Question

find equations of the tangent plane and the normal line to the given surface at the specified point

Answer 1

xy^2z^3=8 (2,2,1)

Answer 2

i know what to do i just dont know how to get started with this one

Answer 3

@freckles could you help real quick with this problem please?

Answer 4

What's the problem, can I help?

Answer 5

its at the top

Answer 6

Do you have wolfram?

Answer 7

\[(xy ^{2z}^3 ) = 8\]

Is this your problem?

Answer 8

no its \[xy ^{2}z ^{3}=8\]

Answer 9

with the point (2,2,1)

Answer 10

yes

Answer 11

What are you trying to find?

Answer 12

the equation of the tangent plane to the given surface at the specified point

Answer 13

You know the equation for the tangent plane?

Answer 14

no...thats what im looking for

Answer 15

z-z\[z - z _{0} = f _{x}(x _{0},y _{0})(x-x _{0}) + f _{y}(x _{0},y _{0})(y-y _{0})\]

Answer 16

okay but i got z^3

Answer 17

z is your point z after you solve for z and take the third root

Answer 18

what about the x and y^2

Answer 19

the cheater's way !!

for \(xy^2z^3=8\) make a level surface \(f = xy^2z^3\) and then compute \(\nabla f = <y^2z^3, 2xyz^3, 3xy^2z^2> \) at \( (2,2,1)\) for the normal vector

Answer 20

okay and what is the next step? @IrishBoy123

Answer 21

well that's it all pretty much done !!

put (x,y,z) = (2,2,1) into that \(\nabla f\) vector and that is the normal \(\vec n\)

the equation for the plane is then \(<x,y,z>.\vec n = const\)

and you have a point on the plane, ie (2,2,1), to get that constant

hope that helps!

Answer 22

thanks @IrishBoy123 couldnt have done it without your help!