1. ## directional derivative

we know that the grad of Ф is (4a+3c)i+(4a-b)j+(2b-2c)k

and directional derivative of Ф in the direction z has magnitude 64.

so wouldnt it be (4a + 3c)0+(4a-b)0+(2b-2c)1=64

how from this equation could i solve for a b and c

we know that the grad of Ф is (4a+3c)i+(4a-b)j+(2b-2c)k

and directional derivative of Ф in the direction z has magnitude 64.

so wouldnt it be (4a + 3c)0+(4a-b)0+(2b-2c)1=64
Yes, and that simplifies to 2b- 2c= 64 or b- c= 32.

how from this equation could i solve for a b and c
Since that is a single equation in two unknown numbers, without more information, you can't.

3. the book gets a=6 b=24 and c=-8 are they just flat out wrong

4. So I got my homework back and I got a point taken off on this problem.

The question was "Find the values of the constants a, b, and c so that the directional derivative of Ф=axy^2+byz+c(z^2)(x^3) at (1,2-1) has a maximum of magnitude 64 in a direction parallel to the z-axis.
I took the gradient of Ф then did the this gradient dot the vector k since the vector k is a unit vector parallel to the z axis. So I did the dot product and got 2b-2c=64. I said that I couldn't find a solution to this equation and that there was no way to solve this problem but this was not the right answer. Can someone work out this problem and see what you get?

5. someone look at this please

6. help me