Solving Grad f and Grad g for Math Problem

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The discussion focuses on solving the mathematical problem involving the function f(x, y) = xye^(-((x-a)^2)-((y-b)^2)) with parameters a=8 and b=5. The user derived the gradients df/dx and df/dy, obtaining grad f(a, b) = (5, 8). The tangent plane equation was determined as 5x + 8y - z = 40. The user also sought to find points where the surface z=f(x, y) is horizontal, concluding that this occurs when df/dx = df/dy = 0.

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electricman
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Hi,

I have some problems to solve this:

a=8 , b=5

Let f (x, y)=xye^(− ((x−a)^2)− ((y−b)^ 2)). In which direction, one will go from the point (a, b) in the definition amount if one wants to that the function values will increase so quickly as possible? Decide an equation for tangent plane to the surface z=f (x, y) in the point (a, b, ab). Use differential to f in order to calculate a close value to f (9a/10, 6b/5). In which points, key planet is to the surface z=f (x, y) horizontal?

I derivated and got:

df/dx = ye^(− ((x−a)^2)− ((y−b)^ 2)) * (-2x+2a) + ye^(− ((x−a)^2)− ((y−b)^ 2))

df/dy = xe^(− ((x−a)^2)− ((y−b)^ 2)) * (-2y+2a) + ye^(− ((x−a)^2)− ((y−b)^ 2))

and grad f(a,b) = (y,x) = (5,8) , Is this correct?
______________________________________________________________

And for the tangent plane:

g(x,y,z)=f(x,y) - z

grad g = (5,8,-1)

And the tangent plane: 5(x-8)+8(y-5)+(1)(z-40)=0 => 5x+8y-z=40 , Is this correct?

_____________________________________________________________

And for the close value:

z=5x+8y-40 close to (8,5). In (7.2,6) is z=5*7,2+8*6-40=44 , Is this correct?

_________________________________________________________

The are horizontal when df/dx = df/dy = 0 , Is this correct? I don't know if i can get a value to this?
 
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electricman said:
df/dx = ye^(− ((x−a)^2)− ((y−b)^ 2)) * (-2x+2a) + ye^(− ((x−a)^2)− ((y−b)^ 2))

(I presume you mean [tex]\frac{\partial}{\partial x}[/tex].)

There should be an 'x' in the first term.

[tex]\frac{\partial f}{\partial x} =[/tex] xye^(− ((x−a)^2)− ((y−b)^ 2)) * (-2x+2a) + ye^(− ((x−a)^2)− ((y−b)^ 2)).

The same missing 'y' in [tex]\frac{\partial f}{\partial y}[/tex].

Correct these first, in order to get the correct answers for the rest of the problems.
 

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