√99以上 f(x y z)=x^2 y^2 z^2 graph 299964-F x y z x 2 y 2 z 2 graph

Z=x^2y^2 WolframAlpha Volume of a cylinder?Subject to the constraint 2x2 (y 1)2 18 Solution We check for the critical points in the interior f x = 2x;f y = 2(y1) =)(0;For all x;y 2C, x 6= y, and 2(0;1), then we say that f(x) is strictly concave Intuitively, the graph of a convex function lies on or below any chord between two points on the graph

Surfaces Part 2

F x y z x 2 y 2 z 2 graph

F x y z x 2 y 2 z 2 graph-Question f1(x,y,z) = x^2 y^2 z^2 −1 = 0 f2(x, y, z) = 2x^2 y^2 − 4z = 0 f3(x,y,z) = 3x^2 −4yz^2 = 0 This system can be concisely represented as F(x) = 0, where F(x) = (f1, f2, f3)T , x=(x,y,z)T and 0 = (0,0,0)T (transpose written because these should be column vectors) Using matlab Starting with the initial condition x0 = (05, 01 2 f(x) 1 2 f(y) = 1 2 f(x) 1 2 f(x) = f(x) But this contradicts (6) Exercise For each function below, determine whether it is convex, strictly convex, strongly convex or none of the above f(x) = (x 1 3x 2)2 9

Sketching Surfaces In 3d

The gradient is =<12, 9 ,16>Z) At any point on the circles of intersection, these two vectors give the normal lines of the tangent planes of the two surfaces The two vectors are orthogonal at all such intersection points, since (x;y;z) 2(x;y;Solution The given surface is the zero level surface of the function F(x;y;z) = x 2y xz 2y2z So, the normal vector to the tangent plane at the point P(1;1;1) is given by rF(1;1;1) We have rF(x;y;z) = h2xy z 2;x2 4yz;2xz 2yi=)rF(1;1;1) = h3;

→n dS, where → F = bxy2,bx2y,(x2 y2)z2 and S is the closed surface bounding the region D consisting of the solid cylinder x2 y2 6 a2 and 0 6 z 6 b Solution This is a problem for which the divergence theorem is ideally suited Calculating the divergence of → F, we get → ∇
Plane z = 1 The trace in the z = 1 plane is the ellipse x2 y2 8 = 1(e) Below is the graph of z = x2 y2 On the graph of the surface, sketch the traces that you found in parts (a) and (c) For problems 1213, nd an equation of the trace of the surface in the indicated plane Describe the graph of the trace 12 Surface 8x 2 y z2 = 9;

Steps to graph x^2 y^2 = 4In the first function, (x, y, z) (x, y, z) represents a point in space, and the function f f maps each point in space to a fourth quantity, such as temperature or wind speed In the second function, (x, y) (x, y) can represent a point in the plane, and t t can represent time3Dplot of x^2y^2z^2=1 Learn more about isosurface;

Surfaces Part 2

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The level curves of f(x,y) = x 2 y 2 are curves of the form x 2 y 2 =c for different choices of c These are circles of radius square root of c Several of them are shown below One can think of the level curve f(x,y)=c as the horizontal crosssection of the graph at height z=c When each level curve f(x,y)=c is plotted at a height of c units above the xyplane, we get the figure3dprinting, solidworks f(0,0,0) is 0, not 1 (the isosurface level), so you only get points drawn completing the cones if there are enough points near the origin that happen to have value 1 But when you switch to linspace(,,), the closest coordinates to the origin are at about 105, leaving a gap of about 21(ii) The Graph Of 3 = X^2 2y^2 2 1 Both (i) And (ii) Are Counter Maps 2 Both (i) And (ii) Are Contour Maps 3 (i) Is A Level Curve, And (ii) Is A Counter Map 4 (ii) Is A Level Curve

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This says that the gradient vector is always orthogonal, or normal, to the surface at a point So, the tangent plane to the surface given by f (x,y,z) = k f ( x, y, z) = k at (x0,y0,z0) ( x 0, y 0, z 0) has the equation, This is a much more general form of the equation of a tangent plane than the one that we derived in the previous sectionGraph y^2z^2=9 y2 z2 = 9 y 2 z 2 = 9 Reorder terms x2 y2 = 9 x 2 y 2 = 9 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 theThe gradient is a vector gradf=((delf)/(delx), (delf)/(dely), (delf)/(delz)) f(x,y,z)=3x^2yy^3z^2 (delf)/(delx)=6xy (delf)/(dely

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This tool graphs z = f (x,y) mathematical functions in 3D It is more of a tour than a tool All functions can be set different boundaries for x, y, and z, to maximize your viewing enjoyment This tool looks really great with a very high detail level, but you may find it more comfortable to use less detail if you want to spin the modelThe graph of a 3variable equation which can be written in the form F(x,y,z) = 0 or sometimes z = f(x,y) (if you can solve for z) is a surface in 3D One technique for graphing them is to graph crosssections (intersections of the surface with wellchosen planes) and/or tracesThe vector f x, f y is very useful, so it has its own symbol, ∇ f, pronounced del f'';

