Let's solve **Problem 1** step by step, as requested: --- ## Problem 1 Let \( f(x, y, z) = x^2 y + 3y + y^2 z \). ### a) Find the gradient, ∇f(x, y, z). **Step 1: Compute Partial Derivatives** - \( \frac{\partial f}{\partial x} = 2xy \) (since \( x^2y \) is the only term with \( x \)) - \( \frac{\partial f}{\partial y} = x^2 + 3 + 2yz \) (derivative of each term with respect to \( y \)) - \( \frac{\partial f}{\partial z} = y^2 \) (only \( y^2z \) has \( z \)) **Gradient:** \[ \boxed{ \nabla f(x, y, z) = (2xy,\, x^2 + 3 + 2yz,\, y^2) } \] **Explanation:** The gradient is found by taking the partial derivative of \( f \) with respect to each variable. --- ### b) Evaluate ∇f at the point (1, 2, -1). **Step 2: Substitute \( x = 1, y = 2, z = -1 \) into the gradient** - \( 2xy = 2 \times 1 \times 2 = 4 \) - \( x^2 + 3 + 2yz = 1^2 + 3 + 2 \times 2 \times (-1) = 1 + 3 - 4 = \) - \( y^2 = 2^2 = 4 \) So, \[ \boxed{ \nabla f(1, 2, -1) = (4,\, ,\, 4) } \] **Explanation:** Substituting the values into each component gives the gradient at the specific point. --- ## **Final Answer:** - **Gradient:** \( \nabla f(x, y, z) = (2xy,\, x^2 + 3 + 2yz,\, y^2) \) - **At (1, 2, -1):** \( \nabla f(1, 2, -1) = (4,\, ,\, 4) \)