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mathstina Group Title

Find the directional derivatives of the following functions at the given points P in the direction of the vectors v. f(x,y,z) =√xyz , P (3,2,6) , v =<-1,-2,2>

  • 2 years ago
  • 2 years ago

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  1. piscez.in Group Title
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    find the derivatives of the function x, y and z seperately, by keeping the other 2 variables constant in each case. Think of a smart way to keep the other 2 variables constant!

    • 2 years ago
  2. mathstina Group Title
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    ok, i tried to work out. is the ans -1?

    • 2 years ago
  3. TuringTest Group Title
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    what is you function?\[f(x,y,z)=\sqrt{xyz}\]or\[f(x,y,z)=\sqrt xyz\]?

    • 2 years ago
  4. mathstina Group Title
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    the first one

    • 2 years ago
  5. TuringTest Group Title
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    \[D_{\vec v}=\nabla f(3,2,6)\cdot\frac{\vec v}{\|\vec v\|}\]I don't think the answer is one but I haven't done it yet...

    • 2 years ago
  6. Algebraic! Group Title
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    is there an advanced equation editor available to super users that has cool features like gradient?

    • 2 years ago
  7. mathstina Group Title
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    negative one

    • 2 years ago
  8. piscez.in Group Title
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    @mathstina @TuringTest is right

    • 2 years ago
  9. piscez.in Group Title
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    @mathstina first fing the derivative of the function with respect to x, by substituting the y and x values

    • 2 years ago
  10. mathstina Group Title
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    ok

    • 2 years ago
  11. TuringTest Group Title
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    oh yeah, I do get -1 :)

    • 2 years ago
  12. mathstina Group Title
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    ok thanks a lot!

    • 2 years ago
  13. piscez.in Group Title
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    @TuringTest @mathstina no its not -1, atleast as per my calculations

    • 2 years ago
  14. TuringTest Group Title
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    I will show my work and you can spot a mistake if you see one...

    • 2 years ago
  15. piscez.in Group Title
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    ok

    • 2 years ago
  16. TuringTest Group Title
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    \[\nabla f=\frac12\langle\sqrt{\frac{yz}x},\sqrt{\frac{xz}y},\sqrt{\frac{xy}z}\rangle\implies\nabla f(3,2,6)=\frac12\langle2,3,1\rangle\]\[\frac{\vec v}{\|\vec v\|}=\frac13\langle-1,-2,2\rangle\]\[\nabla f(3,2,6)\cdot\frac{\vec v}{\|\vec v\|}=\frac16(-2-6+2)=-1\]

    • 2 years ago
  17. mathstina Group Title
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    Great!

    • 2 years ago
  18. piscez.in Group Title
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    @TuringTest @mathstina im sorry, your right, i made a mistake by putting all -ves as +ves and vice versa. Bravo you two

    • 2 years ago
  19. TuringTest Group Title
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    cheers!

    • 2 years ago
  20. mathstina Group Title
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    Its ok. u too got the same ans

    • 2 years ago
  21. piscez.in Group Title
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    i got it as 1 previously, now i have -1 :)

    • 2 years ago
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