Assignment 22

course Mth 272

7/30 8 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: To find the standard form of the equation of the pictured sphere, we need an equation in the form of (x-h)^2 + (y-k)^2 + (z-1)^2 = r^2. To find the center, we need to take the midpoint of (-1,-2,1) and (0,3,3) using the midpoint formula, which we find to be (-.5,.5, 2). Next we can find the diameter of the sphere using those same points in the distance formula, getting a diameter of sqrt30. We can now put our equation together in the standard form. We find that the equation of the sphere is (x+.5)^2 + (y-.5)^2 + (z-2)^2 = (sqrt(30)/2)^2, which simplifies to (x+.5)^2 + (y-.5)^2 + (z-2)^2 = 7.5.

confidence rating: 3

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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: To find the yz-trace, we need to first form an accurate sphere equation and then plug in 0 for x. By rearranging the given equation and then completing the square, we end up with the spherical equation (x-3)^2 + (y-5)^2 + (z+3)^2 = 13. Now we can plug in 0 for x to get the circle equation of the yz-trace. We find that this equation is (y-5)^2 + (z+3)^2 = 4.

confidence rating: 3

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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):

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Self-critique Rating:

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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: As found in the previous solution, the yz-trace is the circle equation (y-5)^2 + (z+3)^2 = 4. This makes the shape of a circle around the sphere and this defines the y-z plane along the y and z axes.

confidence rating: 3

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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: Because we know that the trace has the circular equation of (y-5)^2 + (z+3)^2 = 4 and that this equation is in standard form, we can easily pick out the center and radius. Since (h,k) is the center, we know that the center for this equation is (0, 5, -3). Since r^2 = 4, r = 2.

confidence rating: 3

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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: 3

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

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&#Very good responses. Let me know if you have questions. &#