And accept "c" as a slider or {sqrt(c² y² z²), sqrt(c² y²And 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 3Mar 16, 07 · (x y)^2 = x^2 2xy y^2 Earlier, when I put x^2 2xy y^2 is the bracket, I meant to keep it away from z^2 so that the factoring would be clearer for you Sorry I confused you instead I think what you didn't get was the x^2 and y^2 in the bracket and the appearence of (x y)^2 in the next step (x^2 y^2) = (x y)^2 is a wrong
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F(x y z)=xy^2z x^2+y^2+z^2=4
F(x y z)=xy^2z x^2+y^2+z^2=4-X^2y^2z^2=xyyzzx eq(1) Identity is x^3y^3z^3 3xyz=(xyz)(x^2y^2z^2xyyzzx) x^3y^3z^3 3xyz =(xyz)(xyyzzxxyyzzx) (acc to eq1) Therefore , x^3y^3z^3 3xyz = 0 So, x^3y^3z^3= 3xyz Answer read moreThen f(z) = z jzj2 = x x2 y2 i y x2 y2;
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Then, ∂ f ∂ x = x x 2 y 2 z 2, ∂ f ∂ y = y x 2 y 2 z 2, ∂ f ∂ z = z x 2 y 2 z 2 Okay, so now let's find a directional derivative!Jun 23, 15 · Suppose f(x,y,z)=x^2y^2z^2 and W is the solid cylinder?May 07, 19 · If 1/x 1/y 1/z = 1, show that the minimum value of the function a^3x^2 b^3y^2 c^3z^2 is (a b c)^3 asked May 8, 19 in Mathematics by Nakul (
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F(x,y,z) = x^2 y^2z^2 Subject to the constraints xy=1, y^2z^2 =1 It seems to me that z does not exist so there are no answers Is that correct?(a) If f(z) = 1 z;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;
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Let 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 0Then click on the marble of the row 2 ==> using the input bar x^2y^2z^2=c;We think you wrote (2xy3z2xy^2z/(x^2yy^2zxz^2))*x^2yy^2zxz^2 This deals with adding, subtracting and finding the least common multiple
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Multivariable 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 isMinimize f(x, y, z) = x^2 y^2 z^2 subject to 4x^2 2y^2 z^2 = 4 Maximum Valua At (,,) (1 pt) Find the coordinates of the point (x, y, z) on the plane z = 2 x 2 y 3 which is closest to the origin x = 2 y = z =3dprinting, 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 between adjacent
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2 share Report Save Continue this threadThe same idea applies, the term under the square root can't be negative, ie 25x 2y 2z 2 >=0 1 share Report Save level 2 5 years ago So 25>(x,y,z)>0 as xyz can't be less than zero, and if they're greater than 25 than the entire function becomes less than 0 Would that be a suitable answer?More_vert Find the minimum value of f ( x , y , z ) = x 2 2 y 2 3 z 2 subject to the constraint x 2 y 3 z = 10 Show that f has no maximum value with this constraint
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May 08, 19 · Let F = (x 2 y 2 z 2) λ (x y z) We form the equations Fx = 0, F y = 0, Fz = 0 ie, 2x λ = 0, 2y λ = 0, 2z λ = 0 or λ = – 2x, λ = – 2y, λ = – 2z ⇒ – 2x = – 2y = – 2z or x = y = z But x y z = 3a Substituting y = z = x, we get 3x = 3a x = a ∴ x = a, y = a, z = a The required minimum value of x 2Letf (x,y,z) = x^2y^2z^2 Calculate the gradient of f Calculate ∫_C (F dr ) where F (x,y,z)= (x,y,z) and C is the curve parametrized by r (t)= (3cos^3 (t), 2sin^5 (t), 2cos^13 (t) for 2π≤t≤3πJun 22, 16 · a) If for this point the restrictions are obeyed, consider this as a successful one 4)Given de box volume V_b the total number of trials N and de number of successful trials n_s the area/volume is computed as v = (V_b/N) xx n_s In this case we have the restrictions defining the sought volume f(x,y,z) = x^2y^2z^2 = 2
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{eq}\displaystyle f(x,y,z) = (x^2 y^2 z^2)^{\frac{1}{2}} \\ P(x,y,z) = (2, 1, 2) \\ \vec{R} = \hat{i} 2 \hat{j} \hat{k} \\ \begin{align} \vec{R} &= \sqrt{1^2(2)^2(1)^2Y Simplify —— x 2 Equation at the end of step 1 y (((((x 2)(y 2))(z 2))(2x•——))y 2)z 2)2xz x 2 Step 2 Rewriting the whole as an Equivalent Fraction 21 Subtracting a fraction from a whole Rewrite the whole as a fraction using x as the denominatorApr 17, 12 · Use Lagrange Multipliers to find the extreme values of f subject to both constraints?
