Problem 2
Question
If \(f(x, y)=x y e^{x y}, \Delta x=-0.1\), and \(\Delta y=0.2\), find (a) the increment of \(f\) at \((2,-4)\) and (b) the total differential of \(f\) at \((2,-4)\).
Step-by-Step Solution
Verified Answer
Increment of f at (2,-4) is \( -5.6e^{-8} \). Total differential of f at (2,-4) is also \( -5.6e^{-8} \)
1Step 1 - Compute Partial Derivatives
First, find the partial derivatives of the function with respect to both variables x and y. Given the function: \text{f}(x, y) = x y e^{x y}Find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \):\( \frac{\partial f}{\partial x} = y e^{xy} + x y^2 e^{xy} \)\( \frac{\partial f}{\partial y} = x e^{xy} + x^2 y e^{xy} \)
2Step 2 - Evaluate Partial Derivatives at (2, -4)
Substitute \( x = 2 \) and \( y = -4 \) into each partial derivative computed in Step 1:\( \frac{\partial f}{\partial x} \bigg|_{(2, -4)} = (-4) e^{2(-4)} + (2)(-4)^2 e^{2(-4)} = -4e^{-8} + 32e^{-8} = 28e^{-8} \)\( \frac{\partial f}{\partial y} \bigg|_{(2, -4)} = 2 e^{2(-4)} + (2)^2(-4) e^{2(-4)} = 2e^{-8} - 16e^{-8} = -14e^{-8} \)
3Step 3 - Compute the Increment of f
The increment of \( f \) at \( (2, -4) \) is found by \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) as follows:\( \text{Increment of } f = \frac{\partial f}{\partial x} \bigg|_{(2, -4)} * \Delta x + \frac{\partial f}{\partial y} \bigg|_{(2, -4)} * \Delta y \)Substitute corresponding values:\( \text{Increment of } f = 28e^{-8} * (-0.1) + (-14e^{-8}) * 0.2 = -2.8e^{-8} - 2.8e^{-8} = -5.6e^{-8} \)
4Step 4 - Compute the Total Differential of f
The total differential \( df \) at \( (2, -4) \) is given by:\( df = \frac{\partial f}{\partial x} \bigg|_{(2, -4)} \Delta x + \frac{\partial f}{\partial y} \bigg|_{(2, -4)} \Delta y \)This calculation is exactly the same as the increment calculation. Hence, \( df = -5.6e^{-8} \)
Key Concepts
Total DifferentialPartial DerivativesMultivariable CalculusIncrement of a Function
Total Differential
The concept of the total differential is crucial in multivariable calculus. It deals with how a small change in each of the variables of a function affects the function's overall change.
For a function of two variables, like our function \ f(x, y) = x y e^{xy} \, the total differential is denoted as \( df \).
It is calculated using the partial derivatives with respect to each variable and their corresponding changes (\( \Delta x \) and \( \Delta y \)).
In the context of the exercise, we found the total differential at (2, -4) to compute how the function's value changes with small increments in \( x \) and \( y \).
Mathematically, it is given by the formula:
\[ df = \frac{\partial f}{\partial x} \bigg|_{(2, -4)} \Delta x + \frac{\partial f}{\partial y} \bigg|_{(2, -4)} \Delta y \]. This tells us how the function changes overall when both variables change slightly.
For a function of two variables, like our function \ f(x, y) = x y e^{xy} \, the total differential is denoted as \( df \).
It is calculated using the partial derivatives with respect to each variable and their corresponding changes (\( \Delta x \) and \( \Delta y \)).
In the context of the exercise, we found the total differential at (2, -4) to compute how the function's value changes with small increments in \( x \) and \( y \).
Mathematically, it is given by the formula:
\[ df = \frac{\partial f}{\partial x} \bigg|_{(2, -4)} \Delta x + \frac{\partial f}{\partial y} \bigg|_{(2, -4)} \Delta y \]. This tells us how the function changes overall when both variables change slightly.
Partial Derivatives
Partial derivatives play a significant role in understanding how functions change with respect to each variable independently in multivariable calculus.
