Assignment 24

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course Mth 272

7/21 12:59am

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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:

Equation after rearrange: x^2/1 + y^2/2 - z^2/1

It is an elliptic cone because since z is squared, if you move it over to the left side, it sets the equation equal to 0 and makes z^2/1 negative.

confidence rating #$&*: 3

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

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Your solution:

It is an elliptic cone because it follows the standard form:

(x^2 / a^2) + (y^2 / b^2) - (z^2 / c^2) = 0.

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): Just following the standard form isn’t enough?

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Self-critique Rating: 3

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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:

y = 0

x^2 + 0^2/2 = z^2

sqrt (x^2 = z^2) = +-x = z

Since z = +-x, the shape will be two lines. Example would be to replace z with y. y equaling a constant results in a horizontal line. Since z equals +-x, the result is two vertical lines because z equals two numbers.

confidence rating #$&*: 3

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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): OK

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Self-critique Rating: OK

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

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Your solution:

2^2 = x^2 + (y^2/2)

4/4 = (x^2 + (y^2/2))/4

Result: 1 = ((x^2/4) + (y^2/8))

The resulting equation makes an ellipse since it’s equal to 1.

confidence rating #$&*: 3

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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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Self-critique (if necessary):

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Self-critique rating:

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

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Your solution:

2^2 = x^2 + (y^2/2)

4/4 = (x^2 + (y^2/2))/4

Result: 1 = ((x^2/4) + (y^2/8))

The resulting equation makes an ellipse since it’s equal to 1.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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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Self-critique (if necessary):

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Self-critique rating:

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&#This looks good. Let me know if you have any questions. &#