#$&* course Mth 272 07/21 7:30 Question: `qQuery problem 7.2.52 (was 7.2.48) identify quadric surface z^2 = x^2 + y^2/2.
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Ok ********************************************* Question: `qWhat is the name of this quadric surface, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The name of the surface is an elliptic paraboloid because xz and yz planes are parabolas. The xy is an ellipse. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a f z = c, a constant, then x^2 + y^2/2 = c^2, or x^2 / c^2 + y^2 / (`sqrt(2) * c)^2 = 1. This gives you ellipse with major axis c and minor axis `sqrt(2) * c. Thus for any plane parallel to the x-y plane and lying at distance c from the x-y plane, the trace of the surface is an ellipse. In the x-z plane the trace is x^2 - z^2 = 0, or x^2 = z^2, or x = +- z. Thus the trace in the x-z plane is two straight lines. In the y-z plane the trace is y^2 - z^2/2 = 0, or y^2 = z^2/2, or y = +- z * `sqrt(2) / 2. Thus the trace in the y-z plane is two straight lines. The x-z and y-z traces show you that the ellipses in the 'horizontal' planes change linearly with their distance from the x-y plane. This is the way cones grow, with straight lines running up and down from the apex. Thus the surface is an elliptical cone. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: Ok ********************************************* Question: `qGive the equation of the xz trace of this surface and describe its shape, including a justification for your answer. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y=0 z^2 = x^2 + (y^2)/4 z^2 = x^2 It's a parabola because the term is squared. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a The xz trace consists of the y = 0 points, which for z^2 = x^2 + y^2/2 is z^2 = x^2 + 0^2/2 or just z^2 = x^2. The graph of z^2 = x^2 consists of the two lines z = x and z = -x in the yz plane. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: Ok ********************************************* Question: `qDescribe in detail the z = 2 trace of this surface. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if z=2, 4= x^2 + y^2/2 The shape is an ellipse. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a If z = 2 then z^2 = x^2 + y^2/2 becomes 2^2 = x^2 + y^2 / 2, or x^2 + y^2 / 2 = 4. This is an ellipse. If we divide both sides by 4 we can get the standard form: x^2 / 4 + y^2 / 8 = 1, or x^2 / 2^2 + y^2 / (2 `sqrt(2))^2 = 1. This is an ellipse with major axis 2 `sqrt(2) in the y direction and 2 in the x direction. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat is the name of this quadric surface, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The name of the surface is an elliptic paraboloid because xz and yz planes are parabolas. The xy is an ellipse. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a f z = c, a constant, then x^2 + y^2/2 = c^2, or x^2 / c^2 + y^2 / (`sqrt(2) * c)^2 = 1. This gives you ellipse with major axis c and minor axis `sqrt(2) * c. Thus for any plane parallel to the x-y plane and lying at distance c from the x-y plane, the trace of the surface is an ellipse. In the x-z plane the trace is x^2 - z^2 = 0, or x^2 = z^2, or x = +- z. Thus the trace in the x-z plane is two straight lines. In the y-z plane the trace is y^2 - z^2/2 = 0, or y^2 = z^2/2, or y = +- z * `sqrt(2) / 2. Thus the trace in the y-z plane is two straight lines. The x-z and y-z traces show you that the ellipses in the 'horizontal' planes change linearly with their distance from the x-y plane. This is the way cones grow, with straight lines running up and down from the apex. Thus the surface is an elliptical cone. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: Ok ********************************************* Question: `qGive the equation of the xz trace of this surface and describe its shape, including a justification for your answer. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y=0 z^2 = x^2 + (y^2)/4 z^2 = x^2 It's a parabola because the term is squared. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a The xz trace consists of the y = 0 points, which for z^2 = x^2 + y^2/2 is z^2 = x^2 + 0^2/2 or just z^2 = x^2. The graph of z^2 = x^2 consists of the two lines z = x and z = -x in the yz plane. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: Ok ********************************************* Question: `qDescribe in detail the z = 2 trace of this surface. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if z=2, 4= x^2 + y^2/2 The shape is an ellipse. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a If z = 2 then z^2 = x^2 + y^2/2 becomes 2^2 = x^2 + y^2 / 2, or x^2 + y^2 / 2 = 4. This is an ellipse. If we divide both sides by 4 we can get the standard form: x^2 / 4 + y^2 / 8 = 1, or x^2 / 2^2 + y^2 / (2 `sqrt(2))^2 = 1. This is an ellipse with major axis 2 `sqrt(2) in the y direction and 2 in the x direction. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!