Question

What should the answer be?

(see photo)

Answer 1

Posting photo. Hold on.

Answer 2

Posted.

Answer 3

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

Answer 4

Wolfram Alpha says:

Answer 5

Where did the negative sign inside the parentheses go?

Answer 6

Anyone know Multivariable Calculus?!

Answer 7

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

Answer 8

y?! That's it?

Answer 9

Yes, familiar with gradient ?

Answer 10

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.

Answer 11

Awesome! how about level curves ?

Answer 12

Yes to Level Curves.

Answer 13

then it will be easy, we use the below fact :

`gradient` is perpendicular to the `level curve/surface`

Answer 14

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

Answer 15

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

Answer 16

Ohhhhh

Answer 17

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?

Answer 18

\(f_x, f_y, f_z\) 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

Answer 19

Oops. Yes, you're right. :P

Answer 20

lets work the tangent plane using gradient

Answer 21

\[z=y\cos(x-y)\tag{1}\]
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

Answer 22

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

Answer 23

Oh so we equate it to zero

Answer 24

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.

Answer 25

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

subtract \(z\) both sides

Answer 26

Oh. Duh...

Answer 27

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

Answer 28

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

Answer 29

Thanks! :)