Problem 3 Find $D_{\mathfrak{u}} f$ at \... [FREE SOLUTION]

Chapter 13: Problem 3

Find $D_{\mathfrak{u}} f$ at $P$. $$ \begin{array}{l}{f(x, y)=\ln \left(1+x^{2}+y\right) ; P(0,0)} \\\ {\mathbf{u}=-\frac{1}{\sqrt{10}} \mathbf{i}-\frac{3}{\sqrt{10}} \mathbf{j}}\end{array} $$

Short Answer

Expert verified

The directional derivative $D_{\mathfrak{u}} f$ is $-\frac{3}{\sqrt{10}}$.

Step by step solution

Calculate Gradient of f

First, find the gradient of the function $ f(x, y) = \ln(1+x^2+y) $. The gradient is computed as: $ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) $. Calculate each partial derivative.- $ \frac{\partial f}{\partial x} = \frac{2x}{1+x^2+y} $,- $ \frac{\partial f}{\partial y} = \frac{1}{1+x^2+y} $.Thus, the gradient of $ f $ is $ abla f = \left( \frac{2x}{1+x^2+y}, \frac{1}{1+x^2+y} \right) $.

Calculate Gradient at Point P

Substitute the point $ P(0, 0) $ into the gradient components.- For $ \frac{\partial f}{\partial x} $:$ \frac{2(0)}{1+(0)^2+0} = 0 $- For $ \frac{\partial f}{\partial y} $:$ \frac{1}{1+(0)^2+0} = 1 $Thus, $ abla f(P) = (0, 1) $.

Calculate Directional Derivative $D_{\mathfrak{u}} f$

The directional derivative $ D_{\mathfrak{u}} f $ is given by $ abla f \cdot \mathbf{u} $. Here, $ \mathbf{u} = -\frac{1}{\sqrt{10}} \mathbf{i} - \frac{3}{\sqrt{10}} \mathbf{j} $ is the unit vector. Compute the dot product:\[ D_{\mathfrak{u}} f = (0, 1) \cdot \left( -\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}} \right) \].This simplifies to $-\frac{3}{\sqrt{10}} $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gradient

The gradient of a function is a vector that points in the direction of the greatest rate of increase of a function. For a function of two variables, like $ f(x, y) = \ln(1 + x^2 + y) $, the gradient, denoted as $ abla f $, is given by

The partial derivative with respect to $ x $, $ \frac{\partial f}{\partial x} $
The partial derivative with respect to $ y $, $ \frac{\partial f}{\partial y} $

Substituting these into the gradient vector forms:$ abla f = \left( \frac{2x}{1+x^2+y}, \frac{1}{1+x^2+y} \right) $.
At a specific point, say $ P(0,0) $, substitute $ x = 0 $ and $ y = 0 $

$ \frac{\partial f}{\partial x} = 0 $
$ \frac{\partial f}{\partial y} = 1 $

Thus, the gradient at $ P $ is $ abla f(P) = (0, 1) $.
This means the steepest increase of the function at this point is in the positive $ y $-direction.

Partial Derivatives

Partial derivatives are used to understand how a function changes as one of its variables changes, while keeping other variables constant. Essentially, they are the derivatives of functions with multiple variables.
In our function $ f(x, y) = \ln(1 + x^2 + y) $, there are two partial derivatives:

$ \frac{\partial f}{\partial x} = \frac{2x}{1+x^2+y} $ - This tells us the rate of change of the function with respect to $ x $, treating $ y $ as constant.
$ \frac{\partial f}{\partial y} = \frac{1}{1+x^2+y} $ - This expresses the rate of change with respect to $ y $, keeping $ x $ constant.

These derivatives form the components of the gradient vector, providing essential information about the function's behavior.

Dot Product

The dot product is a key operation in vector calculus and can be seen as a way to multiply two vectors to obtain a scalar. With regard to directional derivatives, it helps determine how a function changes in the direction of a specific vector.

Given two vectors $ \mathbf{a} = (a_1, a_2) $ and $ \mathbf{b} = (b_1, b_2) $, the dot product is: $ \mathbf{a} \cdot \mathbf{b} = a_1 \cdot b_1 + a_2 \cdot b_2 $
For our directional derivative, the gradient $ abla f(P) = (0, 1) $ and the unit vector $ \mathbf{u} = -\frac{1}{\sqrt{10}} \mathbf{i} - \frac{3}{\sqrt{10}} \mathbf{j} $.
The dot product here yields: $ D_{\mathfrak{u}} f = (0, 1) \cdot \left( -\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}} \right) $

This results in $ -\frac{3}{\sqrt{10}} $, describing the rate of change of the function in the specific direction defined by $ \mathbf{u} $.
In essence, the dot product gives us the directional sensitivity of the function at a particular point.

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91影视

Find \(D_{\mathfrak{u}} f\) at \(P\). $$ \begin{array}{l}{f(x, y)=\ln \left(1+x^{2}+y\right) ; P(0,0)} \\\ {\mathbf{u}=-\frac{1}{\sqrt{10}} \mathbf{i}-\frac{3}{\sqrt{10}} \mathbf{j}}\end{array} $$

Short Answer

Step by step solution

Calculate Gradient of f

Calculate Gradient at Point P

Calculate Directional Derivative \(D_{\mathfrak{u}} f\)

Key Concepts

Gradient

Partial Derivatives

Dot Product

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