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Chapter 12: Problem 5

$$ \text { In Problems 1-10, find the gradient } \nabla f \text {. } $$ $$ f(x, y)=x^{2} y /(x+y) $$

Short Answer

Expert verified
The gradient is \( \nabla f = \left(\frac{x^2y + 2xy^2}{(x+y)^2}, \frac{x^3}{(x+y)^2}\right).\)

Step by step solution

01

Understand the Function

We are given a function of two variables, \( f(x,y) = \frac{x^2 y}{x+y} \). The gradient, \( abla f \), is a vector that represents the rate and direction of the fastest increase of the function.
02

Compute the Partial Derivative with respect to x

Using the quotient rule for derivatives, the partial derivative of \( f \) with respect to \( x \) is: \[ \frac{\partial f}{\partial x} = \frac{(2xy)(x+y) - (x^2y)(1)}{(x+y)^2} = \frac{2x^2y + 2xy^2 - x^2y}{(x+y)^2} = \frac{x^2y + 2xy^2}{(x+y)^2}. \]
03

Compute the Partial Derivative with respect to y

Similarly, using the quotient rule, we find the partial derivative of \( f \) with respect to \( y \): \[ \frac{\partial f}{\partial y} = \frac{(x^2)(x+y) - (x^2y)(1)}{(x+y)^2} = \frac{x^3 + x^2y - x^2y}{(x+y)^2} = \frac{x^3}{(x+y)^2}. \]
04

Form the Gradient Vector

The gradient \( abla f \) is the vector of partial derivatives:\[ abla f = \left(\frac{x^2y + 2xy^2}{(x+y)^2}, \frac{x^3}{(x+y)^2}\right). \]

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Key Concepts

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

Partial Derivatives
In calculus, partial derivatives are a way to find the rate of change of a function concerning one of its variables, while keeping others constant. When dealing with a function of multiple variables, partial derivatives allow us to explore how changes in one variable influence the function's value.
  • Consider the function \( f(x, y) \). Here, we can find the partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), which represents the sensitivity of \( f \) to small changes in \( x \).
  • Similarly, \( \frac{\partial f}{\partial y} \) is the partial derivative concerning \( y \), indicating how \( f \) responds to variations in \( y \).
Applying partial derivatives to analyze a function like \( f(x, y) = \frac{x^2 y}{x+y} \) involves differentiating the function with respect to each variable independently, as shown in the original solution. These derivatives are crucial for constructing the gradient vector.
Quotient Rule
The quotient rule is a technique used in calculus to manage differentiation involving division between two functions. Specifically, it applies when the function has the form \( \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions.
  • The formula for the quotient rule is: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).
  • This rule can be particularly helpful when computing partial derivatives of expressions that are fractions, such as \( \frac{x^2 y}{x+y} \).
  • In our example, we apply the quotient rule separately for each variable while keeping the other constant, to find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
This approach ensures you correctly account for the interaction between the numerator and the denominator when differentiating.
Gradient Vector
The gradient vector is a fundamental concept in multivariable calculus, representing the ensemble of partial derivatives of a function. For a function \( f(x, y) \), the gradient is denoted by \( abla f \) and given as:
  • \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)
  • This vector indicates the direction of steepest ascent for the function at a given point.
  • For our function \( f(x, y) = \frac{x^2 y}{x+y} \), the gradient vector combines the partial derivatives derived in the earlier steps: \( abla f = \left( \frac{x^2y + 2xy^2}{(x+y)^2}, \frac{x^3}{(x+y)^2} \right) \).
Thus, the gradient provides essential information about the function's behavior and helps in optimization problems where you seek maximum or minimum values.
Multivariable Functions
Multivariable functions are mathematical expressions that depend on more than one variable. They are central to modeling real-world relationships where outcomes hinge on multiple factors.
Take the function \( f(x, y) = \frac{x^2 y}{x+y} \) as an example. It depends on two variables, \( x \) and \( y \), reflecting how these inputs jointly affect the function's output.
  • Multivariable functions require tools like partial derivatives and gradient vectors for thorough analysis.
  • Understanding these functions allows you to explore how variations in each variable influence the overall function.
Adopting this perspective is essential for disciplines ranging from economics to engineering, where outcomes often rely on multiple variables interacting simultaneously.

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Most popular questions from this chapter

Use the total differential dz to approximate the change in \(z\) as \((x, y)\) moves from \(P\) to \(Q\). Then use a calculator to find the corresponding exact change \(\Delta z\) (to the accuracy of your calculator). See Example \(3 .\) $$ z=x^{2}-5 x y+y ; P(2,3), Q(2.03,2.98) $$

For the function \(f(x, y)=\tan \left(\left(x^{2}+y^{2}\right) / 64\right)\), find the second-order Taylor approximation based at \(\left(x_{0}, y_{0}\right)=(0,0)\). Then estimate \(f(0.2,-0.3)\) using (a) the first-order approximation, (b) the second-order approximation, and (c) your calculator directly.

The temperature \(T\) in degrees Celsius at \((x, y, z)\) is given by \(T=10 /\left(x^{2}+y^{2}+z^{2}\right)\), where distances are in meters. A bee is flying away from the hot spot at the origin on a spiral path so that its position vector at time \(t\) seconds is \(\mathbf{r}(t)=\) \(t \cos \pi t \mathbf{i}+t \sin \pi t \mathbf{j}+t \mathbf{k}\). Determine the rate of change of \(T\) in each case. (a) With respect to distance traveled at \(t=1\). (b) With respect to time at \(t=1\). (Think of two ways to do this.)

Find the directional derivative of \(f(x, y)=e^{-x} \cos y\) at \((0, \pi / 3)\) in the direction toward the origin.

Show that if \(w=f(r-s, s-t, t-r)\) then $$ \frac{\partial w}{\partial r}+\frac{\partial w}{\partial s}+\frac{\partial w}{\partial t}=0 $$
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