#$&* course mthb 272 Question: `qQuery problem 7.1.24 picture of sphere, diam from (-1,-2,1) to (0, 3, 3).What is the standard form of the equation of the pictured sphere?
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Given Solution: `a The midpoint between (-1, -2, 1) and (0, 3, 3) is (-1/2, 1/2, 2). Thus the sphere will have the form (x - (-1/2) )^2 + (y - 1/2)^2 + (z-2)^2 = r^2. r is half the diameter, which is half of `sqrt( (0 - -1)^2 + (3 - -2)^2 + (3 - 1)^2 ) = `sqrt(1+25+4) = `sqrt(30). The radius is therefore `sqrt(30) / 2 and r^2 = 30 / 4 = 15/2. The equation is therefore (x - (-1/2) )^2 + (y - 1/2)^2 + (z-2)^2 = 15/2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery problem 7.1.39 (was 7.1.38) yz-trace of x^2 + y^2 + z^2 - 6x - 10 y + 6 z + 30 = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 0 y^2 - 10 y + z^2 + 6z + 30 = 0 (y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 (y-5)^2 + (z+3)^2 = 4 (y-5)^2 + (z+3)^2 = 2^2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The yz trace is characterized by x = 0. The equation of the y-z trace is therefore y^2 - 10 y + z^2 + 6z + 30 = 0. This is the equation of a circle in the y-z plane. Completing the square we get (y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 or (y-5)^2 + (z+3)^2 = 4 or (y-5)^2 + (z+3)^2 = 2^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qWhat is the equation of the yz-trace of the given sphere and what shape does this equation define in the y-z plane? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 0 y^2 - 10 y + z^2 + 6z + 30 = 0 (y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 (y-5)^2 + (z+3)^2 = 4 (y-5)^2 + (z+3)^2 = 2^2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a `a The yz trace is characterized by x = 0. The equation of the y-z trace is therefore y^2 - 10 y + z^2 + 6z + 30 = 0. This is the equation of a circle in the y-z plane. Completing the square we get (y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 or (y-5)^2 + (z+3)^2 = 4 or (y-5)^2 + (z+3)^2 = 2^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qWhat is the center and what is the radius of the circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 0 y^2 - 10 y + z^2 + 6z + 30 = 0 (y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 (y-5)^2 + (z+3)^2 = 4 (y-5)^2 + (z+3)^2 = 2^2 radius 2, in the y-z plane, centered at (0, 5, -3) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a `a The yz trace is characterized by x = 0. The equation of the y-z trace is therefore y^2 - 10 y + z^2 + 6z + 30 = 0. This is the equation of a circle in the y-z plane. Completing the square we get (y^2 - 10 y + 25 - 25) + (z^2 + 6 z + 9 - 9) + 30 = 0 or (y-5)^2 + (z+3)^2 = 4 or (y-5)^2 + (z+3)^2 = 2^2. We thus have a circle of radius 2, in the y-z plane, centered at (0, 5, -3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course mthb 272 Question: `qQuery problem 7.2.6 intercepts and sketch graph of 2x - y + z = 4.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: `a The x-intercept occurs when y and z are 0, giving us 2x = 4 so x = 2. The y-intercept occurs when x and z are 0, giving us -y = 4 so y = -4. The z-intercept occurs when x and y are 0, giving us z = 4. The intercepts are therefore (2, 0, 0), (0, -4, 0) and (0, 0, 4). These three points form a triangle and this triangle defines the plane 2x - y + z = 4. This plane contains the triangle but extends beyond the triangle, extending infinitely far in all directions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qIf you released a marble on the plane at the point where it intercepts the z axis, it would roll down the incline. When the marble reached the xy plane would it be closer to the x axis or to the y axis? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (0,0,4) to (2,0,0) is the steepest possible path closer to the x axis confidence rating #$&*: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: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: (0,0,4) to (2,0,0), in the x-y plane, has slope 2 and is therefore steeper (0, 0, 4) to (0, -4, 0) in the yz plane has slope of magnitude 1 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The line from (0,0,4) to (2,0,0), in the x-y plane, has slope 2 and is therefore steeper than the line from (0, 0, 4) to (0, -4, 0) in the yz plane, which has slope of magnitude 1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery problem 7.2.34 (was 7.2.30) match y^2 = 4x^2 + 9z^2 with graph Which graph matches the equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qThe graph couldn't be (e). Explain why not. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: e) is set equal to 1 and the needed equation is set equal to 0 one has a constant term the other does not confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The equation for e) is set equal to 1 and the needed equation is set equal to 0. So one has a constant term while the other does not. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qThe graph could not be (c) because the picture shows that the surface is not defined for | y | < 1, while 4x^2 + 9z^2 = .25, for example, is the trace for y = 1/2, and is a perfectly good ellipse. State this in your own words. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y^2 = 4x^2 + 9z^2 y = 1/2 (1/2)^2 = 4x^2 + 9z^2, or 1/4 = 4x^2 + 9z^2 16 x^2 + 36 z^2 = 1 x^2 / [1/4^2] + y^2 / [ 1/6^2] confidence rating #$&*: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:3 ********************************************* 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: y^2 = 4x^2 + 9z^2 z = 0, is y^2 = 4 x^2 y = 2x and y = -2x confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a If y^2 = 4x^2 + 9z^2 then the xy trace, which occurs when z = 0, is y^2 = 4 x^2. This is equivalent to the two equations y = 2x and y = -2x, two straight lines. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qWhat is the shape of the trace of the graph in the plane y = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4 x^2 + 9 z^2 = 1 ellipse major axis 1/2 in the x direction and minor axis 1/3 in the z direction confidence rating #$&*: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:3 ********************************************* Question: `qWhat is the shape of the trace of the graph in the plane x = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y^2 - 9 z^2 = 4 hyperbola with vertices at y = +- 2, z = 0 asymptotes are the lines y = 3z and y = -3z in the plane x = 1 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):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qWhat is the shape of the trace of the graph in the plane z = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y^2 - 4 x^2 = 9 hyperbola with vertices at x = 0 and y = +- 3 asymptotes y = 2x and y = -2x in the plane z = 1 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):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!