Agent_A
  • Agent_A
What should the answer be? (see photo)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Agent_A
  • Agent_A
Posting photo. Hold on.
Agent_A
  • Agent_A
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Posted.

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Agent_A
  • Agent_A
I wonder if this: "-4sin(-4)-2sin(-4)" is the issue....
Agent_A
  • Agent_A
Wolfram Alpha says:
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Agent_A
  • Agent_A
Where did the negative sign inside the parentheses go?
Agent_A
  • Agent_A
Anyone know Multivariable Calculus?!
ganeshie8
  • ganeshie8
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
Agent_A
  • Agent_A
y?! That's it?
ganeshie8
  • ganeshie8
Yes, familiar with gradient ?
Agent_A
  • Agent_A
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.
ganeshie8
  • ganeshie8
Awesome! how about level curves ?
Agent_A
  • Agent_A
Yes to Level Curves.
ganeshie8
  • ganeshie8
then it will be easy, we use the below fact : `gradient` is perpendicular to the `level curve/surface`
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
btw, gradient is just a package of partial derivatives : For a function of 3 variables\(f(x,y,z)\), the gradient is \(\langle f_x,~f_y,~f_z \rangle \)
Agent_A
  • Agent_A
Ohhhhh
Agent_A
  • Agent_A
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?
ganeshie8
  • ganeshie8
\(f_x, f_y, f_z\) 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
Agent_A
  • Agent_A
Oops. Yes, you're right. :P
ganeshie8
  • ganeshie8
lets work the tangent plane using gradient
ganeshie8
  • ganeshie8
\[z=y\cos(x-y)\tag{1}\] 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
ganeshie8
  • ganeshie8
start by finding the partials \(f_x, f_y, f_z\)
Agent_A
  • Agent_A
Oh so we equate it to zero
Agent_A
  • Agent_A
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.
ganeshie8
  • ganeshie8
we have \[z=y\cos(x-y)\] subtract \(z\) both sides
Agent_A
  • Agent_A
Oh. Duh...
Agent_A
  • Agent_A
Hmmm I think I'll read-up on the gradient vector. It sounds interesting.
ganeshie8
  • ganeshie8
try this when free http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-12-gradient/
Agent_A
  • Agent_A
Thanks! :)

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