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course Mth 277
4/16 3Had some trouble understanding what exactly I was supposed to do with the vector questions. I can rework if you let me know exactly where I'm wrong
Application to quadric surfaces
Consider first the surface z = 4 - x^2 - y^2.
What is the x-z trace of this surface (the x-z trace is the graph of the surface, confined to the x-z plane; we confine the graph to the x-z plane by letting y = 0)? What are the coordinates of the vertex, the x = 1 and the x = -1 points of this graph?
Letting y = 0 we have,
z = 4 - x^2, which is parabola shifted up 4 units( right because when x^2 = 0, z =4) so our vertex is (0, 4) in the xz-plane.
Coordinate at x=1 is (1, 3) and the same z value at x =-1, giving us the coordinate (-1, 3)
#$&*Can you sketch a graph of the x-z trace on an x-z coordinate system? Begin by sketching the vertex and the x = 1 and x = -1 points of the parabola.
Can you sketch this graph as it would appear within an x-y-z coordinate system? Begin by sketching the vertex and the x = 1 and x = -1 points of the parabola. With these points to guide you it shouldn't be difficult to make a reasonable sketch.
Yes, in the xz-plane it’s the standard parabola directed down with vertex at (0, 4) because as |x| of x increases z decreases(This is because we have the form z = `alpha - x^2, I mean the reason the parabola is pointed downwards? If we had the form z = x^2 - `alpha the parabola would have to have a direction upwards correct?)
@& Right. Consider the vertex (0, 4) and the points (1, 3) and (-1, 3). These points alone tell you that the parabola opens downward.
The - coefficient of x^2 will always cause the vertex to be the highest point.*@
As to the sketch in the xyz coordinate system the shape would be the elliptic paraboloid because the yz-plane has the same appearance as the xz-plane, so it would look like a teacup of some sort(kind of maybe, may not be the best description)
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What is the x-y trace of this surface?
It would be a circle, which I can see because of seeing the other graphs.
Also I’m pretty sure we find the trace by placing z = 0 which gives us 0 = 4 - x^2 - y^2, which can be rearranged to x^2+y^2 = 4. This is the equation of a circle with 4 = r^2 so r = `sqrt(4) = 2. So in xy-plane we have a circle of radius 2.
@& Right. This is the reason the x-y trace is a circle.*@
So does that mean that the elliptic paraboloid extends down through the xy-plane, or actually thinking back to what we have been working on putting together it actually would have the end that comes to the vertex coming UP through the xy-plane?
So if we wanted to see what’s going on at 1 unit above xy-plane, instead of having z =0 we would set z = 1. This would actually be the process of finding the level planes, I mean by allowing z to take on different values and putting together what’s going on in xy-plane. I knew what was going on with the level curve but never connected it to this process before, I know that sounds not possible to make that connection but in my brain level curves logic = very easy to grasp, seeing random objects in the xy, xz, yz much less xyz planes = overwhelmingly confusing.
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Can you sketch the x-y trace in an x-y coordinate system? Begin by finding and sketching the points where x = 0, and the points where y = 0.
Can you sketch the x-y trace in an x-y-z coordinate system? Begin by finding and sketching the points where x = 0, and the points where y = 0. With these points to guide you it shouldn't be difficult to make a reasonable sketch.
In response to xy-trace and xy-coordinate system, yes I can. Is this done by setting x =0 and z =0 to find y pt, and y =0 and z =0 to find x pt? You didn’t say anything about the z variable but I expect that was because it seems pretty much like common sense that if we are looking to find pt in xy-plane then z must equal 0.
We would have x =2 and y =2 at the pts where y =0 and x =0 respectively, that is if we also set z =0 for both calculations.
@& The x-y trace is the intersection of the surface with the x-y plane. In the x-y plane z = 0, at all points of the plane. So there would be no z coordinate in an xy graph of the trace.
The x-y trace is the curve x^2 + y^2 = 4, as you have shown. The x = 0 points of the xy trace are (0, 1) and (0, -1). The y = 0 points are (1, 0) and (-1, 0). If you plot these four points, they can help guide your sketch of the circle.
Then when you plot these same point in a xyz system, they will help you sketch the same circle, in 3 dimensions instead of two.*@
Ok I guess this answer one of the questions I just asked. Looking at the xy-trace in the xyz-coordinate system we could let z = 0 as well as x or y only at the plane where the x axis, y axis and z axis all come together. Moving up or down from that plane we would have to take into account the change in z in our equation.
Yes I can sketch the trace in the xyz coordinate system, it appears wrt xy-trace we are stacking smaller and smaller circles as we move up towards the z plane that is equal to 4.
If we had an equation where there was no constant, as we had in this equation(the 4). We would have the cylinder shape and -x^2 - y^2 would be the same no matter what the value of z was. Is that what you are referring to when you say we can generalize the form to include z, meaning the equation is set up so that no matter what level of z we are at the expression -x^2-y^2 is the same as the level of z above or below it???
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Go back to your sketches for the plane x = 1. Look at the two-dimensional sketch you made for the y-z plane. Sketch in the unit vectors `j and `k, each vector originating from the point (0, 0).
Find the y = 1 point on your graph, and find the slope of your graph at that point. Sketch a vector with `j component 1, in the direction of the tangent line, originating at the y = 1 point of your curve. What is the `k component of your vector?
m = -1, at (1,2), I’m confused about what I’m supposed to be putting together. Ihave the slope, the two unit vectors at (0,0) w/ 1 running along y axis and the other running along the z axis, have the tang. line at the pt (1,2), believe I know what needs to be done `j vector of component 1 in direction of tang vector(whould that be neg, with the vector moving from the pt(1,2) back at an angle in the direction back towards the y axis??)
Ok well looking through the next few problems maybe this is what you are wanting me to do. I originally drew unit vectors where `k vector moved 1 unit up z axis(from (0,0) to (0,1)) and j vector moved one unit across y axis(from (0,0) to (1,0)), but maybe they were supposed to have length of `k = 3 and `j = `sqrt(3). If so moving to the right one space our fn then equals 0 at(2,0), k would be -2 I believe.
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@& Your curve is z = 3 - y^2.
The slope of that curve at a y value would be the derivative evaluated at that point.
The derivative is z_y = -2 y.
So the slope at the y = 1 point would be -2.
If you were to sketch a vector tangent to the plane, it would have to have a slope of -2.
If the vector has y component `j, then in order to have slope -2, the `k component would have to be -2. So your tangent vector would be `j - 2 `k.*@
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Self-critique (if necessary):
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@& Looks good overall. Make sure you've done the 2- and 3-dimensional sketches correctly. The sketching part of the exercise really helps focus your thinking.
See my last note. You don't appear to have really understood the question about the vectors at that point, though it looks like you did later. If you're not sure you get the whole picture, feel free to submit a revision, marking your insertions with &&&&.