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# quick question on tangent plane watch

1. Let be a polynomial.
Consider the surface

Write down the equation of the tangent plane at the origin

Is this because x and y are the only variables in the polynomial that aren't to the power of 2, 3... etc? Or have I got this wrong?

Thank you
2. Don't you work out the value of the partial derivatives f_x, f_y, f_z at (0,0,0) then use

f_x( x-0) + f_y(y-0) + f_z(z-0) = 0 to get the tangent plane at the origin?

This does give you x+y=0
3. (Original post by vc94)
Don't you work out the value of the partial derivatives f_x, f_y, f_z at (0,0,0) then use

f_x( x-0) + f_y(y-0) + f_z(z-0) = 0 to get the tangent plane at the origin?

This does give you x+y=0
hi, thank you!

i've tried to do as (i think) you're explaining but it doesn't make sense.

i derive and get .

substituting (0, 0, 0), i am left with which obviously isn't right.

i realise that this is from so is that what you've done? i'm probably doing it all wrong!

thank you again!
4. f_x = 1+2x, f_y = 1, f_z = -3z^2.
Evaluate at (0,0,0) gives f_x = 1, f_y = 1, f_z = 0.

Use the equation f_x( x-0) + f_y(y-0) + f_z(z-0) = 0 to get the result.

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