anonymous one year ago check my work

1. anonymous

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

@ganeshie8

3. anonymous

@IrishBoy123

4. IrishBoy123

5. anonymous

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

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

7. anonymous

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

8. anonymous

I just noticed my mistake when I saw your sheet lol

9. IrishBoy123

that must be quite some coincidence