Question

find the equation of the tangent plane surface at the specific points

z=y cos (x - y) (2,2,2)

Answer 1

it's east

Answer 2

diffrentiate the given function with respect to x

Answer 3

then put the value of the diffrential coefficient in the question

Answer 4

for example it can be found in the same way when you deal with the circle chapter.

Answer 5

i dont think so goformit, -thumbs down-

Answer 6

he wants an equation, - gulps-

Answer 7

@Andresfon12
Gradients... you know how to get them? :D

Answer 8

(Y-y)^2 = dy/dx = (X-x)^2 is the tangent equation

Answer 9

ok lets find the equation of the tangent plane at a point for z= f(x,y)
-calls wikipedia-

Answer 10

@Andresfon12
Place all expressions on one side of the equation (it doesn't really matter which side)
and the other side should be 0.

Let's call that function g(x,y,z)
g(x,y,z)=z-y cos(x - y) = 0

Right?

Answer 11

goformit100, you in mit?

Answer 12

so far i know that fx= -(1)sin (x-y) and i think is fy=1 cos (x-y)

Answer 13

goformit, ask for your money back, you suck

Answer 14

I think you better redo that... anyway...
since you've started this way, I think it best to redefine our g :)
g(x,y,z) = -z + y cos(x - y)

Answer 15

Now please find the partial derivatives of g with respect to x, y, and z...

Answer 16

z=-1+y cos (x-y) , x= y cos (x-y) and y = -sin(x-y)

Answer 17

the equation of the tangent plane is
fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)−(z−z0)=0

Answer 18

so we find fx, and fy , the partials with respect to x and y

Answer 19

here i found it on this website with an example
http://www.math.hmc.edu/calculus/tutorials/tangentplanes/

Answer 20

Remember that all you need to define a plane in 3d is a point and a normal vector.

By taking the gradient (the vector formed by the partial derivatives of g)
and evaluating it at the point (2,2), you'll get the normal vector at that point, and since you already have a point, you'll get your plane.

Answer 21

i know that z-2= (x- ) + (y- )

Answer 22

given f(x,y) = y cos (x - y)
fx = y(-sin(x-y))
fy= 1*cos(x-y) + y * (-sin(x-y))

Answer 23

z=y

Answer 24

given f(x,y) = y cos (x - y)
fx = y(-sin(x-y)) = - y sin (x-y)
fy= 1*cos(x-y) + y * (-sin(x-y)) = cos(x-y) - y sin(x-y)
The equation will be
fx (2,2) (x-2) + fy( y - 2) + (z-2) = 0

Answer 25

i knew it product to find fy

Answer 26

can you check my fx, and fy?

Answer 27

product rule*

Answer 28

yes

Answer 29

yes he is right, I get z = 2

Answer 30

because fx(2,2) = 0, and fy(2,2) = 0

Answer 31

- y sin (x-y) =0 (2,2)
cos(x-y) - y sin(x-y)= 1 (2,2)

Answer 32

oh wait, sorry

Answer 33

cos (0)=1

Answer 34

fy(2,2) = 1 , so I get
(1) (y-2) + (z-2) = 0

Answer 35

the equation of tangent plane is
fx (2,2)* (x-2) + fy(2,2) *( y - 2) + (z-2) = 0

Answer 36

0(x-2) + 1(y-2) + z-2 = 0

Answer 37

z-2=1(y-2)+(x-2)=> z-2=y=> z=y+2

Answer 38

sorry made a mistake

Answer 39

i notice

Answer 40

the equation of tangent plane is
fx (2,2)* (x-2) + fy(2,2) *( y - 2) - (z-2) = 0

Answer 41

0 * (x-2) + 1( y-2 ) - (z-2) = 0

Answer 42

z-2=y-2=> z=y

Answer 43

correct

Answer 44

:)

Answer 45

thx @perl