anonymous one year ago ques

1. anonymous

2. anonymous

If the directional derivative of $\phi(x,y,z)=ax^2y+by^2z+cz^2x$ at the point (1,1,1) has maximum magnitude 15 in the direction parallel to the line $\frac{x-1}{2}=\frac{y-3}{-2}=\frac{z}{1}$ find the values of a,b, and c I've calculated $\vec \nabla \phi(1,1,1)=(2a+c)\hat i+(a+2b)\hat j+(b+2c)\hat k$ The given line is parallel to the vector $\vec b=2\hat i-2\hat j+\hat k$ Since $\vec \nabla \phi(1,1,1) \parallel \vec b$ $\implies (2a+c)\hat i+(a+2b)\hat j+(b+2c)\hat k=\lambda2\hat i+-\lambda2\hat j+\lambda \hat k$$\frac{2a+c}{2}=\frac{2b+a}{-2}=\frac{2c+b}{1}=\lambda$ Now using the equation for magnitude, $|\vec \nabla \phi(1,1,1,)|^2=(2a+c)^2+(a+2b)^2+(b+2c)^2=15^2$ I've calculated $\lambda=\pm5$ Thus 3 equations i've found $2a+c=\pm10$$a+2b=\pm10$$b+2c=\pm5$ Now how can I solve this without the annoying determinant method or without having to remember that nasty cross multiply formula

3. anonymous

Sorry 2nd equation is $a+2b=\mp10$

4. anonymous

Meh I ended up using the determinant method anyway, $\left[\begin{matrix}2 & 0&1 \\ 1 & 2&0\\0&1&2\end{matrix}\right]\left[\begin{matrix}a \\ b\\c\end{matrix}\right]=\left[\begin{matrix}\pm10 \\ \mp10\\ \pm5\end{matrix}\right]$ AX=B $\det(A)=9$ $adj(A)=\left[\begin{matrix}4 & 1&-2 \\ -2 & 4&1\\1&-2&4\end{matrix}\right]$ $X=\frac{adj(A)}{\det(A)}B$ $X=\left[\begin{matrix}4/9 & 1/9&-2/9 \\ -2/9 & 4/9&1/9\\1/9&-2/9&4/9\end{matrix}\right]\left[\begin{matrix}\pm10 \\ \mp10\\ \pm5\end{matrix}\right]$ $X=\left[\begin{matrix}\pm\frac{20}{9} \\ \mp\frac{55}{9}\\ \pm\frac{50}{9}\end{matrix}\right]$ $a=\pm\frac{20}{9}$$b=\mp\frac{55}{9}$$c=\pm\frac{50}{9}$

5. ganeshie8

looks good to me!

6. anonymous

Didn't take me as long as I thought

7. IrishBoy123

8. IrishBoy123

@Nishant_Garg i posted that because i took a slightly different tack, but i am not really sure it is any easier admin-wise in terms of cranking out an answer

9. anonymous

Great stuff, I think it would've been just as good!