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anonymous
 one year ago
ques
anonymous
 one year ago
ques

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's your question buddy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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 \[\frac{x1}{2}=\frac{y3}{2}=\frac{z}{1}\] 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 \[\vec b=2\hat i2\hat j+\hat k\] 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry 2nd equation is \[a+2b=\mp10\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Meh 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 \[\det(A)=9\] \[adj(A)=\left[\begin{matrix}4 & 1&2 \\ 2 & 4&1\\1&2&4\end{matrix}\right]\] \[X=\frac{adj(A)}{\det(A)}B\] \[X=\left[\begin{matrix}4/9 & 1/9&2/9 \\ 2/9 & 4/9&1/9\\1/9&2/9&4/9\end{matrix}\right]\left[\begin{matrix}\pm10 \\ \mp10\\ \pm5\end{matrix}\right]\] \[X=\left[\begin{matrix}\pm\frac{20}{9} \\ \mp\frac{55}{9}\\ \pm\frac{50}{9}\end{matrix}\right]\] \[a=\pm\frac{20}{9}\]\[b=\mp\frac{55}{9}\]\[c=\pm\frac{50}{9}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Didn't take me as long as I thought

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2@Nishant_Garg i posted that because i took a slightly different tack, but i am not really sure it is any easier adminwise in terms of cranking out an answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Great stuff, I think it would've been just as good!
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