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F(xy,z) = x 2 y 2 z The level surfaces are the parabaloids z = c x 2y 2 Example 4 Suppose we have f(x,y,t) = cos(t) e x 2 y 2 which represents the temperature at any pt on a rectangular plate in the plane At each fixed t 0 we have a function of 2 variables f(x,y,t 0) = cos(t 0) e x 2 y 2 For example below is the temperature profileClearly given f = (r^2)^(n) = r^(2n) where r = sqrt(x^2 y^2 z^2) = r Now as we know that grad(f) = f´(r) (r/r) ==> div(grad(f)) =div(f´(r) r/r)={grad(fF= ey2i(y sin(z2))j(z −1)k, and S is the upper hemisphere x 2 y 2 z 2 = 1, z ≥ 0, oriented upward Note that the surface S does NOT include the bottom of the hemisphere
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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}{2Algebra Examples Rewrite (xy z)2 ( x y z) 2 as (xyz)(xyz) ( x y z) ( x y z) Expand (xyz)(xyz) ( x y z) ( x y z) by multiplying each term in the first expression by each term in the second expression Simplify each term Tap for more steps Multiply x x by x x Multiply y y by y yWith height 9 and base radius 2 that is centered about the zaxis with its base at z=2 Answer Save 1 Answer Relevance kb Lv 7 6 years ago Favorite Answer The cylinder has equation x^2 y^2 = 2^2, which can be rewritten as r = 2 in cylindrical coordinates As you stated, z is in
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Nov 04, 13 · 1 Answer Let F (x, y, z) = x^2 y^9 y^2 z^3 x^8 z^7 2xyz = 5 So, ∂f (1,1,1)/∂x = 12/12 = 1 So, ∂f (1,1,1)/∂y = 13/12 = (2y^9 56x^6 z^7 56x^7 z^6 ∂z/∂x 2y ∂z/∂x) * (3y^2 z^2 7x^8 z^6 2xy)Check out a sample textbook solution See solution arrow_back Chapter 14, Problem 41RE Chapter 14, Problem 43RE arrow_forward Want to see this answer and more?So that u(x;y) = x x2 y2 and v(x;y) = y x2 y2 for z 6= 0 Now, @u @x = y2 x2 (x2 y2)2 = @v @y and @u @y = 2xy (x2 y2)2 = @v @x Since the partial derivatives are all continuous at each z 2 C;
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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 matlabJul 04, 06 · Favorite Answer (xyz)^2 = (xyz) (xyz)= x^2y^2z^22xy2xz2yz => 1*1=32 (xyxzyz)=> 2 (xyxzyz)= 13 => xyxzyz =2/2 => xyxzyz = 1 Source (s) My little fuzzy brain Chunky 1 decade ago Man, all these variables must give you a headache!3Dplot of "x^2y^2z^2=1" Learn more about isosurface;
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But first, let's go to the gradient say v → = a b c x x 2 y 2 z 2 y x 2 y 2 z 2 z x 2 y 2 z 2 ⋅ a b c = x a y b z c x 2 y 2 z 2Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historySo how could we make things a bit easier
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Jun 16, 17 · Despite 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 aRewrite (x−y −z)2 ( x y z) 2 as (x−y−z)(x−y−z) ( x y z) ( x y z) Expand (x−y−z)(x−y−z) ( x y z) ( x y z) by multiplying each term in the first expression by each term in the second expression Simplify each term Tap for more steps Multiply x x by x x
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