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anonymous
 one year ago
check my work
anonymous
 one year ago
check my work

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find 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, \[\vec \nabla \phi(1,2,1)=(4a+3c)\hat i+(4ab)\hat j+(2c)\hat k\] 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 \[\therefore (4a+3c)=\lambda0=0\] \[(4ab)=\lambda0=0\] \[2c=\lambda z\] \[\vec \nabla \phi(1,2,1)=\sqrt{(4a+3c)^2+(4ab)^2+(2c)^2}=64\] \[\vec \nabla \phi(1,2,1)=\sqrt{\lambda^2.z^2}=64\] \[\implies \lambda.z=\pm64\] \[\lambda=\pm \frac{64}{z}\] \[2c=\lambda.z=\pm64\]\[c=\mp8\] \[\therefore a=\frac{3c}{4}=\frac{\pm24}{4}=\pm6\] \[b=4a=\pm24\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh shoot, I missed one term!! but still my answers are holding good, magic lol \[2b2c\] I've written as 2c How did I not get the wrong answer??amazing

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2good point, i should have made them \(\pm\) cos it could point up or down the z axis....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Didn't you see my mistake?I still got the right answer though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just noticed my mistake when I saw your sheet lol

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2that must be quite some coincidence
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