#$&* course MTH 272 Resubmitted on 12/17 about 2:43 pm ***This is a resubmission***023.
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Given Solution: `a The x-intercept occurs when y and z are 0, giving us 2x = 4 so x = 2. The y-intercept occurs when x and z are 0, giving us -y = 4 so y = -4. The z-intercept occurs when x and y are 0, giving us z = 4. The intercepts are therefore (2, 0, 0), (0, -4, 0) and (0, 0, 4). These three points form a triangle and this triangle defines the plane 2x - y + z = 4. This plane contains the triangle but extends beyond the triangle, extending infinitely far in all directions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qIf you released a marble on the plane at the point where it intercepts the z axis, it would roll down the incline. When the marble reached the xy plane would it be closer to the x axis or to the y axis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Slope from z int. to y int. is 1 and slope from z int. to x int. is 2. Therefore the triangle is steepest towards the x axis side and the marble would come down closer to the x axis. confidence rating #$&*: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):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* 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: Slope from z int. to y int. is 1 and slope from z int. to x int. is 2. Therefore the triangle is steepest towards the x axis and the climb would be steepest towards the x axis. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The line from (0,0,4) to (2,0,0), in the x-y plane, has slope 2 and is therefore steeper than the line from (0, 0, 4) to (0, -4, 0) in the yz plane, which has slope of magnitude 1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 7.2.34 (was 7.2.30) match y^2 = 4x^2 + 9z^2 with graph Which graph matches the equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All the variables are squared and the equation is equal to zero when all variables are grouped to the same side. There is one negative variable so this is an elliptic cone (B) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qThe graph couldn't be (e). Explain why not. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When rearranged, it is equal to zero. To be a hyperboloid of two sheets (E), it would need to be equal to 1. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The equation for e) is set equal to 1 and the needed equation is set equal to 0. So one has a constant term while the other does not. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qThe graph could not be (c) because the picture shows that the surface is not defined for | y | < 1, while 4x^2 + 9z^2 = .25, for example, is the trace for y = 1/2, and is a perfectly good ellipse. State this in your own words. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At y=(1/2) (1/2)^2 = 4x^2 +9z^2 1 = x^2/(1/4)^2 + z^2/(1/6)^2 This shows an ellipse in the xz plane with the points (-1/4,0,0)(0,0,1/6)(1/4,0,0)(0,0,-1/6) In the picture for (C) at y=(1/2) it shows near the following points (-3,0,0)(0,0,3)(3,0,0)(0,0,-3) confidence rating #$&*: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):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qThe graph couldn't be (c). Explain why not. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At y=(1/2) (1/2)^2 = 4x^2 +9z^2 1 = x^2/(1/4)^2 + z^2/(1/6)^2 This shows an ellipse in the xz plane with the points (-1/4,0,0)(0,0,1/6)(1/4,0,0)(0,0,-1/6) In the picture for (C) at y=(1/2) it shows near the following points (-3,0,0)(0,0,3)(3,0,0)(0,0,-3) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a In the plane y = 1/2 the trace of y^2 = 4x^2 + 9z^2 is found by substituting y = 1/2 into this equation. We obtain (1/2)^2 = 4x^2 + 9z^2, or 1/4 = 4x^2 + 9z^2. Multiplying both sides by 4 we get the 16 x^2 + 36 z^2 = 1, which can be expressed as x^2 / [1/4^2] + y^2 / [ 1/6^2]. This is the standard form of an ellipse with major axis 1/4 in the x direction and minor axis 1/6 in the y direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qThe trace of this graph exists in each of the coordinate planes, and is an ellipse in each. The graph of the given equation consists only of a single point in the xz plane, since there y = 0 and 4x^2 + 9z^2 = 0 only if x = z = 0. Explain why the xy trace is not an ellipse. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Z=0 for xy trace so Y^2=4x^2 and y=+/- 2x these are linear functions which show the outer two edges of the cone. confidence rating #$&*: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):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat is the shape of the trace of the graph in the plane y = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1^2=4x^2+9z^2 1=x^2/(1/2)^2 + z^2/(1/3)^2 This is an ellipse in the plane y=1 with points +/- 1/2 on the x axis and +/- 1/3 on the z axis confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a In the plane y = 1 the trace of y^2 = 4x^2 + 9z^2 becomes 4 x^2 + 9 z^2 = 1, which is an ellipse. In standard form the ellipse is x^2 / [ 1 / 2^2 ] + z^2 / [ 1 / 3^2 ] = 1, so has major axis 1/2 in the x direction and minor axis 1/3 in the z direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qWhat is the shape of the trace of the graph in the plane x = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y^2=4*1^2+9z^2 Y^2-9z^2=4 Y^2/(1/2)^2 - z^2/(2/3)^2 = 1 This is a hyperbola in the plane x=1 vertex +/- 2 along y axis M=((2/3)-0)/(2-0 = 1/3 so the asymptotes have the equations y= +/- (1/3) z confidence rating #$&*: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. STUDENT QUESTION #### From our form of equation, or after solving for x = 1, how is the hyperbola exactly found? I see where this is in the text, but just not getting this exactly clear, at least at this very point. INSTRUCTOR RESPONSE The ellipse and hyperbolas corresponding to the equation +- x^2 / a^2 +- y^2 / b^2 = 1 are all constructed based on the rectangular 'box' bounded by the lines x = +- a and y = +- b. The x and y axes. and these two lines, are plotted in the figure below. The same figure with the ellipes x^2 / a^2 + y^2 / b^2 = 1: Note that the intercepts (a, 0), (-a, 0), (0, b), (0, -b) lie on the graph of the ellipse. If you plug the coordinates of any of these points into the equation you get 1 = 1. The original 'box' with the lines y = b/a * x and y = -b/a * x. The same, with the hyperbola x^2 / a^2 - y^2 / b^2 = 1 Note that of the intercepts (a, 0), (-a, 0), (0, b), (0, -b) of the box, only (a, 0) and (-a, 0) lie on the graph of the hyperbola. If you plug in the coordinates of (0, +- b) you get -1 = 1. Furthermore if | y | < b/a * | x | the left-hand side is negative, so the graph is excluded completely from the corresponding region of the plane. For large x and y, the 1 on the right-hand side becomes insignificant and the graph approaches one of the lines y = +- b/a * x. The same, with the hyperbola -x^2 / a^2 + y^2 / b^2 = 1 Note that of the intercepts (a, 0), (-a, 0), (0, b), (0, -b) of the box, only (0, b) and (0, -b) lie on the graph of the hyperbola. If you plug in the coordinates of (+- a, 0) you get -1 = 1. Furthermore if | y | > b/a * | x | the left-hand side is negative, so the graph is excluded completely from the corresponding region of the plane. For large x and y, the 1 on the right-hand side becomes insignificant and the graph approaches one of the lines y = +- b/a * x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):???You have the asymptotes y=+/-3z and I had y=+/-(1/3)z. I solved for m using the two point and plugged into y=mx+b. Is this wrong??? ------------------------------------------------ Self-critique 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):???Why is x=0 a vertex??? ------------------------------------------------ Self-critique Rating:3