Question

check my work

Answer 1

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
\[\therefore (4a+3c)=\lambda0=0\]
\[(4a-b)=\lambda0=0\]
\[-2c=\lambda z\]
\[|\vec \nabla \phi(1,2,-1)|=\sqrt{(4a+3c)^2+(4a-b)^2+(-2c)^2}=64\]
\[|\vec \nabla \phi(1,2,-1)|=\sqrt{\lambda^2.z^2}=64\]
\[\implies \lambda.z=\pm64\]
\[\lambda=\pm \frac{64}{z}\]
\[-2c=\lambda.z=\pm64\]\[c=\mp8\]
\[\therefore a=\frac{-3c}{4}=\frac{\pm24}{4}=\pm6\]
\[b=4a=\pm24\]

Answer 2

@ganeshie8

Answer 3

@IrishBoy123

Answer 4

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

Answer 5

good point, i should have made them \(\pm\) cos it could point up or down the z axis....

Answer 6

Didn't you see my mistake?I still got the right answer though

Answer 7

I just noticed my mistake when I saw your sheet lol

Answer 8

that must be quite some coincidence