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course Mth 277
query_09_7*********************************************
Question: Identify the quadric surface 4y = (z^2)/4 - (x^2)/9.
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Your solution:
4y = (z^2)/4 - (x^2)/9
y=(z^2)/16 - (x^2)/36
Lets say that z=1
y=1/16-x^2/36
This conic section has the shape of a hyperbole.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):OK
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Self-critique rating:OK
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The cross-sections perpendicular to the y axis are indeed hyperbolas.
What about the cross-sections perpendicular to the x and the z axes?
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Question: Identify the quadric surface given by the equation 8z^2 = (1/8) + (x^2)/9 + (y^2). Describe the traces in planes parallel to the coordinate planes (and sketch the graph).
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Your solution:
8z^2=(1/8)+(x^2)/9+(y^2)
z^2=1+8(x^2)/9+8(y^2)
1=-8(x^2)/9-8(y^2)+z^2
z=1
1=-8(x^2)/9-8(y^2)+1
0=-8(x^2)/9-8(y^2)
z=2
1=-8(x^2)/9-8(y^2)+4
-3=-8(x^2)/9-8(y^2)
1=8(x^2)/27+8(y^2)/3
z=3
1=-8(x^2)/9-8(y^2)+9
-8=-8(x^2)/9-8(y^2)
1=(x^2)/9+(y^2)
From this, we can see that the semi axis are increasing as z increases through 1 to 3.
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These cross sections are ellipses and your description is valid.
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1=-8(x^2)/9-8(y^2)+z^2
x=1
1=-8/9-8(y^2)+z^2
17/9=-8(y^2)+z^2
1=-72/17(y^2)+9/17(z^2)
x=2
1=-32/9-8(y^2)+z^2
41/9=-8(y^2)+z^2
1=-72/41(y^2)+9/41(z^2)
x=3
1=-72/9-8(y^2)+z^2
9=-8(y^2)+z^2
1=-8/9(y^2)+1/9(z^2)
The same thing is happeining to the semi axis in this case. As x is increased, the semi axis is also increasing.
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These cross sections are not ellipses.
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1=-8(x^2)/9-8(y^2)+z^2
y=1
1=-8(x^2)/9-8+z^2
9=-8(x^2)/9+z^2
1=-8(x^2)/81+z^2/9
y=2
1=-8(x^2)/9-32+z^2
33=-8(x^2)/9+z^2
1=-8(x^2)/297+z^2/33
y=3
1=-8(x^2)/9-72+z^2
73=-8(x^2)/9+z^2
1=-8(x^2)/657+z^2/73
Once again, the y values show a similar trend. AS the y values are increased so is the semi axis values.
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These cross sections are not ellipses either.
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You're on the right track but you haven't identified the surface, and while your equations of the cross sections are correct you have misidentified some of them.
A more detailed description would benefit you.
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Shows that it is centered at origin.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: Describe the quadric surface given by the equation ((x-3)^2)/2 - ((y-1)^2)/4 - (z^2-2)/9 = 4.
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Your solution:
((x-3)^2)/2-((y-1)^2)/4-(z^2-2)/9=4
((x-3)^2)/8-((y-1)^2)/16-(z^2-2)/36=1
From here we can see the shape is centered at (3,1,2)
Based on the equation it seems to be a hyperbola. However, I believe it is not fully connected based on the fact two of the components are negative.
Just based on the signs, you can tell where the shape is going to be the ellipse or end point of the hyperbola. I believe the ellipses occur in the xcompnent.
This is because when you plug in a number for x, both of the other components are the same sign which fits the formula for an ellipse.
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Good insight, but you still haven't identified the surface.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):OK
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Self-critique rating:OK
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Question: Describe the curve intersection of the two quadric surfaces 4z = (y^2)/9 - (x^2)/16 and (x^2)/4 + 2(y^2) - 4(z^2)/3 = 1.
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Your solution:
I am going to start by setting the equations equal to eachother.
-4z+(y^2)/9-(x^2)/16=(x^2)/4+2(y^2)-4(z^2)/3-1
1=5/16(x^2)+(y^2)-4/3(z^2)+4z
We can see while pluggin in for z that the shape turns out to be an ellipse.
While plugging in for x or y I am unsure of what shape it gives. Since there is a z^2 and a z it is getting me somewhat confused on the shape.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):OK
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Self-critique rating:OK
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You would complete the square on the z terms in order to see how the graph shifts relative to one centered at the origin.
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You're on the right track, but it would be to your benefit to try to be a little more specific.
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