A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

ques

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What's your question buddy

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    looks good to me!

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Didn't take me as long as I thought

  7. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    1 Attachment
  8. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  10. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.