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Example 1 The graph of z = f ( x, y) as a surface in 3 space can be regarded as the level surface w = 0 of the function w ( x, y, z) = z − f ( x, y) Example 2 Spheres x 2 y 2 z 2 = r 2 can be interpreted as level surfaces w = r 2 of the function w = x 2 y 2 z 2Math Input NEW Use textbook math notation to enter your math Try itTwo Model Examples Example 1A (Elliptic Paraboloid) Consider f R2!R given by f(x;y) = x2 y2 The level sets of fare curves in R2Level sets are f(x;y) 2R 2 x y2 = cg The graph of fis a surface in R3Graph is f(x;y;z) 2R3 z= x2 y2g Notice that (0;0;0) is a local minimum of f

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{0,1,2}, f(x) = (bx/3c,x − 3
1 f Z → Z, f(x) = 2x1 2 f R → R, f(x) = 2x1 3 f Z×Z → Z, f(x,y) = 2x2 y 4 f Z×Z∗ → Q, f(x,y) = x/y 5 f N → N ×In the negative z direction through the part of the surface z=g(x,y)=16x^2y^2 that lies above the xy plane (see the figure below) For this problem

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Let {eq}f(x,y,z)=x^2y^2z^2 {/eq} and let S be the level surface defined by f(x,y,z) = 4 (a) Find an equation for the plane tangent to S at {eq}P_{0}(1,1,2)4 Determine f x, f y, and f z for the following functions (a) f(x;y;z) = 1 p x2 y2 z2 Solution f x = x=(x2 y2 z2)3=2 The expression for f y (respectively f z) is the same, but interchange the roles of y (respectively z) and x (b) f(x;;y;z) = yzln(xy) Solution f x = ((yz)=(xy))(y) = yz=x f y = zln(xy) yz(1=xy)(x) = zlnxy z f z = yln(xy) 5The graph of something like z = f(x;y) is a surface in threedimensional space Such graphs are usually quite di–cult to draw by hand Since z = f(x;y) is a function of two variables, if we want to difierentiate we have to decide whether we are difierentiating with

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A sphere is the graph of an equation of the form x 2 y 2 z 2 = p 2 for some real number p The radius of the sphere is p (see the figure below) Ellipsoids are the graphs of equations of the form ax 2 by 2 cz 2 = p 2, where a, b, and c are all positiveCalculate surface integral ∬ S f (x, y, z) d S, ∬ S f (x, y, z) d S, where f (x, y, z) = z 2 f (x, y, z) = z 2 and S is the surface that consists of the piece of sphere x 2 y 2 z 2 = 4 x 2 y 2 z 2 = 4 that lies on or above plane z = 1 z = 1 and the disk that is enclosed by intersection plane z = 1 z = 1 and the given sphere (FigureHere, x means the cross product Note, one may have to multiply the normal vector r_u x r_v by 1 to get the correct direction Example Find the flux of the vector field <y,x,z>

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Example 5 X and Y are jointly continuous with joint pdf f(x,y) = (e−(xy) if 0 ≤ x, 0 ≤ y 0, otherwise Let Z = X/Y Find the pdf of Z The first thing we do is draw a picture of the support set (which in this case is the firstSetting \(f (x, y, z) = z^2 – x^2 y^2 = C\text{,}\) we can write a single implicit equation for each level surface For \(C = 2\text{,}\) we will plot the implicit equation \(z^2 x^2 y^2 = 2\) For \(C = 2\text{,}\) we will plot the implicit equation \(z^2 x^2 y^2 = 2\) To do this,Now f(x,y) = x2 k2x2 = (p x2 k2x2)2 So the crosssection is the "same" parabola as in the xz and yz planes, namely, the height is always the distance from the origin squared This means that f(x,y) = x2 y2 can be formed by starting with z = x2 and rotating this curve around the z axis Finally, picking a value z = k, at whatpoints does f(x,y) = k?

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Bx/3c), where bxc = greatest integer less than or equal to x Solution 1 Onetoone It is not onto because theSelect a few x x values, and plug them into the equation to find the corresponding y y values The x x values should be selected around the vertex Tap for more steps Substitute the x x value − 1 1 into f ( x) = √ − x 1 f ( x) = x 1 In this case, the point is ( − 1, ) ( 1, )It is also called the gradient of f Example 1451 Find the slope of z = x 2 y 2 at ( 1, 2) in the direction of the vector 3, 4 We first compute the gradient at ( 1, 2) ∇ f = 2 x, 2 y , which is 2, 4 at ( 1, 2)

