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1 (Exercise 12) Find the maximum and minimum of f(x;y;z) = x4 y4 z4 subject to the constraint x 2y2 z = 1 Solution We have ∇f(x;y;z) = 4x3;4y3;4z3 = 2 x;2 y;2 z = ∇g(x;y;z) Case 1 If all of x;y;z ̸= 0, we can divide 4x3 = 2 x, 4y3 = 2 y, 4z3 = 2 z by 4x;4y;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)Are expressed in the usual manner except that the independent variable z= xiyis complex Thus f(z) has a real part u(x,y) and an imaginary part v(x,y) f(z) = u(x,y) iv(x,y) (12) Extra difficulties appear in differentiating and integrating such functions becausezvaries in a plane and not on a line
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Z ∞ −∞ f(x,y)dydx = Z 1 0 Z 2 0 (cx2 xy 3)dydx = 2c 3 1 3, so c = 1 (b) Draw a picture of the support set (a 1by2 rectangle), and intersect it with the set {(x,y) x y ≥ 1}, which is the region above the line y = 1 − x See figure above, right To compute the probability, we double integrate the joint density over thisWenigstens eine der Zahlen und von 0 verschieden ist Bitte wenden! Ex 42, 9 By using properties of determinants, show that 8 (x&x2&yz@y&y2&zx@z&z2&xy) = (x – y) (y – z) (z – x) (xy yz zx) Solving LHS 8 (𝑥&𝑥^2&𝑦𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦) Applying R1→ R1 – R2 = 8 (𝑥−𝑦&𝑥^2−𝑦^2&𝑦𝑧−𝑥𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦) Ex
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X2 y 2= (x y) 2xy 2xy gilt xy x2y2 2 für x;y 0 mit Gleichheit nur für x= y Allgemein gilt n p x 1 x n x 1 x n n für beliebige x 1;;x n 0 mit Gleichheit nur für x 0 = xyxzyz 3 2 (xyz) = 8 3 10 ;SOLUTIONS TO PROBLEMS FROM ASSIGNMENT 2 Problems 132d and 133d Statement Find general solutions of yu xy 2u x= xusing ODE techniques, as well as its particular solution satisfying the side conditions u(x;1) = 0 and u(0;y) = 0Factorizar x^2xyy^2 Cookies y Privacidad Este sitio web utiliza cookies para garantizar que obtenga la mejor experiencia en nuestro sitio web
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Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!Find the work carried out by the following vector fields• Field F(x, y, z) = which causes a particle to move along the curve , for 0 ≤ t ≤ 1• The force F(x,y,z) = < x/ x^2 y^2 z^2, y/ x^2 y^2 z^2 , z/ x^2 y^2 z^2> that makes move an object from point (1, 0, 0) to point (3, 4, 12) The partial derivatives of #z=f(x,y)=xy^23x^2y^22x2# are #\frac{\partial z}{\partial x}=y^26x2# and #\frac{\partial z}{\partial y}=2xy2y=2y(x1)# Setting these equal to zero gives a system of equations that must be solved to find the critical points #y^26x2=0, 2y(x1)=0# The second equation will be true if #y=0#, which will lead to the first equation becoming #6x2=0
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(d) f(x;y) = 2xy y2 8x 4y fonksiyonunun 1 x 2 ile 1 y 1 in belirledi gi R b olgesi uzerindeki en buy uk ve en ku˘ cuk de gerini bulunuz 14) S nav notlar g(x;y;z) = 10f(x;y;z Ex 25, 13 If x y z = 0, show that x3 y3 z3 = 3xyz We know that x3 y3 z3 3xyz = (x y z) (x2 y2 z2 xy yz zx) Putting x y z = 0, x3 y3 z3 3xyz = (0) (x2 y2 z2 xy yz zx) x3 y3 z3 3xyz = 0 x3 y3 z3 = 3xyz Hence proved Show More Ex 25 Ex 25, 1F(r r ˆ)dA = Z 2ˇ 0 Zp 2 0 hx;y;z4i(r r ˆ)dˆd = Z 2ˇ 0 Z p 2 0 ˆ 2 p 2 ˆ5 8 dˆd = 2ˇ Z 2 0 ˆ2 2 p 2 ˆ5 8 dˆ = 1 3 1 6 = 1 6 1784 Use Stokes' Theorem to evaluate ZZ S curlFdS when F(x;y;z) = x2y3zi sin(xyz)j xyzk, and Sis the part of the cone y2 = x2 z2 that lies between the planes y= 0 and y= 3, oriented in the direction
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Plot x^2 3y^2 z^2 = 1 Natural Language;And 4z to get x2 = y2 = z2 = 2 Since x2 y2 z2 = 3 2 = 1, we get = 2 3 and thus each of x;y;z is p1 3 You can write the Lagrangian as $$\mathcal L(x,y,z,\lambda)=x^22y^23z^2\lambda(x^2y^2z^21)$$ So we can get the gradient and putting it equal zero
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= −2y −g0(x) =⇒ g0(x) = 0 =⇒ g = const = α, say Hence f(z) = x2 −y2 2ixy iα = (xiy)2 iα = z2 iα Examples of Analytic Functions (i) f(z) = z is analytic in the whole of C Here u = x, v = y, and the Cauchy–Riemann equations are satisfied (1 = 1;X2yz=0, 2xyz=1, 3xy2z=5 \square!Factorizar x^22xyy^2 x2 2xy y2 x 2 2 x y y 2 Verificar el término medio multiplicando 2ab 2 a b y comparando el resultado con el término medio de la expresión original 2ab = 2⋅ x⋅y 2 a b = 2 ⋅ x ⋅ y Simplifica 2ab = 2xy 2 a b = 2 x y Factorizar utilizando la regla del trinomio del cuadrado perfecto, a2 2abb2 = (ab
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2x2yz = 1 First we rearrange the equation of the surface into the form f(x,y,z)=0 z=x^22xyy^2 x^22xyy^2z = 0 And so we define our surface function, f, by f(x,y,z) = x^22xyy^2z 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 fClick here👆to get an answer to your question ️ If the system of equations 2x y z = 0, x 2y z = 0, lambda x y 2z = 0 has infinitely many solutions and f(x) be a continuous function such that f(5 x) f(x) = 2 , then int 0^2lambdaf (x) dx is equal toQ = {(x,y) 0 ≤ x ≤ 1,0 ≤ y ≤ 1 − x} Consequently, Z C F dr = Z 1 0 Z 1−x 0 2x 4(2 − 2x − 2y) − 2y dydx = Z 1 0 Z 1−x 0 (−6x − 10y 8)dydx = Z 1 0 (x2 − 4x 3)dx = 4/3 (b) F(x,y,z) = xiy j(x2 y2)k, C is the boundary of the part of the paraboloid z = 1 − x 2− y in the first octant Solution The curl of F
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Take the square root of both sides of the equation x^ {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=0Math Input NEW Use textbook math notation to enter your math Try it22 Limits and continuity The absolute value measures the distance between two complex numbers Thus, z 1 and z 2 are close when jz 1 z 2jis smallWe can then de ne the limit of a complex function f(z) as follows we write
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