1. The problem: Find the directional derivative of the function $f(x,y) = x^2y + 3y^2$ at the point $(1,2)$ in the direction of the vector $\mathbf{v} = (3,4)$. 2. Formula: The directional derivative of $f$ at point $\mathbf{a}$ in the direction of a unit vector $\mathbf{u}$ is given by: $$D_{\mathbf{u}}f(\mathbf{a}) = \nabla f(\mathbf{a}) \cdot \mathbf{u}$$ where $\nabla f(\mathbf{a})$ is the gradient vector of $f$ at $\mathbf{a}$. 3. Important rules: - The direction vector must be a unit vector, so normalize $\mathbf{v}$ by dividing by its magnitude. - The gradient vector is composed of partial derivatives: $\nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$. 4. Intermediate work: - Compute partial derivatives: $$\frac{\partial f}{\partial x} = 2xy$$ $$\frac{\partial f}{\partial y} = x^2 + 6y$$ - Evaluate at $(1,2)$: $$\frac{\partial f}{\partial x}(1,2) = 2 \times 1 \times 2 = 4$$ $$\frac{\partial f}{\partial y}(1,2) = 1^2 + 6 \times 2 = 1 + 12 = 13$$ - Gradient vector at $(1,2)$: $$\nabla f(1,2) = (4, 13)$$ - Normalize $\mathbf{v} = (3,4)$: $$|\mathbf{v}| = \sqrt{3^2 + 4^2} = 5$$ $$\mathbf{u} = \left(\frac{3}{5}, \frac{4}{5}\right)$$ - Compute directional derivative: $$D_{\mathbf{u}}f(1,2) = (4, 13) \cdot \left(\frac{3}{5}, \frac{4}{5}\right) = \frac{12}{5} + \frac{52}{5} = \frac{64}{5} = 12.8$$ 5. Explanation: The directional derivative measures the rate of change of the function $f$ at the point $(1,2)$ in the direction of the vector $\mathbf{v}$. We first find the gradient vector which points in the direction of greatest increase. Then, by projecting this gradient onto the unit direction vector, we find the rate of change in that specific direction. Final answer: The directional derivative of $f$ at $(1,2)$ in the direction of $\mathbf{v} = (3,4)$ is $12.8$.