Find the curvature of r = (t, t^2, t^3)at point (1, 1, 1) Find the limit of lim_

Question

Find the curvature of r = (t, t^2, t^3)at point (1, 1, 1) Find the limit of lim_(x, y) rightarrow (0, 0) 6x^3 y/2x^4 + y^4 Given r = x^2 + y^5, z^3, x = uve^2s, y = u^2 - v^2s, and z = sin(uvs^2) then find partial differential r/partial differential/s using the chain rule Given f(x, y) = x^2/2 + 3y^3 + 9y^2 - 3xy - 9x + 9y find its local max, local min and saddle point if they exist Find the directional derivative of f(x, y, z) = x^2y + x middot squareroot 1 + z at the point P(1, 2, 3) in the direction of v^vector = 2i^vector + j^vector - 2k^vector Find the point on the plane x + 2y + 3z = 4 that is closed to the origin, Find the minimum value of f(x, y, z) = 2x^2 + y^2 + 3z^2 subject to the constraint 2x - 3y - 4z = 49

Explanation / Answer

Solution(8)

We need to minimize the square-of-the-distance function f(x, y) = x^2 + y^2 + z^2 with z = (4/3) - (x/3) - (2y/3). That is, the function

f(x, y) = x^2 + y^2 + {(4/3) - (x/3) - (2y/3)}^2

Computing partials:

fx = 2x - (2/3) ((4/3) - (x/3) - (2y/3)) = 0 => 2x - {8/9 - 2x/9 - 4y/9} = 0 => 20x/9 + 4y/9 - 8/9 = 0 => 5x + y = 2

fy = 2y - 2*(2/3) ((4/3) - (x/3) - (2y/3)) = 0 => 2y - {16/9 - 4x/9 - 8y/9} = 0 => 4x/9 + 26y/9 -16/9 = 0 => 2x+13y = 8

from above two equations;

5x + y = 2 -----(1) and 2x + 13y = 8 ----(2)

eq^n (1) multiply by 2 and eq^n (2) multiply by 5 , then substract eq^n(2) - eq^n (1);

10x + 2y = 4 and 10x + 65y = 40

63y = 36 => y = 4/7

and 5x + 4/7 = 2 => x = 2/7

Finally, z =(4/3) - (x/3) - (2y/3) = 4/3 - 2/21 - 8/21 = 6/7

(x, y, z) = (2/7, 4/7, 6/7)

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