course Mth 272 7/30 8:45 pm 023.
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Given Solution: `a 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIf 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because two lines have endpoints at z = 4, the marble could roll down both of them. However, if the marble rolled down to the point (2,0,0) it would be closer to the x-axis, and if it rolled down to the point (0, -4,0), it would be closer to the y-axis. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This makes more sense after reading the given answer. Next time I will assume the marble will take the steepest path before solving. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qIf 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From our graph, we can easily see that the line from (2,0,0) has a much steeper slope than the line from (0, -4,0). So the climb from the x intercept would be steeper. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The line from (0,0,4) to (2,0,0), in the x-y plane, has slope 2 and is therefore steeper than the line from (0, 0, 4) to (0, -4, 0) in the yz plane, which has slope of magnitude 1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery problem 7.2.34 (was 7.2.30) match y^2 = 4x^2 + 9z^2 with graph Which graph matches the equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We find that the equation has the graph of G, the elliptic cone. We can figure this out easily by rearranging the equation to 0 = 4x^2 – y^2 + 9z^2. Because all variables are squared, only y is negative, and the equation equals 0, the equation fits the standard model for an elliptic cone. Graph G also proves to be correct because the elliptic cone is situated on the y-axis. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qThe graph couldn't be (e). Explain why not. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Graph E is a graph of a hyperboloid of two sheets. This equation cannot have graph E because it equals 0, not 1, and two variables are positive, it does not have the two negative variables that E requires. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The equation for e) is set equal to 1 and the needed equation is set equal to 0. So one has a constant term while the other does not. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qThe 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph cannot be graph C because abs(y) < 1 is not defined on graph C and the xy-trace of the equation would not give us the xy-trace of graph C, an ellipse. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qThe graph couldn't be (c). Explain why not. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph cannot be graph C because graph C is that of an ellipsoid. As shown in the previous answer, the xy-traces of the equation and graph C are different. The equation for an ellipsoid also requires that all the coefficients in the equation be positive, and that the equation equal 1, and our equation does neither. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qThe 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because the graph of our equation has only one point in the xz plane, there is no way its xy-trace could be an ellipse. Shown in red, graph G has two straight lines as its traces. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the shape of the trace of the graph in the plane y = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the shape of the trace where y = 1, we need to plug in that value. When we plug in 1 for y, our equation is 1 = 4x^2+9z^2. This is an ellipse equation, making the trace an ellipse. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the shape of the trace of the graph in the plane x = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the trace in the plane x = 1, we need to plug this value into our equation. When we plug this in, we find we end up with the equation 4 = y^2 – 9z^2. This is a hyperbola equation, so the trace is in the shape of a hyperbola. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the shape of the trace of the graph in the plane z = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the shape of the trace when z =1, we need to plug in that value. When we plug z = 1 into our equation, we find that our equation is 9 = y^2 – 4x^2. This is a hyperbola equation, making the shape of our trace a hyperbola. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. "