Assignment 23

course Mth 272

ߏʭfЉРVysassignment #023

023.

Applied Calculus II

11-09-2008

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18:30:40

List the intercepts of the graph.

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RESPONSE -->

2x= 4

x=2

-y= 4

y=-4

z= 4

(2,0,0), (0, -4, 0), (0,0, 4)

confidence assessment: 3

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18:32:45

Describe the graph of the plane.

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RESPONSE -->

It is triangular with the intercepts listed previously, and the triangle is on a vertically-sloped 3D plane. The base of the triangle is towards the xy plane.

I have no idea how to describe the graph of a plane without showing it to you. Please let me know how I could describe this graph.

confidence assessment: 3

You've pretty much described it. Compare with the following (especially the last line):

The x-intercept occurs when y and z are 0, giving us 2x = 4 so x = 2.

The y-intercept occurs when x and z are 0, giving us -y = 4 so y = -4.

The z-intercept occurs when x and y are 0, giving us z = 4.

The intercepts are therefore (2, 0, 0), (0, -4, 0) and (0, 0, 4).

These three points form a triangle and this triangle defines the plane 2x - y + z = 4. This plane contains the triangle but extends beyond the triangle, extending infinitely far in all directions.

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18:33:12

If you released a marble on the plane at the point where it intercepts the z axis, it would roll down the incline. When the marble reached the xy plane would it be closer to the x axis or to the y axis?

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RESPONSE -->

Closer to the x-axis because its intercept is closer to where the marble was released.

confidence assessment: 3

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18:33:23

If you were climbing the plane straight from your starting point to the point for the plane intercepts the z axis, with your climb be steeper if you started from the x intercept or from the y intercept?

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RESPONSE -->

x-intercept

confidence assessment: 3

** The marble would travel the steepest possible path. The line from (0,0,4) to (2,0,0), in the xz plane, is steeper than the line from (0, 0, 4) to (0, -4, 0) in the yz plane. So the marble would reach the xy plane closer to the x axis than to the y axis. **

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18:34:07

Which graph matches the equation?

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RESPONSE -->

G, the equation matches that of an elliptic cone. There is one negative coefficient.

confidence assessment: 3

** In the plane y = 1/2 the trace of y^2 = 4x^2 + 9z^2 is found by substituting y = 1/2 into this equation. We obtain (1/2)^2 = 4x^2 + 9z^2, or 1/4 = 4x^2 + 9z^2.

Multiplying both sides by 4 we get the 16 x^2 + 36 z^2 = 1, which can be expressed as x^2 / [1/4^2] + y^2 / [ 1/6^2].

This is the standard form of an ellipse with major axis 1/4 in the x direction and minor axis 1/6 in the y direction. **

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18:34:48

The graph couldn't be (e). Explain why not.

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RESPONSE -->

E is a hyperboloid of two sheets, whereas G is an elliptic cone, and an elliptic cone only has one positive coefficient.

confidence assessment: 3

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18:37:37

The graph couldn't be (c). Explain why not.

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RESPONSE -->

For an ellipsoid graph the coefficients must all be equal. This is not the case. The graph cannot be ellipsoid.

confidence assessment: 3

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18:38:21

The trace of this graph exists in each of the coordinate planes, and is an ellipse in each. The graph of the given equation consists only of a single point in the xz plane, since there y = 0 and 4x^2 + 9z^2 = 0 only if x = z = 0. Explain why the xy trace is not an ellipse.

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RESPONSE -->

If all are zeroes, the xy trace will just be a single point. A single point cannot form an elipse.

confidence assessment: 1

** If y^2 = 4x^2 + 9z^2 then the xy trace, which occurs when z = 0, is y^2 = 4 x^2. This is equivalent to the two equations y = 2x and y = -2x, two straight lines. **

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18:39:11

What is the shape of the trace of the graph in the plane y = 1?

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RESPONSE -->

4x^2 + 9z^2= 1

circle

confidence assessment: 1

** In the plane y = 1 the trace of y^2 = 4x^2 + 9z^2 becomes 4 x^2 + 9 z^2 = 1, which is an ellipse.

In standard form the ellipse is

x^2 / [ 1 / 2^2 ] + z^2 / [ 1 / 3^2 ] = 1,

so has major axis 1/2 in the x direction and minor axis 1/3 in the z direction. **

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18:40:26

What is the shape of the trace of the graph in the plane x = 1?

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RESPONSE -->

When x=1, the equation looks a lot like examples in the book which revealed a parabolic shape. I am not exactly sure how to confirm this, but from reading the chapter I am going to go with parabolic for this one.

confidence assessment: 1

** In the plane x = 1 the trace of y^2 = 4x^2 + 9z^2 is

y^2 - 9 z^2 = 4,

which is a hyperbola with vertices at y = +- 2, z = 0 (i.e., at points (1, +-2, 0) since x = 1); the asymptotes are the lines y = 3z and y = -3z in the plane x = 1. **

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18:40:37

What is the shape of the trace of the graph in the plane z = 1?

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RESPONSE -->

Parabolic

confidence assessment: 2

** In the plane z = 1 the trace of y^2 = 4x^2 + 9z^2 is

y^2 - 4 x^2 = 9,

a hyperbola with vertices at x = 0 and y = +- 3 (i.e., at points (0, +- 3, 1) ) and asymptotes y = 2x and y = -2x in the plane z = 1. **

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18:41:07

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

confidence assessment:

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Good work overall. See my notes, though, related to some of the traces.