Question

Find the gradient of the function and the maximum value of the directional derivative at the given point. $$ \frac{\text { Function }}{g(x, y)=\ln \sqrt[3]{x^{2}+y^{2}}} \frac{\text { Point }}{(1,2)} $$

Short answer

To find the gradient of function \(g(x, y)=\ln \sqrt[3]{x^{2}+y^{2}}\) and the maximum value of directional derivative at (1,2), calculating the partial derivatives of the function is necessary. These are computed and evaluated at the given point to yield the gradient vector, \(\nabla g(1, 2)\). The magnitude of this vector represents the maximum value of the directional derivative.

Step-by-step solution

Compute the partial derivatives

The gradient of a function is a vector that is composed of the function's first order partial derivatives. Hence to find the gradient of \(g(x, y)=\ln \sqrt[3]{x^{2}+y^{2}}\), the first order partial derivatives with respect to \(x\) and \(y\) should be calculated:\n Partial derivative with respect to \(x\) uses the derivative rule for natural logarithms along with the chain rule: \(\frac{\partial g(x,y)}{\partial x} = \frac{1}{\sqrt[3]{x^{2}+y^{2}}}\frac{1}{3}\left(\frac{2x}{\sqrt[3]{(x^2+y^2)^2}}\right)\) \n Similarly compute \(\frac{\partial g(x,y)}{\partial y}\) by changing \(x\) to \(y\) in the above equation.

Evaluate the partial derivatives at the given point

The given point is (1,2). Substitute \(x=1\) and \(y=2\) into the partial derivatives obtained in Step 1. That is, \(\frac{\partial g(1,2)}{\partial x}\) and \(\frac{\partial g(1,2)}{\partial y}\). This yields the gradient at the given point.

Compute the gradient vector

The gradient of a function is a vector of its partial derivatives. This will be \(\nabla g(x, y) = \left< \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right>\). The gradient vector at point (1, 2) is \(\nabla g(1, 2) = \left< \frac{\partial g(1, 2)}{\partial x}, \frac{\partial g(1, 2)}{\partial y} \right>\) which was computed in the previous step.

Find the Maximum Value of the Directional Derivative

The maximum value of the directional derivative corresponds to the magnitude of the gradient. This gives the rate of steepest increase of the function from the given point. It is calculated as \(\| \nabla g(1, 2) \| = \sqrt{ \left( \frac{\partial g(1,2)}{\partial x} \right)^2 + \left( \frac{\partial g(1,2)}{\partial y} \right)^2 }\).