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Chapter 13: Problem 16

Computing gradients Compute the gradient of the following functions and evaluate it at the given point \(P\). $$h(x, y)=\ln \left(1+x^{2}+2 y^{2}\right) ; P(2,-3)$$

Short Answer

Expert verified
Answer: The gradient of the function at the point P(2,-3) is (鈭噃(2,-3) = (4/23, -12/23).

Step by step solution

01

Given function and point

We are given the function, $$h(x, y) = \ln \left(1+x^{2}+2 y^{2}\right)$$ and we have to find the gradient of h at the point P(2, -3).
02

Compute partial derivatives

We first compute partial derivatives of h with respect to x and y: $$\frac{\partial h}{\partial x} = \frac{1}{1+x^{2}+2 y^{2}} \cdot \frac{\partial}{\partial x}(1+x^{2}+2 y^{2}) = \frac{2x}{1+x^{2}+2 y^{2}}$$ $$\frac{\partial h}{\partial y} = \frac{1}{1+x^{2}+2 y^{2}} \cdot \frac{\partial}{\partial y}(1+x^{2}+2 y^{2}) = \frac{4y}{1+x^{2}+2 y^{2}}$$
03

Evaluate partial derivatives at P

Now we plug the point P(2, -3) into the partial derivatives, $$\frac{\partial h}{\partial x}(2,-3) = \frac{2(2)}{1+(2)^{2}+2 (-3)^{2}} = \frac{4}{1+4+18} = \frac{4}{23}$$ $$\frac{\partial h}{\partial y}(2,-3) = \frac{4(-3)}{1+(2)^{2}+2 (-3)^{2}} = \frac{-12}{1+4+18} = \frac{-12}{23}$$
04

Find the gradient at P

The gradient is given by the vector formed by the partial derivatives: $$\nabla h(2,-3) = \left(\frac{\partial h}{\partial x}(2,-3), \frac{\partial h}{\partial y}(2,-3) \right) = \left(\frac{4}{23}, \frac{-12}{23}\right)$$ The gradient of the given function at the point P(2,-3) is: $$\nabla h(2,-3) = \left(\frac{4}{23}, \frac{-12}{23}\right)$$

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