Let's go through each part step-by-step carefully. --- ### Given: \[f(x, y) = x e^{x^2 - y}\] and point: \[P = (2, 3)\] --- ## (a) Calculate \(\nabla f\) The gradient vector \(\nabla f\) is composed of the partial derivatives: \[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] ### Step 1: Find \(\frac{\partial f}{\partial x}\) Given \(f(x, y) = x e^{x^2 - y}\) Using the product rule: \[ \frac{\partial f}{\partial x} = e^{x^2 - y} + x \cdot \frac{\partial}{\partial x} e^{x^2 - y} \] Since \(e^{x^2 - y}\) is a composite function: \[ \frac{\partial}{\partial x} e^{x^2 - y} = e^{x^2 - y} \cdot 2x \] Thus: \[ \frac{\partial f}{\partial x} = e^{x^2 - y} + x \cdot e^{x^2 - y} \cdot 2x = e^{x^2 - y} (1 + 2x^2) \] ### Step 2: Find \(\frac{\partial f}{\partial y}\) \[ \frac{\partial f}{\partial y} = x \cdot \frac{\partial}{\partial y} e^{x^2 - y} \] Since: \[ \frac{\partial}{\partial y} e^{x^2 - y} = e^{x^2 - y} \cdot (-1) = - e^{x^2 - y} \] Therefore: \[ \frac{\partial f}{\partial y} = x \cdot (- e^{x^2 - y}) = - x e^{x^2 - y} \] --- ### **Answer for (a):** \[ \boxed{ \nabla f(x, y) = \left( e^{x^2 - y} (1 + 2x^2),\ - x e^{x^2 - y} \right) } \] --- ## (b) Calculate \(\nabla f_P\) Evaluate the gradient at \(P = (2, 3)\): First, compute \(e^{x^2 - y}\) at \(x=2\), \(y=3\): \[ x^2 - y = 4 - 3 = 1 \] \[ e^{1} = e \] Now, plug into the gradient: \[ \nabla f(2, 3) = \left( e^{1} (1 + 2 \cdot 2^2),\ - 2 e^{1} \right) \] Calculate each component: - \(1 + 2 \cdot 4 = 1 + 8 = 9\) - \(- 2 e^{1} = - 2 e\) So, \[ \nabla f_P = ( e \cdot 9,\ - 2 e ) = (9 e,\ - 2 e) \] --- ### **Answer for (b):** \[ \boxed{ \nabla f_P = (9 e,\ - 2 e) } \] --- ## (c) Calculate \(\|\nabla f_P\|\) The magnitude: \[ \|\nabla f_P\| = \sqrt{(9 e)^2 + (- 2 e)^2} = e \sqrt{81 + 4} = e \sqrt{85} \] --- ### **Answer for (c):** \[ \boxed{ \|\nabla f_P\| = e \sqrt{85} } \] --- ## (d) Find the rate of change of \(f\) at \(P\) in the **direction of** \(\nabla f_P\) The rate of change of \(f\) in the direction of a unit vector \(\mathbf{u}\) is: \[ D_{\mathbf{u}} f = \nabla f_P \cdot \mathbf{u} \] But when the direction is exactly \(\nabla f_P\), the unit vector is: \[ \mathbf{u} = \frac{\nabla f_P}{\|\nabla f_P\|} = \frac{(9 e, - 2 e)}{e \sqrt{85}} = \left( \frac{9}{\sqrt{85}}, - \frac{2}{\sqrt{85}} \right) \] The rate of change in this direction: \[ D_{\mathbf{u}} f = \nabla f_P \cdot \mathbf{u} = \|\nabla f_P\| \times \left\| \mathbf{u} \right\| \times \cos 0^\circ = \|\nabla f_P\| \times 1 = e \sqrt{85} \] **Alternatively**, directly: \[ \boxed{ \text{Rate of change} = \nabla f_P \cdot \frac{\nabla f_P}{\|\nabla f_P\|} = \|\nabla f_P\| } \] --- ### **Answer for (d):** \[ \boxed{ \boxed{ e \sqrt{85} } } \] --- ## (e) Find the rate of change of \(f\) in the direction making a \(45^\circ\) angle with \(\nabla f_P\) The rate of change in a direction making an angle \(\theta\) with \(\nabla f_P\): \[ D_{\mathbf{v}} f = \|\nabla f_P\| \cos \theta \] Given \(\theta = 45^\circ\): \[ \cos 45^\circ = \frac{\sqrt{2}}{2} \] Thus: \[ D_{\mathbf{v}} f = e \sqrt{85} \times \frac{\sqrt{2}}{2} = \frac{e \sqrt{85} \sqrt{2}}{2} \] Simplify: \[ D_{\mathbf{v}} f = e \times \frac{\sqrt{170}}{2} \] --- ### **Final answer for (e):** \[ \boxed{ \frac{e \sqrt{170}}{2} } \] --- ## **Summary:** | Part | Result | |---------|---------| | (a) | \(\nabla f(x, y) = \left( e^{x^2 - y} (1 + 2x^2),\ - x e^{x^2 - y} \right)\) | | (b) | \(\nabla f_P = (9 e,\ - 2 e)\) | | (c) | \(\|\nabla f_P\| = e \sqrt{85}\) | | (d) | Rate of change in \(\nabla f_P\) direction: \(\boxed{ e \sqrt{85} }\) | | (e) | Rate of change at 45° angle: \(\boxed{ \frac{e \sqrt{170}}{2} }\) | --- Let me know if you'd like further clarification!