directional derivative q

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• Oct 26th 2009, 04:44 PM
pjb08005
directional derivative q
ok so heres the question: find the directional derivative of the function f(x,y,z)=xy+yz+zx at the P(3,-3,4) in the direction of Q(7,1,7)...
Here's what I have done:
found partial derivatives and added them to give the gradient
(y+z)i+(x+z)j+(y+x)k
plugging in P gives: (-3+4)i+(3+4)j+(-3+3)K = i+7j
since v is not a unit vector, i took the magnitude of Q and got sqrt(99)
I got: 7/sqrt(99)i+1/sqrt(99)j+7/sqrt(99)
so the gradient times the Q gives 14/sqrt(99)
but for some reason webassign says this answer is incorrect, can someone please help, I appreciate it
• Oct 27th 2009, 01:54 PM
BobP
Doesn't the phrase ' in the direction of Q ' mean, in the direction of the vector from P to Q ?
That is, in the direction of the vector 4i + 4j + 3k ?
• Oct 27th 2009, 03:05 PM
Plato
Quote:

Originally Posted by pjb08005
ok so heres the question: find the directional derivative of the function f(x,y,z)=xy+yz+zx at the P(3,-3,4) in the direction of Q(7,1,7)...

$\overrightarrow {PQ} = \left\langle {4,4,3} \right\rangle$
Now the directional derivative is $\frac{{\nabla f(3, - 3,4) \cdot \overrightarrow {PQ} }}{{\left\| {\overrightarrow {PQ} } \right\|}}$
• Oct 28th 2009, 06:01 PM
pjb08005
thank you both for your help.