course Mth 272 7/31 2:30 pm 024.
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the name of this quadric surface, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This quadratic surface is an elliptic cone. We can tell that tis is an elliptic cone from the equation and from its traces. If z, the axis the elliptic cone is centered around, was a constant, the xy-trace would be an ellipse. Both the xz and yz-traces would consist of two straight lines, x = +-z and y = +-z. All three of these traces are characteristics of an elliptic cone. 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: ********************************************* Question: `qGive the equation of the xz trace of this surface and describe its shape, including a justification for your answer. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The equation for the xz-trace is that of two straight lines. We can find this by plugging in the quantity of 0 for y, making our equation x^2 = z^2. If we take the square root of this equation, we end up with x = +-z, two straight lines. 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: ********************************************* Question: `qDescribe in detail the z = 2 trace of this surface. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the z=2 trace of this surface, we need to plug in 2 for z and solve. When we plug in 2 for z, our equation is 4 = x^2 +y^2/2. This then becomes 1= (x^2+y^2/2)/4 = x^2/4 + y^2/8. This makes the shape of our trace an ellipse because it is in the standard form of an ellipse equation. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.
course Mth 272 7/31 2:30 pm 024.
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the name of this quadric surface, and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This quadratic surface is an elliptic cone. We can tell that tis is an elliptic cone from the equation and from its traces. If z, the axis the elliptic cone is centered around, was a constant, the xy-trace would be an ellipse. Both the xz and yz-traces would consist of two straight lines, x = +-z and y = +-z. All three of these traces are characteristics of an elliptic cone. 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: ********************************************* Question: `qGive the equation of the xz trace of this surface and describe its shape, including a justification for your answer. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The equation for the xz-trace is that of two straight lines. We can find this by plugging in the quantity of 0 for y, making our equation x^2 = z^2. If we take the square root of this equation, we end up with x = +-z, two straight lines. 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: ********************************************* Question: `qDescribe in detail the z = 2 trace of this surface. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the z=2 trace of this surface, we need to plug in 2 for z and solve. When we plug in 2 for z, our equation is 4 = x^2 +y^2/2. This then becomes 1= (x^2+y^2/2)/4 = x^2/4 + y^2/8. This makes the shape of our trace an ellipse because it is in the standard form of an ellipse equation. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.