1. **State the problem:** Find the gradient and the equation of the tangent plane to the surface given by $$z = 3x^2 - xy$$ at the point $$(1,2,1)$$. 2. **Recall the formula for the gradient:** The gradient vector of a function $$f(x,y)$$ is given by $$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$$. It points in the direction of the greatest rate of increase of the function. 3. **Find partial derivatives:** - $$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(3x^2 - xy) = 6x - y$$ - $$\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(3x^2 - xy) = -x$$ 4. **Evaluate the gradient at the point $$(1,2)$$:** - $$\frac{\partial z}{\partial x}(1,2) = 6(1) - 2 = 4$$ - $$\frac{\partial z}{\partial y}(1,2) = -1$$ So, $$\nabla z(1,2) = (4, -1)$$. 5. **Equation of the tangent plane:** For a surface $$z = f(x,y)$$, the tangent plane at $$(x_0,y_0,z_0)$$ is given by: $$ z - z_0 = f_x(x_0,y_0)(x - x_0) + f_y(x_0,y_0)(y - y_0) $$ Substitute values: $$ z - 1 = 4(x - 1) - 1(y - 2)$$ 6. **Simplify the equation:** $$ z - 1 = 4x - 4 - y + 2$$ $$ z = 4x - y - 1$$ **Final answer:** - Gradient at $$(1,2)$$ is $$\nabla z = (4, -1)$$. - Equation of the tangent plane is $$z = 4x - y - 1$$.