1. **State the problem:** Find the directional derivative of the function $f(x,y) = x^2 y$ at the point $P = (4,6)$ in the direction of the vector $\vec{v} = 4\vec{i} - 3\vec{j}$. 2. **Recall the formula:** The directional derivative of $f$ at $P$ in the direction of a unit vector $\hat{u}$ is given by $$D_{\hat{u}} f = \nabla f(P) \cdot \hat{u}$$ where $\nabla f(P)$ is the gradient of $f$ at $P$ and $\hat{u}$ is the unit vector in the direction of $\vec{v}$. 3. **Find the gradient $\nabla f(x,y)$:** $$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$$ Calculate each partial derivative: $$\frac{\partial f}{\partial x} = 2xy$$ $$\frac{\partial f}{\partial y} = x^2$$ 4. **Evaluate the gradient at $P = (4,6)$:** $$\nabla f(4,6) = (2 \cdot 4 \cdot 6, 4^2) = (48, 16)$$ 5. **Find the unit vector $\hat{u}$ in the direction of $\vec{v} = 4\vec{i} - 3\vec{j}$:** Calculate the magnitude: $$|\vec{v}| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ Unit vector: $$\hat{u} = \frac{1}{5}(4, -3) = \left( \frac{4}{5}, -\frac{3}{5} \right)$$ 6. **Calculate the directional derivative:** $$D_{\hat{u}} f = \nabla f(4,6) \cdot \hat{u} = (48,16) \cdot \left( \frac{4}{5}, -\frac{3}{5} \right) = 48 \cdot \frac{4}{5} + 16 \cdot \left(-\frac{3}{5}\right) = \frac{192}{5} - \frac{48}{5} = \frac{144}{5} = 28.8$$ **Final answer:** The directional derivative of $f$ at $P$ in the direction of $\vec{v}$ is $\boxed{28.8}$. This shows the previous answer 288 was incorrect because the vector $\vec{v}$ was not normalized before computing the directional derivative.