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

1. Agent_A

Posting photo. Hold on.

2. Agent_A

3. Agent_A

Posted.

4. Agent_A

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

5. Agent_A

Wolfram Alpha says:

6. Agent_A

7. Agent_A

Anyone know Multivariable Calculus?!

8. 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

9. Agent_A

y?! That's it?

10. ganeshie8

11. 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.

12. ganeshie8

Awesome! how about level curves ?

13. Agent_A

Yes to Level Curves.

14. ganeshie8

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

15. 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

16. ganeshie8

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

17. Agent_A

Ohhhhh

18. 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?

19. 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

20. Agent_A

Oops. Yes, you're right. :P

21. ganeshie8

lets work the tangent plane using gradient

22. 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

23. ganeshie8

start by finding the partials $$f_x, f_y, f_z$$

24. Agent_A

Oh so we equate it to zero

25. 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.

26. ganeshie8

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

27. Agent_A

Oh. Duh...

28. Agent_A

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

29. ganeshie8
30. Agent_A

Thanks! :)