Question

Please help me check my work. I needed to find the tangent plane and normal line of the function f(x,y) = x^3*y at point (1,2,f(1,2)).

Answer 1

I wrote for the plane

z = f(1,2) + a (x - 1) + b (y - 2)

where a and b are the partial derivatives of f(x,y) at the point (1,2)

f(1,2) = 2
a = 3x^2 y at the point (1,2) = 6
b = x^3 at the point (1,2) = 1

z = 2 + 6 (x - 1) + (y - 2)

Then I chose two points on the plane:

(1,2,2) and (2,-4,2) and subtracted the second from the first getting:

(-1,6,0), so (-1,6) is parallel to the plane, and the normal line goes through (1,2,2) and is normal to it.

r: (x,y,z) = (1,2,2) + lambda (6,1,0)

Answer 2

I would just rewrite the surface as f(x,y)=z
then create a new surface function in R3
g(x,y,z)=x^3y-z
find grad(g) at x=1, y=2 (which from the function we know is also at z=2)
that is your normal vector n

Answer 3

looks like you got the normal vector to be n=(6,1,0) but I got n=(6,1,-1)

Answer 4

\[g(x,y,z)=x^3y-z\]\[\nabla g=\langle3x^2y,x^3,-1\rangle\]\[\nabla g(1,2,2)=\langle6,1,-1\rangle=\vec n\]the given point on the surface is\[\vec P_0=\langle1,2,2\rangle\]and let\[\vec P=\langle x,y,z\rangle\]now you can use\[\vec n\cdot\vec P_0=\vec n\cdot\vec P\]to get the equation of the plane

Answer 5

so your plane is right, but I think your normal line is wrong
use\[\vec r(t)=\vec P_0+t\vec n\]to find the equation of the line

Answer 6

Thanks.

Answer 7

welcome :)