Question

Find the equation of the tangent plane to the graph of z=xy^2 at the point (1,1,1)

Answer 1

Start with finding the gradient of your function.

Answer 2

We haven't done that yet

Answer 3

Okie doke, not a problem. Have you done partial derivatives yet?

Answer 4

Yes

Answer 5

Okay, first thing's first. Let's move that z over to the other side and let

\[f(x) = xy^2 - z\]

Answer 6

The gradiant of f is just going to be:

(\[(f _{x}, f _{y},f _{z})\]

Answer 7

so...
\[(y ^{2},2yx,-1)\]

Answer 8

Good, now plug in your point and this will give you your vector for the tangent plane.

Answer 9

Just use (1,1,1)

Answer 10

Oh, I see, yes, but your 2nd coordinate should be 2. 2(1)(1)

Answer 11

my bad... (1,2,-1)

Answer 12

Okay, this is the vector for the tangent plane. You've have a point on that plane and a vector.

Where do we go from here?

Answer 13

Some sort of linearization?

Answer 14

Something like that.

We know the general equation for this plane will have:

<1,2,-1>

So:

x + 2y - z = d

Plug your point in (1,1,1) in to that to get your d.

Answer 15

So d=2

Answer 16

Cool, so there's your equation:

x + 2y - z = 2

Answer 17

In general, every sort of problem like this uses these steps.

Answer 18

That was actually really simple! Thank you!

Answer 19

No problem, good luck. The hardest part about this level of calculus is just visualizing what your looking for. The math is pretty much just plug and chug.

:D

Good luck!

Answer 20

Thanks!