What should the answer be?
(see photo)

- Agent_A

What should the answer be?
(see photo)

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- schrodinger

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- Agent_A

Posting photo. Hold on.

- Agent_A

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- Agent_A

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## More answers

- Agent_A

I wonder if this: "-4sin(-4)-2sin(-4)" is the issue....

- Agent_A

Wolfram Alpha says:

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- Agent_A

Where did the negative sign inside the parentheses go?

- Agent_A

Anyone know Multivariable Calculus?!

- 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

y?! That's it?

- ganeshie8

Yes, familiar with gradient ?

- 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

Awesome! how about level curves ?

- Agent_A

Yes to Level Curves.

- ganeshie8

then it will be easy, we use the below fact :
`gradient` is perpendicular to the `level curve/surface`

- 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

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

Ohhhhh

- 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

\(f_x, f_y, f_z\) 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

Oops. Yes, you're right. :P

- ganeshie8

lets work the tangent plane using gradient

- 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

start by finding the partials \(f_x, f_y, f_z\)

- Agent_A

Oh so we equate it to zero

- 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

we have \[z=y\cos(x-y)\]
subtract \(z\) both sides

- Agent_A

Oh. Duh...

- Agent_A

Hmmm I think I'll read-up on the gradient vector. It sounds interesting.

- ganeshie8

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

- Agent_A

Thanks! :)

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