course MTH 272
......!!!!!!!!...................................Applied Calculus II
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Asst # 24
07-06-2006
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16:26:55
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**** Query problem 7.2.52 (was 7.2.48) identify quadric surface z^2 = x^2 + y^2/4.
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16:30:32
z^2 = x^2 + y^2/4
sqrt (z^2) = sqrt (x^2) + sqrt (y^2 / 4)
z = x + y/2
This fits the equation for an elliptic paraboloid: z= x^2 / a^2 + y^2 / b^2
** If z = c, a constant, then x^2 + y^2/2 = c^2, or x^2 / c^2 + y^2 / (`sqrt(2) * c)^2 = 1. This gives you ellipse with major axis c and minor axis `sqrt(2) * c. Thus for any plane parallel to the x-y plane and lying at distance c from the x-y plane, the trace of the surface is an ellipse.
In the x-z plane the trace is x^2 - z^2 = 0, or x^2 = z^2, or x = +- z. Thus the trace in the x-z plane is two straight lines.
In the y-z plane the trace is y^2 - z^2/2 = 0, or y^2 = z^2/2, or y = +- z * `sqrt(2) / 2. Thus the trace in the y-z plane is two straight lines.
The x-z and y-z traces show you that the ellipses in the 'horizontal' planes change linearly with their distance from the x-y plane. This is the way cones grow, with straight lines running up and down from the apex. Thus the surface is an elliptical cone. **
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16:30:33
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**** What is the name of this quadric surface, and why?
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16:31:35
The name of this quadric surface is an elliptic paraboloid because the xy-trace is an ellipse, while the xz and yz-planes are parabolas.
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16:31:36
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**** Give the equation of the xz trace of this surface and describe its shape, including a justification for your answer.
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16:35:30
Set y=0
z^2 = x^2 + (y^2)/4
z^2 = x^2 + 0/4
z^2 = x^2 +0
z^2 = x^2
This is a parabola because it contains a squared term.
A parabola is square in one variable, linear in the other.
z^2 = x^2 is equivalent to the two linear equations
z = x and
z = -x,
which give you two straight lines.
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16:35:31
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**** Describe in detail the z = 2 trace of this surface.
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16:39:34
2^2 = x^2 + (y^2)/4
4 = x^2 + (y^2)/4
I'm not really sure what to do from here. Maybe you could take the square root of both sides, leaving you with:
sqrt(4) = sqrt(x^2) + sqrt(y^2 /4)
Careful. You can take the square root of both sides, but you can't take the square root of every term. sqrt(a^b + b^2) is not a + b; the simplest way to see this is to see that the square of a + b is not a^2 + b^2 but a^2 + 2 a b + b^2.
but then that leaves you with all of the terms being either positive or negative
+,- 2 = +,- x + +,- (y/2)
I don't think this is the right answer, so maybe you don't take the square root, and it leaves you with:
2^2 = (x-0)^2 + (y/2 - 0)^2, which would be an ellipse with a center of (0,0) and a radius of 2.
The equation (which should read 2^2 = x^2 + (y^2)/2) is an ellipse
This is an ellipse. If we divide both sides by 4 we can get the standard form:
x^2 / 4 + y^2 / 8 = 1, or x^2 / 2^2 + y^2 / (2 `sqrt(2))^2 = 1.
This is an ellipse with major axis 2 `sqrt(2) in the y direction and 2 in the x direction.
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16:39:37
Your response has been entered.
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Again you have the right ideas; you just need a little practice distinguishing and identifying circles, hyperbolas, parabolas and ellipses.