For our given function \( f(x, y) = x y e^{xy} \), the partial derivative with respect to \( x \) tells how \( f \) changes as \( x \) changes, while keeping \( y \) constant. Similarly, the partial derivative with respect to \( y \) tells how \( f \) changes as \( y \) changes, while keeping \( x \) constant.
After differentiating, we found:
\( \frac{\partial f}{\partial x} = y e^{xy} + x y^2 e^{xy} \)
\( \frac{\partial f}{\partial y} = x e^{xy} + x^2 y e^{xy} \)
These expressions allow us to compute the total changes and differentials at any point, as demonstrated in the exercise.
For our given function \( f(x, y) = x y e^{xy} \), the partial derivative with respect to \( x \) tells how \( f \) changes as \( x \) changes, while keeping \( y \) constant. Similarly, the partial derivative with respect to \( y \) tells how \( f \) changes as \( y \) changes, while keeping \( x \) constant.
After differentiating, we found:
\( \frac{\partial f}{\partial x} = y e^{xy} + x y^2 e^{xy} \)
\( \frac{\partial f}{\partial y} = x e^{xy} + x^2 y e^{xy} \)
These expressions allow us to compute the total changes and differentials at any point, as demonstrated in the exercise.
Multivariable Calculus
Multivariable calculus extends single-variable calculus concepts to functions of several variables. This field studies how functions with more than one input variable change and behave.
In the given exercise, moving from single-variable calculus to functions like \( f(x, y) \) involves understanding changes in two dimensions. Multivariable calculus provides the tools to explore how these functions change in different directions.
The total differential and partial derivatives are fundamental concepts, helping us to analyze how changes in inputs influence the function value. They allow us to predict function behavior in multi-dimensional spaces, which is invaluable for fields such as physics, engineering, and economics.
In the given exercise, moving from single-variable calculus to functions like \( f(x, y) \) involves understanding changes in two dimensions. Multivariable calculus provides the tools to explore how these functions change in different directions.
The total differential and partial derivatives are fundamental concepts, helping us to analyze how changes in inputs influence the function value. They allow us to predict function behavior in multi-dimensional spaces, which is invaluable for fields such as physics, engineering, and economics.
Increment of a Function
The increment of a function shows how much the function's value changes due to changes in the input variables. In our exercise, we denote it by \( \Delta f \).
For small changes in \( x \) and \( y \) (given as \( \Delta x = -0.1 \) and \( \Delta y = 0.2 \)), we calculate the increment using the partial derivatives.
Using the formula:
\( \text{Increment of } f = \frac{\partial f}{\partial x} \bigg|_{(2, -4)} \Delta x + \frac{\partial f}{\partial y} \bigg|_{(2, -4)} \Delta y \)
we found the increment for our specific point (2, -4).
In simple terms, the increment tells us how the function's value changes when we slightly tweak its input values.
For small changes in \( x \) and \( y \) (given as \( \Delta x = -0.1 \) and \( \Delta y = 0.2 \)), we calculate the increment using the partial derivatives.
Using the formula:
\( \text{Increment of } f = \frac{\partial f}{\partial x} \bigg|_{(2, -4)} \Delta x + \frac{\partial f}{\partial y} \bigg|_{(2, -4)} \Delta y \)
we found the increment for our specific point (2, -4).
In simple terms, the increment tells us how the function's value changes when we slightly tweak its input values.
Other exercises in this chapter
Problem 1
Let the function \(f\) of two variables \(x\) and \(y\) be the set of all ordered pairs of the form \((P, z)\) such that \(z=(x+y) /(x-y)\). Find: (a) \(f(-3,4)
View solution Problem 2
In Exercises 1 through 8 , do each of the following: (a) Find \(D_{11} f(x, y)\); (b) find \(D_{22} f(x, y) ;\) (c) show that \(D_{12} f(x, y)=D_{21} f(x, y) .\
View solution Problem 2
Let the function \(g\) of three variables, \(x, y\), and \(z\), be the set of all ordered pairs of the form \((P, w)\) such that \(w=\) \(\sqrt{4-x^{2}-y^{2}-z^
View solution Problem 3
In Exercises 1 through 8 , do each of the following: (a) Find \(D_{11} f(x, y)\); (b) find \(D_{22} f(x, y) ;\) (c) show that \(D_{12} f(x, y)=D_{21} f(x, y) .\
View solution