Compute answers using Wolfram's breakthrough technology &WolframAlpha brings expertlevel knowledge and capabilities to the broadest possible range of people—spanning all professions and education levelsAnswer 4@2z @x 2 = @2z @y 4 Form a PDE from the relation z = f 1(ax by) f 2(cx dy) Answer bd@ 2z @x2 (ad bc) @ z x y ac @2z @y2 = 0 5 Form a partial di erential equation from the relation z = f 1(2x 3y) f 2(4x 5y)Answer 15@ 2z @x2 22 @ 2z x y 8 @ z @y2 = 0 6 Form a partial di erential equation by eliminating the functions from z = f 1(x2 3y) f 2(x2 3y) Answer 9@ 2z @x 2 9 x @z 4x2 @2z @y = 0
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Y , we get p = ( x y ) f ' ( x 2 y 2 ) 2x f ( x 2 y 2 ) q = ( x y ) f ' ( x 2 yNov 18, 14x y = z c = = log( ) log log v x y z c 2 1 2 ( ) The general solution is given by F(u,v) = 0 F(x1 y1,(x y)z1) = 0 2solve x2 (y z) y2 (z x)q = z2 (x y) solution Auxiliary equations are given by dz dy dx 2 ( ) y 2 ( z x ) z2 (x y) x y z = = 45Click here👆to get an answer to your question ️ If x^2 y^2 z^2 xy yz zx = 0 then the value of x yz
Minimize the function f(x, y, z)=x^{2}y^{2}z^{2} subject to the constraints x2 y3 z=6 and x3 y9 z=9 Video Transcript So the question is gonna look a little bit different We instead of having one constraints, we're gonna find extreme valueFactors of x2y2z2xyyzzx, Factors of x^2y^2z^2xyyzzx, Factors of a2b2c2abbcca,Factors of a^2b^2c^2abbcca, Factors of p2q2r2pqqrpr(a)3 Find the PDE of all spheres whose centre lie on the (i) z axis (ii) xaxis 4 Form the partial differential equations by eliminating the arbitrary functions in the following cases (i) z = f (x y) (ii) z = f (x2 –y2) (iii) z = f (x2 y2 z2) (iv) f(xyz, x y z) = 0 (v) F (xy z2, x y z) = 0
Professionals For math, science, nutrition, historyCheck out a sample textbook solution See solution arrow_back Ch 145 Suppose that the equation F(x, y, z) = 0 Ch 145 EquationSince 0 = u xy u x = (u y u) x, we can integrate at once with respect to xto obtain u yu= f(y)This is a rst order linear \ODE in the variable y Introducing the integrating factor = exp R 1dy = ey, it becomes @y (e yu) = ef(y) Integrating with respect to ythis time yields
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Jun 01, 19Verify GDT for vector F = (x 2 yz)vector i (y 2 zx)j (z 2 xy)k taken over the rectangular parallelepiped 0 ≤x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ cClick here👆to get an answer to your question ️ If the electrostatic potential is given by ϕ = ϕ 0(x^2 y^2 z^2) where ϕ 0 is constant, then the charge density of the given potential would beLet f(x,y,z) =x2−y2−z2 f ( x, y, z) = x 2 − y 2 − z 2 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 P 0(1,−1,−2) P 0
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Knowledgebase, relied on by millions of students &Compute answers using Wolfram's breakthrough technology &Ie, the surfaces orthogonal to the system 244) are the surfaces generated by the integral curves of the equations dx ˆ f ˆ x = dy ˆ f ˆ y = dz ˆ f ˆ z (248) Example 271 Find the surface which intersects the
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The linear PDE 247) is therefore the general PDE determining the surfaces orthogonal to members of the system 244);Jul 25, 11∂x 3 ∂x ∂y ∂x∂y 2 ∂yAns Auxiliary equation m 3 − 2m 2 − 4 m 8 = 0 m = 2,2,−2 Solution is z = f 1 ( y 2 x ) xf 2 ( y 2 x ) f 3 ( y − 2 x ) Part B(1)(i) Form a partial differential equation by eliminating arbitrary functions fromz = xf ( 2 x y ) g ( 2 x y )(ii) Solve p 2 y (1 x 2 ) = qx 2(2)(i) Solve x( zThe problem is I have to find all the possible combination of integers (x, y, z) that will satisfy the equation x^2 y^2 z^2 = N when you are given an integer N You have to find all the unique tuples (x, y, z) For example, if one of the tuple is (1, 2, 1), then (2, 1, 1) is not unique anymore
