anonymous
  • anonymous
ques
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
What's your question buddy
anonymous
  • 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
anonymous
  • anonymous
Sorry 2nd equation is \[a+2b=\mp10\]

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anonymous
  • 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 \[\det(A)=9\] \[adj(A)=\left[\begin{matrix}4 & 1&-2 \\ -2 & 4&1\\1&-2&4\end{matrix}\right]\] \[X=\frac{adj(A)}{\det(A)}B\] \[X=\left[\begin{matrix}4/9 & 1/9&-2/9 \\ -2/9 & 4/9&1/9\\1/9&-2/9&4/9\end{matrix}\right]\left[\begin{matrix}\pm10 \\ \mp10\\ \pm5\end{matrix}\right]\] \[X=\left[\begin{matrix}\pm\frac{20}{9} \\ \mp\frac{55}{9}\\ \pm\frac{50}{9}\end{matrix}\right]\] \[a=\pm\frac{20}{9}\]\[b=\mp\frac{55}{9}\]\[c=\pm\frac{50}{9}\]
ganeshie8
  • ganeshie8
looks good to me!
anonymous
  • anonymous
Didn't take me as long as I thought
IrishBoy123
  • IrishBoy123
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IrishBoy123
  • 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
anonymous
  • anonymous
Great stuff, I think it would've been just as good!

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