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Graph the surface f (x,y,z) = c I have a function f (x,y,z) = x^2 y^2 z^2 and I'd like to graph the surface defined by the equation f (x,y,z) = 1 When I type S x^2 y^2 z^2 = 1 into the input bar, this works perfectly;3;0i Thus, the equation of the tangent plane at (1;1;1) is 3(x 1) 3(y 1) = 0 =)x y= 0;Definition 2 The graph of a function f with the two variables x and y is the surface z = f(x,y) formed by the points (x,y,z) in xyzspace with (x,y) in the domain of the function and z = f(x,y) For a point (x,y) in the domain of the function, its value f(x,y) at (x,y) is determined by moving

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1) is a critical point The second derivative test f xx = 2;f yy = 2;f xy = 0 shows this a local minimum withGraph of z = f(x,y) New Resources Quiz Graphing Exponential Functions (Transformations Included)4 Find the volume and centroid of the solid Ethat lies above the cone z= p x2 y2 and below the sphere x 2y z2 = 1, using cylindrical or spherical coordinates, whichever seems more appropriate Recall that the centroid is the center of mass of the solid

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Answer (1 of 3) It's the equation of sphere The general equation of sphere looks like (xx_0)^2(yy_0)^2(zz_0)^2=a^2 Where (x_0,y_0,z_0) is the centre of the circle and a is the radious of the circle It's graph looks like Credits This 3D Graph is created @ code graphing calculatorFirst, remember that graphs of functions of two variables, z = f (x,y) z = f ( x, y) are surfaces in three dimensional space For example, here is the graph of z =2x2 2y2 −4 z = 2 x 2 2 y 2 − 4 This is an elliptic paraboloid and is an example of a quadric surface We saw several of these in the previous sectionMultivariable calculus If $F=(x^2,y^2,z^2),S=\{x^2y^2z^2=1,z\geq 0\}$, evaluate $\iint_S F dS$ Mathematics Stack Exchange I'm having trouble computing this In spherical coordinates we get$$\int_0^{2\pi}\int_0^{\pi/2}\sin^4\theta\cos^3\phi\sin^4\theta\sin^3\phi\cos^3\theta\sin\theta d\theta d\phi \tag 1$$which is

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The graph of a function f(x;y) = 8 x2 y) So, one surface we could use is the part of the surface z= 8 x 2 yinside the cylinder x2 y = 1 (right picture) 4 x y z x y z Let's call this surface Sand gure out how it should be oriented The original curve was parameterizedPiece of cake Unlock StepbyStep Natural Language Math Input2 We can describe a point, P, in three different ways Cartesian Cylindrical Spherical Cylindrical Coordinates x = r cosθ r = √x2 y2 y = r sinθ tan θ = y/x z = z z = z Spherical Coordinates x = ρsinφcosθ ρ = √x2 y2 z2 y = ρsinφsinθ tan θ = y/x z = ρcosφ cosφ = √x2 y2 z2 z

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Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack ExchangeDespite the fact that the same variables appear in every term, this expression does not factorize into one term There is no common factor, and no common bracket EAch term can be factored by differemce of squares xy(x^2y^2) yz(y^2z^2) zx(z^2x^2) =xy(xy)(xy) yz(yz)(yz) zx(zx)(zx) The only other option would be to multiply out the brackets and try aF(x;y)=œ xy(x2−y2) x2y2 (x;y)≠(0;0) 0 (x;y)=(0;0) Note fis continuous, (by computing lim(x;y)→(0;0) of the formula above, eg using polar coorinates) (a) Find f x and f y when (x;y)≠(0;0) Away from (0;0);fcan be di erentiated using the formula de ning it, as @f @x (x;y)= (x2 y2)y(x2 −y2)2x2y−2x2y(x2 −y2) (x 2y)2;

Calculate The Double Integral Over S Of F X Y Z Ds For Y 7 Z 2 0 Less Than Or Equal To X Less Than Or Equal To 7 0 Less

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S is defined as a sphereZ) = x2 y z2 = 0 for all such points since this last equality just says that the point lies on the coneThe yz plane creates a parabola in the downward direction x (1) = 1 = y z^2 the xz plane creates a hyperbole y (1) = 1 = x^2 z^2 We know that this creates a hyperbolic paraboloid (xy plane creates a parabola up, xy creates parabola down, shaped by a hyperbole from the top saddle like figure) the only hyperbolic paraboloid is graph V

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→ F = h∂x,∂y,∂zi
A quick video about graphing 3d for those who never done it before Pause the video and try itIf f (x, y, z) = x sin yz, (a) find the gradient of f and (b) find the directional derivative of f at (1, 3, 0) in the direction of v = i 2 j – k Solution (a) The gradient of f is ∇f (x, y, z) = 〈f x(x, y, z), f y(x, y, z), f z(x, y, z)〉 = 〈sin yz, xz cos yz, xy cos yz〉

Graphs Of Surfaces Z F X Y Contour Curves Continuity And Limits

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This means that x;y2 and f(x) = f(y) f(z);8z2 (6) But consider z= xy 2 By convexity of , we have z2 By strict convexity, we have f(z) = f x y 2 <

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