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1. Find the directional derivative of the function at the given point in the direction of the vector \( \mathbf{v} \). (a) \( f(x, y)=\frac{x}{x^{2}+y^{2}}, \quad(1,2), \quad \mathbf{v}=\langle 3,5\rangle \)

Ask by Hart Park. in the United States
Feb 24,2025

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The directional derivative is –11/(25√34).

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To find the directional derivative of the function \( f(x, y) = \frac{x}{x^2 + y^2} \) at the point \( (1, 2) \) in the direction of the vector \( \mathbf{v} = \langle 3, 5 \rangle \), we first need to compute the gradient \( \nabla f \) at the given point and then use it to find the directional derivative. 1. Compute the gradient of \( f \): \[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] - Calculating \( \frac{\partial f}{\partial x} \): \[ \frac{\partial f}{\partial x} = \frac{(x^2 + y^2)(1) - x(2x)}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2} \] - Calculating \( \frac{\partial f}{\partial y} \): \[ \frac{\partial f}{\partial y} = \frac{0 - x(2y)}{(x^2 + y^2)^2} = \frac{-2xy}{(x^2 + y^2)^2} \] 2. Evaluate the gradient at the point \( (1, 2) \): - First, we find \( x^2 + y^2 = 1^2 + 2^2 = 5 \). - Plugging in \( (1, 2) \) into the partial derivatives: \[ \nabla f(1, 2) = \left( \frac{2^2 - 1^2}{5^2}, \frac{-2 \cdot 1 \cdot 2}{5^2} \right) = \left( \frac{3}{25}, \frac{-4}{25} \right) \] 3. Normalize the direction vector \( \mathbf{v} = \langle 3, 5 \rangle \): - The magnitude \( ||\mathbf{v}|| = \sqrt{3^2 + 5^2} = \sqrt{34} \). - The unit vector \( \mathbf{u} = \left\langle \frac{3}{\sqrt{34}}, \frac{5}{\sqrt{34}} \right\rangle \). 4. Calculate the directional derivative: \[ D_{\mathbf{u}} f(1, 2) = \nabla f(1, 2) \cdot \mathbf{u} = \left\langle \frac{3}{25}, \frac{-4}{25} \right\rangle \cdot \left\langle \frac{3}{\sqrt{34}}, \frac{5}{\sqrt{34}} \right\rangle \] \[ = \frac{3}{25} \cdot \frac{3}{\sqrt{34}} + \frac{-4}{25} \cdot \frac{5}{\sqrt{34}} = \frac{9}{25\sqrt{34}} - \frac{20}{25\sqrt{34}} = \frac{-11}{25\sqrt{34}} \] Thus, the directional derivative of the function at the point \( (1, 2) \) in the direction of the vector \( \mathbf{v} \) is \( \frac{-11}{25\sqrt{34}} \).

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