Directional derivatives and gradients

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SUMMARY

The discussion focuses on finding points where the magnitude of the gradient (nabla Z) of the function z = 3x² - y² equals 6. The user correctly calculated the partial derivatives, dz/dx = 6x and dz/dy = -2y, but encountered issues when solving the equations derived from the magnitude condition. The correct approach involves setting up the equation √(36x² + 4y²) = 6, leading to the relationships 36x² + 4y² = 36. The user ultimately found x = ±1 and y = ±3, but these results were incorrect, prompting a request for clarification on the method used.

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Mdhiggenz
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1. Homework Statement [/b]
Let z=3x2-y2. Find all points at which magnitude(nabla Z)=6

First things first

I took the partials dz/dx and dz/dy

dz/dx=6x

dx/dy=-2y

I know that √(36x2+4y2)=6 or (36x2+4y2)=36

Then using the above relation I solved for each variable getting
1.y=√(9-x2)

2.x=√(1-1/9y2)

We already have the relation from the partials

A.-2y=0

B.6x=0
So I plugged in 1 into A and 2 into B to get the following

x=+-1 and y=+-3

However the answers don't check out

Where did I go wrong?
 
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Mdhiggenz said:
I took the partials dz/dx and dz/dy
dz/dx=6x
dx/dy=-2y
∂z/∂y
(36x2+4y2)=36
Then using the above relation I solved for each variable getting
1.y=√(9-x2)
No - try that again.
 
Mdhiggenz said:
1. Homework Statement [/b]
Let z=3x2-y2. Find all points at which magnitude(nabla Z)=6

First things first

I took the partials dz/dx and dz/dy

dz/dx=6x

dx/dy=-2y

I know that √(36x2+4y2)=6 or (36x2+4y2)=36

Then using the above relation I solved for each variable getting
1.y=√(9-x2)

2.x=√(1-1/9y2)

We already have the relation from the partials

A.-2y=0

B.6x=0
So I plugged in 1 into A and 2 into B to get the following

x=+-1 and y=+-3

However the answers don't check out

Where did I go wrong?

Why do you try to set dz/dx = 0, and/or dz/dy = 0?

RGV
 

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