Question

This just went over my head.. HELP!

Answer 1

The tangent plane to the surface z = x^3 + 4x^2y −2xy^3 at the point (x, y, z) =
(1, 1, 3) contains which of the following lines:

(A) x = 1 − t, y = 1 + 2t, z = 3 − 13t,
(B) x = 1 − 3t, y = 1 + 4t, z = 3 − 32t,
(C) x = 1 + t, y = 1 + 2t, z = 3 + 8t,
(D) x = 1 + 3t, y = 1 + 3t, z = 3 + 17t,
(E) x = 1 + 4t, y = 1 − 4t, z = 3 + 46t.

Answer 2

any idea how to find the tangent plane ?
(using partial derivatives)

Answer 3

not sure, right now. trying to figure this number out has left me thoroughly confused =/

Answer 4

I would write the surface as
\[ f(x,y,z) = x^3 + 4x^2y −2xy^3 - z = 0 \]
and compute \[\nabla f \]

Answer 5

you mean i take the partial derivatives of f wrt x, y and z separately?

Answer 6

yes, to get the 3 components of the gradient.
then evaluate it at (1,1,3)

Answer 7

okay, i'll just do that

Answer 8

just to see if i'm on the right track,
df/dx = 3x^2 + 8xy - 2y^3
df/dy = 4x^2 - 6xy^2 ?

Answer 9

yes

Answer 10

and df/dz is just -1?

Answer 11

yes.
now evalute each partial at (1,1,3) to get just numbers

Answer 12

i get df/dx= 13
df/dy = -2
and df/dz = -1

Answer 13

sorry, df/dx is 9

Answer 14

yes, and that represents the normal to the tangent plane
<9,-2,-1> * <x,y,z> = d
or
9x -2y - z = d
we know point (1,1,3) is in the plane, so use that to find the constant d

Answer 15

d = 4?

Answer 16

now you have the equation of the tangent plane
9x -2y - z = 4

You can test each line, by replacing x,y and z with the parametric info
and simplify. if you get 4, you found a line in this plane

Answer 17

okay, thank you so much!