessary):

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Self-critique rating #$&*

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

I actually like this section

Good work. See my note(s).