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Agent_A

  • one year ago

What should the answer be? (see photo)

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  1. Agent_A
    • one year ago
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    Posting photo. Hold on.

  2. Agent_A
    • one year ago
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  3. Agent_A
    • one year ago
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    Posted.

  4. Agent_A
    • one year ago
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    I wonder if this: "-4sin(-4)-2sin(-4)" is the issue....

  5. Agent_A
    • one year ago
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    Wolfram Alpha says:

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  6. Agent_A
    • one year ago
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    Where did the negative sign inside the parentheses go?

  7. Agent_A
    • one year ago
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    Anyone know Multivariable Calculus?!

  8. ganeshie8
    • one year ago
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    wolfram gives a much simpler looking equation for tangent plane : \[z=y\] http://www.wolframalpha.com/input/?i=tangent+plane+y*cos%28x-y%29+at+%28-2%2C-2%2C-2%29

  9. Agent_A
    • one year ago
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    y?! That's it?

  10. ganeshie8
    • one year ago
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    Yes, familiar with gradient ?

  11. Agent_A
    • one year ago
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    Got it. No I am not familiar with the gradient. I re-did my solution and got the right answer. My problem was in the Trigonometry and Alegbra. Thanks, @ganeshie8! Can you tell me a bit about this gradient? I think that my teacher will be covering that in this week's lecture.

  12. ganeshie8
    • one year ago
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    Awesome! how about level curves ?

  13. Agent_A
    • one year ago
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    Yes to Level Curves.

  14. ganeshie8
    • one year ago
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    then it will be easy, we use the below fact : `gradient` is perpendicular to the `level curve/surface`

  15. ganeshie8
    • one year ago
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    so the gradient vector at a particular point of the function `f(x,y,z)` gives the normal vector to the tangent plane at that particular point

  16. ganeshie8
    • one year ago
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    btw, gradient is just a package of partial derivatives : For a function of 3 variables\(f(x,y,z)\), the gradient is \(\langle f_x,~f_y,~f_z \rangle \)

  17. Agent_A
    • one year ago
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    Ohhhhh

  18. Agent_A
    • one year ago
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    So whenever I've been solving for \[f_x , f_y , and/or f_z\] those have always been the gradients, and when you consolidate them in to the brackets, they become the gradient vector?

  19. ganeshie8
    • one year ago
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    \(f_x, f_y, f_z\) etc... are called partial derivatives of the function \(f(x,y,z)\) gradient is a package of all the partial derivatives gradient is a vector

  20. Agent_A
    • one year ago
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    Oops. Yes, you're right. :P

  21. ganeshie8
    • one year ago
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    lets work the tangent plane using gradient

  22. ganeshie8
    • one year ago
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    \[z=y\cos(x-y)\tag{1}\] think of it as a level surface of \(f(x,y,z)=y\cos(x-y)-z\) at level \(0\) : \[y\cos(x-y)-z=0\tag{2}\] then the gradient vector will be perpendicular to this level surface

  23. ganeshie8
    • one year ago
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    start by finding the partials \(f_x, f_y, f_z\)

  24. Agent_A
    • one year ago
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    Oh so we equate it to zero

  25. Agent_A
    • one year ago
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    Forgive me if I'm asking stuff that I should know. I'm taking a sped-up class. We go through 2 chapters per day.

  26. ganeshie8
    • one year ago
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    we have \[z=y\cos(x-y)\] subtract \(z\) both sides

  27. Agent_A
    • one year ago
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    Oh. Duh...

  28. Agent_A
    • one year ago
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    Hmmm I think I'll read-up on the gradient vector. It sounds interesting.

  29. ganeshie8
    • one year ago
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    try this when free http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-12-gradient/

  30. Agent_A
    • one year ago
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    Thanks! :)

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

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