Question

Find the tangent plane for the function f(x, y) = x^2 + y^3 at the point (2, 1).​

Answer 1

Compute the gradient of $f$ at (2, 1).

$\nabla f(x,y) = \langle 2x, 3y^2 \rangle \implies \nabla f(2, 1) = \langle 4, 3 \rangle$

Then the tangent plane to $f(x,y)$ at (2, 1) has equation

$T(x,y) = f(2, 1) + \nabla f(2,1) \cdot \langle x-2, y-1\rangle = \boxed{4x + 3y - 6}$