Assignment 23

course Mth 272

......!!!!!!!!...................................

11:34:31
Query problem 7.2.6 intercepts and sketch graph of 2x - y + z = 4.

......!!!!!!!!...................................

RESPONSE -->

confidence assessment:

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

......!!!!!!!!...................................

11:34:35
List the intercepts of the graph.

......!!!!!!!!...................................

RESPONSE -->
2x - 0 + 0 = 4
x = 2
0 - y + 0 = 4
y = -4
0 - 0 + 2 = 4
z = 4
(2,0,0)(0,-4,0) (0,0,4)
confidence assessment: 3

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

......!!!!!!!!...................................

11:34:50
Describe the graph of the plane.

......!!!!!!!!...................................

RESPONSE -->
The graph is a triangle with intercepts at (2,0,0)(0,-4,0) (0,0,4)
confidence assessment: 3

The intersections with the coordinate planes do form triangles. The equation defines a plane, and that triangle is part of the plane.

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

......!!!!!!!!...................................

11:35:02

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?

......!!!!!!!!...................................

RESPONSE -->

It would be closer to the x axis.

confidence assessment: 3

Good.

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.

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

......!!!!!!!!...................................

11:35:12

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?

......!!!!!!!!...................................

RESPONSE -->

It would be steeper from the x intercept

confidence assessment: 3

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

......!!!!!!!!...................................

11:35:15

Query problem 7.2.34 (was 7.2.30) match y^2 = 4x^2 + 9z^2 with graph

......!!!!!!!!...................................

RESPONSE -->

confidence assessment:

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

......!!!!!!!!...................................

11:35:26

Which graph matches the equation?

......!!!!!!!!...................................

RESPONSE -->

Graph matches hyperboloid of One sheet with an axis of y.

confidence assessment: 2

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

......!!!!!!!!...................................

11:35:36

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

......!!!!!!!!...................................

RESPONSE -->

Equation doesn't have two negatives.

confidence assessment: 2

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

......!!!!!!!!...................................

11:35:44

The graph could not be (c) because the picture shows that the surface is not defined for | y | < 1, while 4x^2 + 9z^2 = .25, for example, is the trace for y = 1/2, and is a perfectly good ellipse. State this in your own words.

......!!!!!!!!...................................

RESPONSE -->

confidence assessment:

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.

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

......!!!!!!!!...................................

11:35:53

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

......!!!!!!!!...................................

RESPONSE -->

Coefficients are not equal and non zero.

confidence assessment: 2

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

......!!!!!!!!...................................

11:36:01

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.

......!!!!!!!!...................................

RESPONSE -->

confidence assessment:

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.

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

......!!!!!!!!...................................

11:36:10

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

......!!!!!!!!...................................

RESPONSE -->

ellipse

confidence assessment: 2

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.

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

......!!!!!!!!...................................

11:36:19

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

......!!!!!!!!...................................

RESPONSE -->

hyperbola

confidence assessment: 2

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.

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

......!!!!!!!!...................................

11:36:28

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

......!!!!!!!!...................................

RESPONSE -->

hyperbola

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.

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

......!!!!!!!!...................................

11:36:30

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

......!!!!!!!!...................................

RESPONSE -->

confidence assessment:

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

Assignment 23

The intersections with the coordinate planes do form triangles. The equation defines a plane, and that triangle is part of the plane.

Good.

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.

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.

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.

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.

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.

&#

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

Assignment 23

The intersections with the coordinate planes do form triangles. The equation defines a plane, and that triangle is part of the plane.

Good. ** 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. **

** 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. **

** 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. **

** 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. **

** 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. **

&#

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

Good.

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.

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.

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.

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.

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.