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course mth 272

|XK܋ӂCm~Jцaݮ\wassignment #022

022.

Applied Calculus II

07-27-2008

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09:33:44

Query problem 7.1.24 picture of sphere, diam from (-1,-2,1) to (0, 3, 3).

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RESPONSE -->

ok, first you use the midpoint formula to find the center of the sphere...

h, k, L, = ( (-1 + 0 / 2), (-2 + 3 / 2), (1 + 3 / 2)

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

then, use the distance formula to find radius

r = 'sqrt ( (-1 - 0)^2 + (-2 - 3)^2 + (1 - 3)^2 )

='sqrt (1 + 25 + 30)

='sqrt(30)

plug into standard equation form

(x + 1/2)^2 + (y - 1/2)^2 + (z - 2)^2 = 30

confidence assessment: 2

You got sqrt(30) from the two given points, which form a diameter. The radius is sqrt(30) / 2 and the squared radius is 30/4 = 15/2.

Otherwise excellent work.

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09:34:08

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

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RESPONSE -->

previously answered

confidence assessment: 2

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09:34:29

How did you find the center of the sphere?

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RESPONSE -->

using the midpoint formula

confidence assessment: 3

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09:34:41

How did you find the radius of the sphere?

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RESPONSE -->

using the distance formula

confidence assessment: 3

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09:54:47

Query 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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RESPONSE -->

first you need to obtain the standard equation of the sphere by completing the square.

which yields

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

= (x-3)^2 + (y-5)^2 +(z+3)^2 = 13

center of the sphere is then (3,5,-3)

radius = 'sqrt(13)

now for that equation you want to let x=0 to find the yz-trace

(0-3)^2 + (y-5)^2 +(z+3)^2 = 13

9 + (y-5)^2 +(z+3)^2 = 13

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

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

so, this is a circle of radius 2

confidence assessment: 2

Very good. It's a little more work to first find the standard equation of the sphere, but it's very good to recognize and work this out and provides you with a broader context.

** 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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09:55:10

What 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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RESPONSE -->

just answered

confidence assessment: 2

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09:55:37

What is the center and what is the radius of the circle?

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RESPONSE -->

previously answered

confidence assessment: 2

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09:55:49

What is the standard form of the equation of the circle?

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RESPONSE -->

previoiusly answered

confidence assessment: 2

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09:56:29

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

I feel I have a good, solid understanding of the material in this section.

confidence assessment: 3

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&#Good responses. See my notes and let me know if you have questions. &#