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anonymous

  • one year ago

check my work

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  1. anonymous
    • one year ago
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    Find the values of constants a, b and c so that the maximum value of the directional derivative of \[\phi(x,y,z)=axy^2+byz+cz^2x^3\] at (1,2,-1) is 64 in the direction parallel to the z axis Now, \[\vec \nabla \phi(1,2,-1)=(4a+3c)\hat i+(4a-b)\hat j+(-2c)\hat k\] Vector equation of z axis, \[\vec z=0 \hat i+0 \hat j+z \hat k\] Now for the vector to be parallel to z axis it's corresponding components must be a scalar multiple \[\therefore (4a+3c)=\lambda0=0\] \[(4a-b)=\lambda0=0\] \[-2c=\lambda z\] \[|\vec \nabla \phi(1,2,-1)|=\sqrt{(4a+3c)^2+(4a-b)^2+(-2c)^2}=64\] \[|\vec \nabla \phi(1,2,-1)|=\sqrt{\lambda^2.z^2}=64\] \[\implies \lambda.z=\pm64\] \[\lambda=\pm \frac{64}{z}\] \[-2c=\lambda.z=\pm64\]\[c=\mp8\] \[\therefore a=\frac{-3c}{4}=\frac{\pm24}{4}=\pm6\] \[b=4a=\pm24\]

  2. anonymous
    • one year ago
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    @ganeshie8

  3. anonymous
    • one year ago
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    @IrishBoy123

  4. IrishBoy123
    • one year ago
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  5. anonymous
    • one year ago
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    oh shoot, I missed one term!! but still my answers are holding good, magic lol \[2b-2c\] I've written as -2c How did I not get the wrong answer??amazing

  6. IrishBoy123
    • one year ago
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    good point, i should have made them \(\pm\) cos it could point up or down the z axis....

  7. anonymous
    • one year ago
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    Didn't you see my mistake?I still got the right answer though

  8. anonymous
    • one year ago
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    I just noticed my mistake when I saw your sheet lol

  9. IrishBoy123
    • one year ago
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    that must be quite some coincidence

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spraguer (Moderator)
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