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Math Help - directional derivative q

  1. #1
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    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
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  2. #2
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    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 ?
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  3. #3
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    Quote Originally Posted by pjb08005 View Post
    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\|}}
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  4. #4
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    thank you both for your help.
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