1. May 24, 2007

1. The problem statement, all variables and given/known data

For the function f(x, y, z) = z^4(x^2 − y^2), find the coordinates of
those points in the plane z = 1 at which the direction of the greatest rate of
change of f is parallel to the vector i − 2j + 3k.

2. Relevant equations

3. The attempt at a solution

i just want to verify that i am approaching this question correctly.

1.First, i substitute, z =1 into the equation.
2.so i get f(x,y,z) = x^2- y^2.
3. Then i find grad f(x,y,z). cos i need the direction of greatest rate of change. (i mean i find grad f of the equation x^2-y^2.)
4. To find the coodinates parallel to the vector i-2j+3k, i substitute
these values into my grad f equation.

MY WORKING:

As z= 1
f(x,y,z) = x^2 - y^2

grad f = 2xi - 2yj

Subsitute, x =1,y=-2 ->2xi-2yj

Therefore coordinates are: (2,4,0)

Could someone please tell me if this is correct.

Tanx

Last edited: May 24, 2007
2. May 24, 2007

### Dick

Take the gradient BEFORE you substitute for z! The gradient of x^2-y^2 is NEVER parallel to the given vector.

3. May 24, 2007

So i'll get grad f = (2xz^4)i-(2yz^4)j - (4z^3 x^2 - 4z^3 y ^2)k

And i substitute, x = 1, y=-2 and z=3 into grad f:

and i get, the coordinates, (162, 324,972). <----Is this right????

does that fact that it is on the plane z= 1 hav anything to do with the solution?

Thanx

4. May 24, 2007

### Dick

Take the gradient, then substitute z=1. 2xi-2yj+4(x^2-y^2)k. Now try to find values of x and y that make this parallel to your given vector. Do you know how to determine if two vectors are parallel. BTW - I can't find any such x and y. There may be no solution.

Last edited: May 24, 2007
5. May 24, 2007

### D H

Staff Emeritus
Dick, check the sign on your gradient. I get $2x\hat i - 2y\hat j + 4(x^2-y^2)\hat k$. There are still no solutions.

6. May 24, 2007

### Dick

Check. Thanks.