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If U = F ( Y − X X Y , Z − X X Z ) , Show that X 2 ∂ U ∂ X Y 2 ∂ U ∂ Y Z 2 ∂ U ∂ Z = 0 Applied Mathematics 1 Sum If u =`f ( (yx)/ (xy), (zx)/ (xz)),` show that `x^2 (delu)/ (delx)y^2 (delu)/ (dely)z^2 (delu)/ (delz)=0` Advertisement Remove all adsF2(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)Z−c) = 0 (e) f(xyz) = xyz 4Find the general solution of the following partial differential equations (a) cos(xy)psin(xy)q= z (b) pyqx= xyz 2 (x 2 −y 2 ) (c) (x 2 2y 2 )p−xyq= xz,
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If cos(xyz) = 1 x 2 y 2 z 2, find ∂ z ∂ x and ∂ z ∂ y check_circle Expert Solution Want to see the full answer?Find the slopeintercept form of the equation of the line with the given propertiesForm a PDE by eliminating the arbitrary function (i) $\displaystyle F(xyz^2,xyz)=0 $ (ii) $\displaystyle F(xyz,x^2y^2z^2)=0 $ How do I proceed for any of these?
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See the answer Show transcribed image text Expert Answer 100% (1 rating) Previous question Next question Transcribed Image Text from this Question x^2 y^2 z^2 = 0 in a 3D graph Get more help from Chegg1) z = f(x 2 y2) Differentiating z partially w rt x and y, f x y y y z f x y x q x z p '(2 2)2 , '(2 2)2 p /q = x / y or y p –x q=0 as the pde (2 ) z = f ( x ct ) g (x ct) Differentiating z partially with respect to x and t, '( ) '( ), ( ) ( ) 2 2 f x ct g x ct x z f x ct g x ct x z Thus the pde is (3 ) x y z = f(xJul 13, 10The graph of w = f(x, y, z) is the set of ordered quadruples (x, y, z, w) such that w = f(x, y, z) Such a graph requires four dimensions three for the domain and one for the range Example w = x 2 y 2 z 2 Here the domain is all of R 3, and the range is {w w >= 0}
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Aug 12, 11Take df/dx df/dy df/dz = 0 f(x^2y^2,zxy) = 0 >Answer to Find the work done by the force field F(x, y, z) = \{x y^2, yz^2,z x^2\} on a particle moving along the line segment from (0,0,1) for Teachers for Schools for Working ScholarsMar 27, 21Show that the integral surface of the PDE x(y 2 f)f x y(x 2 f) = (x 2 y 2 )f which passes through the curve xf =a 3 , y = 0 is given by the form f 3 (x 3 y) 2 = a 9 (x y) 3 , where a is a constant Show that a local solution of the partial differential equation f t af x =f
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Partial Differential Equations 11 aaaaa 673 111 INTRODUCTION A relation between the variables (including the dependent one) and the partial differential coefficients of the dependent variable with the two or more independent variables is called F(x y z, x2 y2 z2) = 0Stack Exchange network consists of 176 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 ExchangeX^2 y^2 z xy = 0 then the pde is 2x y 2y x 1 = 0 x y 1 = 0
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Feb 23, 17 x3y4z = 0 First we rearrange the equation of the surface into the form f(x,y,z)=0 x^22z^2 = y^2 x^2 y^2 2z^2 = 0 And so we have our function f(x,y,z) = x^2 y^2 2z^2 In order to find the normal at any particular point in vector space we use the Del, or gradient operator grad f(x,y,z) = (partial f)/(partial x) hat(i) (partial f)/(partial y) hat(j) (partial f)/(partial zMath 311 Spring 13 Solutions to Assignment # 3 Completion Date Wednesday May 15, 13 Question 1 p 56, #10 (a) Use the theorem of Sec 17 to show that limTo show that `z(x,y)=ln(x^2y^2)` is a solution of the pde `{partial^2 z}/{partial x^2}{partial^2 z}/{partial y^2}=0` , we need to calculate the left side of the pde using the function `z(x,y
