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)). 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

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)). 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

anonymous · Answer

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)

anonymous · Answer

looks good.  You can skip the step of using points in the plane to find a line and then find a line normal to that line...  the gradient is normal to the surface (and therefore normal to the tangent plane)   Notice how the components of the Normal line you found are the same as the values of fx anf fy that you found earlier...