course Mth 272 August 13, 2009 at 9:18 AM Question: `qQuery problem 7.2.6 intercepts and sketch graph of 2x - y + z = 4.
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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: it would reach the x axis first because (0,0,4) is closer to (2,0,0) than (0,-4,0). 2 is closer to 0, the intersection of the 3 lines than -4. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: the slope from the x axis is 4/2=2 and the slope of the y axis is 4/4=1, so it is steeper on the x axis. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: it matches (g) in the textbook, which is a set of cones, whose tips touch at the point 0 on the y axis. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qThe graph couldn't be (e). Explain why not. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the graph of (e) is set equal to 1. This is obtained based on the graph starting at 1 and -1. For the graph to correspond to the given equation, it should intersect at 0. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 of © is the shape of an oval, which intersects on the x,y, and z axis. For it to correspond with the given equation, it must intersect at 0 for all. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: this is the exact same as the last question. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: only an xy trace would make the shape two dimensional. Therefore, the graph would be an “X” shape, which would intersect at (0,0). confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: the equation would then be 4x^2+9z^2=1, which is an ellipse confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: the equation is then y^2-9z^2=4, which is a hyperbola parallel to the yz-plane confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: -4x^2+y^2=9 which is a hyperbola that is parallel to the xy-plane confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 "