#$&* course Mth 272 11/10 1 pm EST 022.
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Given Solution: `a The midpoint between (-1, -2, 1) and (0, 3, 3) is (-1/2, 1/2, 2). Thus the sphere will have the form (x - (-1/2) )^2 + (y - 1/2)^2 + (z-2)^2 = r^2. r is half the diameter, which is half of `sqrt( (0 - -1)^2 + (3 - -2)^2 + (3 - 1)^2 ) = `sqrt(1+25+4) = `sqrt(30). The radius is therefore `sqrt(30) / 2 and r^2 = 30 / 4 = 15/2. The equation is therefore (x - (-1/2) )^2 + (y - 1/2)^2 + (z-2)^2 = 15/2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ‘OK’ ********************************************* 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 0 = y^2 + z^2 – 10y + 6z + 30 = 0 Rewriting: = y^2 - 10y + z^2 + 6z + 30 = 0 (y^2 – 10y +25) + (x^2 +6z +9) = -30 = -30 +25+9 = 4 R = [sqrt]4 = 2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The yz trace is characterized by x = 0. The equation of the y-z trace is therefore y^2 - 10 y + z^2 + 6z + 30 = 0. This is the equation of a circle in the y-z plane. Completing the square we get (y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 or (y-5)^2 + (z+3)^2 = 4 or (y-5)^2 + (z+3)^2 = 2^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ‘OK” ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y^2 – 10y +z^2 + 6z + 30 = 0 (Equation) The equation defines that of a circle in the y-z plane. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a `a The yz trace is characterized by x = 0. The equation of the y-z trace is therefore y^2 - 10 y + z^2 + 6z + 30 = 0. This is the equation of a circle in the y-z plane. Completing the square we get (y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 or (y-5)^2 + (z+3)^2 = 4 or (y-5)^2 + (z+3)^2 = 2^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ‘OK’ ********************************************* Question: `qWhat is the center and what is the radius of the circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Radius of the circle: 2 Center: (0, 5 , -3) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a `a The yz trace is characterized by x = 0. The equation of the y-z trace is therefore y^2 - 10 y + z^2 + 6z + 30 = 0. This is the equation of a circle in the y-z plane. Completing the square we get (y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 or (y-5)^2 + (z+3)^2 = 4 or (y-5)^2 + (z+3)^2 = 2^2. We thus have a circle of radius 2, in the y-z plane, centered at (0, 5, -3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ‘OK’ ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. All ok here, text makes it very clear. "