Question

ques

Answer 1

What's your question buddy

Answer 2

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

Answer 3

Sorry 2nd equation is
\[a+2b=\mp10\]

Answer 4

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
\[\det(A)=9\]
\[adj(A)=\left[\begin{matrix}4 & 1&-2 \\ -2 & 4&1\\1&-2&4\end{matrix}\right]\]
\[X=\frac{adj(A)}{\det(A)}B\]
\[X=\left[\begin{matrix}4/9 & 1/9&-2/9 \\ -2/9 & 4/9&1/9\\1/9&-2/9&4/9\end{matrix}\right]\left[\begin{matrix}\pm10 \\ \mp10\\ \pm5\end{matrix}\right]\]
\[X=\left[\begin{matrix}\pm\frac{20}{9} \\ \mp\frac{55}{9}\\ \pm\frac{50}{9}\end{matrix}\right]\]
\[a=\pm\frac{20}{9}\]\[b=\mp\frac{55}{9}\]\[c=\pm\frac{50}{9}\]

Answer 5

looks good to me!

Answer 6

Didn't take me as long as I thought

Answer 7

@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

Answer 8

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