course Mth 272 IЋڒWFχassignment #022
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18:16:57 Query problem 7.1.24 picture of sphere, diam from (-1,-2,1) to (0, 3, 3).
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RESPONSE --> Midpoint= (-1/2, 1/2, 2) (-1 + 1/2)^2 + (-2 - 1/2)^2 + (1-2)^2 = r^2 .25 + 6.25 + 1 = r^2 r= 2.74 (x+1/2)^2 + (y - 1/2)^2 + (z- 2)^2 = 2.74 confidence assessment: 3 ** You should have used the exact expression for the diameter. Otherwise everything looks great. Compare with the following: 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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18:17:07 What is the standard form of the equation of the pictured sphere?
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RESPONSE --> see previous confidence assessment: 3
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18:17:41 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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18:17:54 How did you find the radius of the sphere?
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RESPONSE --> see previous confidence assessment: 3
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18:23:06 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 --> yz trace x=0 y^2 + z^2 - 10y + 6z = -30 (y^2 - 10y + 25) + (z^2 + 6z + 9) 34+ -30= 4 (y-5)^2 + (z+3)^2 = 4= 2^2 radius= 2 confidence assessment: 3
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18:23:12 What is the center and what is the radius of the circle?
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RESPONSE --> 2 confidence assessment: 3
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18:24:37 What is the standard form of the equation of the circle?
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RESPONSE --> (x-3)^2 + (y-5)^2 + (z+ 3)^2 = sqrt(13) confidence assessment: 2
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18:25:33 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I feel unusually confident with this assignment. But, considering it is the 7th chapter in Calc 2, I am sure I didn't do some things right. Please let me know how thigns went. confidence assessment: 3
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