Calculus III

To find the directional derivative of the function \( f(x,y) = xy^2 + x^2y^3 \) at the point (1,2) in the direction of the vector \( \mathbf{a} = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \), we need to follow these steps: 1. Calculate the gradient of the function \( f(x,y) \). 2. Normalize the direction vector \( \mathbf{a} \) if it is not already a unit vector. 3. Compute the dot product of the gradient and the normalized direction vector. ### Step 1: Calculate the Gradient of \( f(x,y) \) The gradient of \( f(x,y) \) is a vector of partial derivatives with respect to each variable. We calculate the partial derivatives as follows: \[ \begin{align*} \frac{\partial f}{\partial x} &= \frac{\partial}{\partial x} (xy^2 + x^2y^3) \\ &= y^2 + 2xy^3, \\ \frac{\partial f}{\partial y} &= \frac{\partial}{\partial y} (xy^2 + x^2y^3) \\ &= 2xy + 3x^2y^2. \end{align*} \] So the gradient vector \( \nabla f \) is: \[ \nabla f = \left( y^2 + 2xy^3, 2xy + 3x^2y^2 \right). \] ### Step 2: Normalize the Direction Vector \( \mathbf{a} \) The direction vector \( \mathbf{a} \) is already a unit vector since its components are \( \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \) and the sum of the squares of its components is 1: \[ \left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2} + \frac{1}{2} = 1. \] Therefore, we do not need to normalize \( \mathbf{a} \); it is already a unit vector. ### Step 3: Compute the Dot Product Now we evaluate the gradient at the point (1,2): \[ \nabla f(1,2) = \left( 2^2 + 2 \cdot 1 \cdot 2^3, 2 \cdot 1 \cdot 2 + 3 \cdot 1^2 \cdot 2^2 \right) = \left( 4 + 16, 4 + 12 \right) = (20, 16). \] The directional derivative of \( f \) at (1,2) in the direction of \( \mathbf{a} \) is the dot product of \( \nabla f(1,2) \) and \( \mathbf{a} \): \[ D_{\mathbf{a}}f(1,2) = \nabla f(1,2) \cdot \mathbf{a} = (20, 16) \cdot \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right). \] Calculating the dot product, we get: \[ D_{\mathbf{a}}f(1,2) = 20 \cdot \frac{1}{\sqrt{2}} + 16 \cdot \frac{1}{\sqrt{2}} = \frac{20}{\sqrt{2}} + \frac{16}{\sqrt{2}} = \frac{36}{\sqrt{2}}. \] To simplify, we can multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \) to rationalize the denominator: \[ D_{\mathbf{a}}f(1,2) = \frac{36}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{36\sqrt{2}}{2} = 18\sqrt{2}. \] Thus, the directional derivative of \( f(x,y) \) at the point (1,2) in the direction of the vector \( \mathbf{a} \) is \( 18\sqrt{2} \).

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Find the directional derivative of \( f(x,y) = xy^2 + x^2y^3 \) at the point (1,2) in the direction of vector \( \mathbf{a} = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \).

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