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X^2y^2z^2xyyzzx=0 multiplying the RHS and LHS by 2 we get , 2 x^2y^2z^2xyyzzx =0 or, (xy)^2(yz)^2(zx)^2=0 since in LHS there areF(x, y, z, p, q ) = 0 Example 5 Obtain the partial differential equation by eliminating „f„from z = ( xy ) f ( x 2 y 2 ) Let us now consider the equation z = (xy ) f(x 2 y 2) _____ (1) Differentiating (1) partially wrt x &A) \( \Large \phi \left(xyz, \frac{y}{z}\right)=0\) B) \( \Large \phi \left(\frac{y}{z},\frac{y}{x^{2}y^{2}z^{2}}\right) =0\) C) \( \Large \phi \left(\frac{y}{2
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Apr 28, 12Let F = (y2 z3,x3 z2,xz) Evaluate RR ∂W FdS for each of the following regions W A x^2 y^2 ≤ z ≤ 3 B x^2y^2 ≤z≤3, x≥0 C x^2y^2 ≤z≤3, x≤0 Yahoo Answers is shutting down on May 4th, 21 (Eastern Time) and beginning April th, 21 (Eastern Time) the Yahoo Answers website will be in readonly modeLetting mathS/math denote the surface of integration, we need to compute the surface integral math\displaystyle \iint_S \textbf{F} \cdot d\textbf{S} \tag*{}/math in two different ways Using the Divergence Theorem (letting mathR/math dX 2z2 dS, where Sis the part of the cone z2 = x2 y between the planes z= 1 and z= 3 The widest point of Sis at the intersection of the cone and the plane z= 3, where x2 y2 = 32 = 9;
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2 Lecture 1 { PDE terminology and Derivation of 1D heat equation Today † PDE terminology † Classiflcation of second order PDEs † Derivation of 1D heat equation Next † Boundary conditions † Derivation of higher dimensional heat equations Review † Classiflcation of conic section of the form Ax2 Bxy Cy2 DxEy F = 0;Knowledgebase, relied on by millions of students &Its thinnest point is where x 2 y = 12 = 1 Thus, Sis the portion of the surface z= p x2 y2 over the region D= f(x;y) 1 x2 y2 9g So ZZ S x2z2 dS = ZZ D
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And with u(x, 0) = f(x) and ∂u / ∂y (x, 0) = g(x) for all values of x Even more phenomena are possible For instance, the following PDE , arising naturally in the field of differential geometry , illustrates an example where there is a simple and completely explicit solution formula, but with the free choice of only three numbers and notAug 29, 15(delw)/(delx) = x/sqrt(x^2 y^2 z^2) (delw)/(dely) = y/sqrt(x^2 y^2 z^2) (delw)/(delz) = z/sqrt(x^2 y^2 z^2) Since you're dealing with a multivariable function, you must treat x, y, and z as independent variables and calculate the partial derivative of w, your dependent variable, with respect to x, y, and z When you differentiate with respect to x, you treat y and zApr 17, 16Let f be a twicedifferentiable function so that f(1)=2 and f(3)=7 Which must be true for the function f on the interval 1<
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Professionals For math, science, nutrition, historyUsing ,multipliers x,y,z we get (xdx ydy zdz)/x(x^2 y^2 z^2) (xdx ydy zdz)/x(x^2 y^2 z^2) = dx/(2xz) 2(xdx ydy zdz)/(x^2 y^2 z^2) = dz/z Integrating, log(x^2 y^2 z^2) = logz logc2 (x^2 y^2 z^2) = zc2 Hence the required solution is f(c1,c2) = 0 = f(y/z, (x^2 y^2 z^2)/z) = 0X^{2}y^{2}z^{2}=0 Subtract z^{2} from both sides y^{2}x^{2}z^{2}=0 Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{b±\sqrt{b^{2}4ac}}{2a}, once they are put in standard form ax^{2}bxc=0
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