Homework 3 Model Solution Han
Knowledgebase, relied on by millions of students &Parameterize the surface z = x 2 2 y 2 over the triangular region R enclosed by the lines y = 32 x / 3, y = 1 and x = 0 as shown in Figure 1555 (a) Solution While the region R in this example is very similar to the region R in the previous example, and our method of parameterizing the surface is fundamentally the same, it will
X^2+y^2+z^2=16 graph
X^2+y^2+z^2=16 graph-By using Pythagoras you would end up with the equation given where the 4 is in fact r^2 To obtain the plot points manipulate the equation as below Given x^2y^2=r^2 >How to plot 3 dimensional graph for x^2 y^2 = 1?
X 2 Y 2 16 Find The Foci And Vertices Of The Ellipse Youtube
This is a circle of radius 4 centred at the origin Given x^2y^2=16 Note that we can rewrite this equation as (x0)^2(y0)^2 = 4^2 This is in the standard form (xh)^2(yk)^2 = r^2 of a circle with centre (h, k) = (0, 0) and radius r = 4 So this is a circle of radius 4 centred at the origin graph{x^2y^2 = 16 10, 10, 5, 5}De ned by z= x 2 y2 and x 2y2 z2 = 7 at the point ( 1;1;2) Hint Think about the geometry of the gradient vectors You don't have to parametrize the curve to do this problem Solution The surface z= x2 y2 can be written as the level surface F(x;y;z) = x 2 y z= 0;F (x,y,z) = 0 x z f y k p R where p is a unit vector normal to the region R and ∇f
Since the surface is in the form x = f ( y, z) x = f ( y, z) we can quickly write down a set of parametric equations as follows, x = 5 y 2 2 z 2 − 10 y = y z = z x = 5 y 2 2 z 2 − 10 y = y z = z The last two equations are just there to acknowledge that we can choose y y and z z to be anything we want them to beExample Find the volume of the solid region above the cone z2 = 3(x2 y2) (z ≥ 0) and below the sphere x 2 y 2 z 2 = 4 Soln The sphere x 2 y 2 z 2 = 4 in spherical coordinates is ρ = 2In Figure 1118 (a), we show part of the graph of the equation x 2 y 2 = 1 by sketching 3 circles the bottom one has a constant zvalue of 15, the middle one has a zvalue of 0 and the top circle has a zvalue of 1 By plotting all possible zvalues, we get the surface shown in Figure 1118 (b) This surface looks like a "tube," or a
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Graph x^2y^2=16 x2 y2 = 16 x 2 y 2 = 16 This is the form of a circle Use this form to determine the center and radius of the circle (x−h)2 (y−k)2 = r2 ( x h) 2 ( y k) 2 = r 2 Match the values in this circle to those of the standard form The variable r r represents the radius of the circle, h h represents the xoffset from the origin, and k k represents the yoffset from originAnswer (1 of 8) Assuming you're only working with real numbers Rearange to get that x^2y^2=0^2 This is a circle of radius 0 cenetered the orgin But if our circle is of radius 0 and at the origin, that must mean one thing the graph is just the origin So
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