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Agent_A
 one year ago
What should the answer be?
(see photo)
Agent_A
 one year ago
What should the answer be? (see photo)

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Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Posting photo. Hold on.

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1I wonder if this: "4sin(4)2sin(4)" is the issue....

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Where did the negative sign inside the parentheses go?

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Anyone know Multivariable Calculus?!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2wolfram gives a much simpler looking equation for tangent plane : \[z=y\] http://www.wolframalpha.com/input/?i=tangent+plane+y*cos%28xy%29+at+%282%2C2%2C2%29

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes, familiar with gradient ?

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Got it. No I am not familiar with the gradient. I redid 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
 one year ago
Best ResponseYou've already chosen the best response.2Awesome! how about level curves ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2then it will be easy, we use the below fact : `gradient` is perpendicular to the `level curve/surface`

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so 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
 one year ago
Best ResponseYou've already chosen the best response.2btw, 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
 one year ago
Best ResponseYou've already chosen the best response.1So 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
 one year ago
Best ResponseYou've already chosen the best response.2\(f_x, f_y, f_z\) 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
 one year ago
Best ResponseYou've already chosen the best response.1Oops. Yes, you're right. :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2lets work the tangent plane using gradient

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[z=y\cos(xy)\tag{1}\] think of it as a level surface of \(f(x,y,z)=y\cos(xy)z\) at level \(0\) : \[y\cos(xy)z=0\tag{2}\] then the gradient vector will be perpendicular to this level surface

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2start by finding the partials \(f_x, f_y, f_z\)

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Oh so we equate it to zero

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Forgive me if I'm asking stuff that I should know. I'm taking a spedup class. We go through 2 chapters per day.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2we have \[z=y\cos(xy)\] subtract \(z\) both sides

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.1Hmmm I think I'll readup on the gradient vector. It sounds interesting.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2try this when free http://ocw.mit.edu/courses/mathematics/1802multivariablecalculusfall2007/videolectures/lecture12gradient/
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