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

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- Apr 6th 2010, 08:57 PMleinadwerdnadirectional 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 - Apr 7th 2010, 03:05 AMHallsofIvy
- Apr 7th 2010, 06:07 AMleinadwerdna
the book gets a=6 b=24 and c=-8 are they just flat out wrong

- Apr 9th 2010, 12:10 PMleinadwerdna
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? - Apr 10th 2010, 06:07 AMleinadwerdna
someone look at this please

- Apr 12th 2010, 08:57 AMleinadwerdna
help me