24 open query

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course Mth 272

11/10 2 pm EST

024.

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Question: `qQuery problem 7.2.52 (was 7.2.48) identify quadric surface z^2 = x^2 + y^2/2.

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Your solution:

X^2 + y^2/2 – x^2 = 0

= Elliptic Cone

X^2/a^2 + y^2/b^2 – z^2/c^2 = 0

Equated to zero and not one.

confidence rating #$&*: 3

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Given Solution:

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Self-critique (if necessary):

#### No solution given.

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self-critique rating #$&*: 3

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Question: `qWhat is the name of this quadric surface, and why?

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Your solution:

X^2 + y^2/2 – x^2 = 0

= Elliptic Cone

X^2/a^2 + y^2/b^2 – z^2/c^2 = 0

Equated to zero and not one.

confidence rating #$&*: 3

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Given Solution:

`a f 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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Self-critique (if necessary):

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self-critique rating #$&*: ‘OK’

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Question: `qGive the equation of the xz trace of this surface and describe its shape, including a justification for your answer.

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Your solution:

x^2 + y^2/2 – z^2 = 0

x^2 = -y2/2 + z^2

x^2 = z^2

confidence rating #$&*: 2

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Given Solution:

`a The xz trace consists of the y = 0 points, which for z^2 = x^2 + y^2/2 is z^2 = x^2 + 0^2/2 or just z^2 = x^2.

The graph of z^2 = x^2 consists of the two lines z = x and z = -x in the yz plane.

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Self-critique (if necessary):

#### How can I successfully construct the graph, and also solve for those two lines in the yz plane?

If z^2 = x^2, then z = +- sqrt(x^2) = +- x.

z = x is a solution to the equation, and z = -x is a solution.

Each gives you a line in the x z plane.

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self-critique rating #$&*: 2

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Question: `qDescribe in detail the z = 2 trace of this surface.

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Your solution:

Z = 2

4 = x^2 + y^2/2

#### No self-critique attached, so I will add my doubts here.

So, diving by 4 allows us to equate to 1, and presents standard form of equation.

However, how did you move into this?

“This is an ellipse with major axis 2 `sqrt(2) in the y direction and 2 in the x direction.”

The standard form will be

x^2 / 4 + y^2 / 8 = 1.

So the semimajor axes are sqrt(4) = 2, and sqrt(8) = 2 sqrt(2).

confidence rating #$&*: 1

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Given Solution:

`a If z = 2 then z^2 = x^2 + y^2/2 becomes 2^2 = x^2 + y^2 / 2, or x^2 + y^2 / 2 = 4.

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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&#Your work looks good. See my notes. Let me know if you have any questions. &#

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