Chapter 14: Problem 5
Find the gradient of the function. $$f(x, y, z)=y z^{2} /\left(1+x^{2}\right)$$
Short Answer
Expert verified
\( \nabla f = \left( \frac{-2yxz^2}{(1+x^2)^2}, \frac{z^2}{1 + x^2}, \frac{2yz}{1 + x^2} \right) \)
Step by step solution
01
Understand the Problem
We need to find the gradient of the function \( f(x, y, z) = \frac{y z^2}{1 + x^2} \), which is a vector containing the partial derivatives of \( f \) with respect to each variable \( x \), \( y \), and \( z \).
02
Compute Partial Derivative with Respect to x
The partial derivative of \( f \) with respect to \( x \) is computed by treating \( y \) and \( z \) as constants. Using the quotient rule, we find: \[ f_x = \frac{\partial}{\partial x}\left(\frac{y z^2}{1 + x^2}\right) = \frac{-2yxz^2}{(1+x^2)^2} \]
03
Compute Partial Derivative with Respect to y
We find the partial derivative of \( f \) with respect to \( y \) by treating \( x \) and \( z \) as constants, leading to a straightforward differentiation: \[ f_y = \frac{\partial}{\partial y}\left(\frac{y z^2}{1 + x^2}\right) = \frac{z^2}{1 + x^2} \]
04
Compute Partial Derivative with Respect to z
The partial derivative of \( f \) with respect to \( z \) is found by treating \( x \) and \( y \) as constants: \[ f_z = \frac{\partial}{\partial z}\left(\frac{y z^2}{1 + x^2}\right) = \frac{2yz}{1 + x^2} \]
05
Combine Partial Derivatives Into Gradient Vector
The gradient vector is composed of the partial derivatives we calculated: \[ abla f = \left( f_x, f_y, f_z \right) = \left( \frac{-2yxz^2}{(1+x^2)^2}, \frac{z^2}{1 + x^2}, \frac{2yz}{1 + x^2} \right) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept in calculus, especially when dealing with functions of multiple variables. To understand them, imagine a function that depends on several variables. For instance, in our function \( f(x, y, z) = \frac{y z^2}{1 + x^2} \), the output depends on \( x \), \( y \), and \( z \). When we compute the partial derivative of this function with respect to \( x \), it means we measure how the function changes as \( x \) changes, while keeping \( y \) and \( z \) constant. This is similar for the partial derivatives with respect to \( y \) and \( z \).
- Computing \( f_x \): This tells us the rate of change of \( f \) with \( x \), considering \( y \) and \( z \) fixed.
- Computing \( f_y \): This gives us the change in \( f \) as \( y \) changes, assuming the other variables remain constant.
- Computing \( f_z \): This shows how \( f \) alters with changes in \( z \), with \( x \) and \( y \) constant.
Multivariable Calculus
Multivariable calculus extends the principles of single-variable calculus to functions of more than one variable. In a world where many phenomena are influenced by several factors, this branch of calculus is incredibly useful. It allows us to explore and understand functions like \( f(x, y, z) = \frac{y z^2}{1 + x^2} \).
In multivariable calculus, we handle functions that depend on multiple independent variables. Instead of analyzing changes along a line, we consider surfaces or even higher-dimensional analogues. Some important concepts in multivariable calculus include:
In multivariable calculus, we handle functions that depend on multiple independent variables. Instead of analyzing changes along a line, we consider surfaces or even higher-dimensional analogues. Some important concepts in multivariable calculus include:
- Gradient: A vector that describes the direction and rate of fastest increase of a scalar field.
- Divergence and Curl: These involve analyzing vector fields and provide ways to understand rotations and expansions.
- Double and Triple Integrals: Extends the idea of integration to compute areas and volumes in higher dimensions.
Quotient Rule
The quotient rule is a technique in calculus used to differentiate functions expressed as the ratio of two functions. In terms of formulas, if you have a function \( g(x) = \frac{u(x)}{v(x)} \), where both \( u \) and \( v \) are differentiable, the derivative \( g'(x) \) is given by:\[\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x) v(x) - u(x) v'(x)}{v(x)^2}.\]
In our original exercise, to find the partial derivative of \( f(x, y, z) = \frac{y z^2}{1 + x^2} \) with respect to \( x \), we applied the quotient rule:
In our original exercise, to find the partial derivative of \( f(x, y, z) = \frac{y z^2}{1 + x^2} \) with respect to \( x \), we applied the quotient rule:
- The numerator was \( y z^2 \), and its derivative with respect to \( x \) is 0, since \( y \) and \( z \) have been treated as constants.
- The denominator \( 1 + x^2 \) differentiates to \( 2x \).
- Substituting these into the formula gives the partial derivative \( f_x \).