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Mobile App Chapter 2: Problem 20
Show that the following functions have exact differentials: (a) \(x^{2} y+3 y^{2}\), (b) \(x \cos x y\), (c) \(x^{3} y^{2}\), (d) \(t\left(t+e^{3}\right)+s\).
Short Answer
Expert verified
All four functions have exact differentials, as for each function, \( \frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}\) or \( \frac{\partial F_t}{\partial s} = \frac{\partial F_s}{\partial t}\), satisfying the condition for exactness.
Step by step solution
01
Understanding Exact Differentials
A function has an exact differential if it can be written as the derivative of another function with respect to one variable, holding the other variables constant. This can be expressed as the function being the total differential of a potential function. For example, if a function has the form F(x, y), an exact differential has the form \(dF = F_x dx + F_y dy\text{, where } F_x = \frac{\partial F}{\partial x}\text{ and } F_y = \frac{\partial F}{\partial y}\). The function will have an exact differential if \( \frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}\).
02
Verifying Exactness for Function (a) \(x^{2} y+3y^{2}\)
Differentiate \(x^{2} y+3y^{2}\) with respect to x to find \(F_x\), and similarly differentiate it with respect to y to find \(F_y\): \begin{align*} F_x &= \frac{\partial}{\partial x}(x^{2} y+3y^{2}) = 2xy, \ F_y &= \frac{\partial}{\partial y}(x^{2} y+3y^{2}) = x^{2} + 6y. \end{align*} Now check for exactness: \begin{align*} \frac{\partial F_x}{\partial y} &= 2x, \ \frac{\partial F_y}{\partial x} &= 2x. \end{align*} Since \( \frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}\), the function \(x^{2} y+3y^{2}\) has an exact differential.
03
Verifying Exactness for Function (b) \(x \cos x y\)
Differentiate \(x \cos xy\) with respect to x to get \(F_x\) and with respect to y to get \(F_y\): \begin{align*} F_x &= \frac{\partial}{\partial x}(x \cos xy) = \cos xy - xy\sin xy, \ F_y &= \frac{\partial}{\partial y}(x \cos xy) = -x^2\sin xy. \end{align*} Verify exactness: \begin{align*} \frac{\partial F_x}{\partial y} &= -x^2\sin xy, \ \frac{\partial F_y}{\partial x} &= -x^2\sin xy. \end{align*} Since \( \frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}\), the function \(x \cos xy\) has an exact differential.
04
Verifying Exactness for Function (c) \(x^{3} y^{2}\)
Differentiate \(x^{3} y^{2}\) with respect to x to find \(F_x\), and with respect to y to find \(F_y\): \begin{align*} F_x &= \frac{\partial}{\partial x}(x^{3} y^{2}) = 3x^{2}y^{2}, \ F_y &= \frac{\partial}{\partial y}(x^{3} y^{2}) = 2x^{3}y. \end{align*} Check for exactness: \begin{align*} \frac{\partial F_x}{\partial y} &= 6x^{2}y, \ \frac{\partial F_y}{\partial x} &= 6x^{2}y. \end{align*} Since \(\frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}\), the function \(x^{3} y^{2}\) has an exact differential.
05
Verifying Exactness for Function (d) \(t(t+e^{3})+s\)
Since function (d) involves t and s, differentiate \(t(t+e^{3})+s\) with respect to t to obtain \(F_t\) and with respect to s to obtain \(F_s\): \begin{align*} F_t &= \frac{\partial}{\partial t}(t(t+e^{3})+s) = 2t + e^{3}, \ F_s &= \frac{\partial}{\partial s}(t(t+e^{3})+s) = 1. \end{align*} Verify exactness: \begin{align*} \frac{\partial F_t}{\partial s} &= 0, \ \frac{\partial F_s}{\partial t} &= 0. \end{align*} Since \(\frac{\partial F_t}{\partial s} = \frac{\partial F_s}{\partial t}\), the function \(t(t+e^{3})+s\) has an exact differential.
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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 the backbone of multivariable calculus, serving as a fundamental tool for analyzing the behavior of functions with several variables. When dealing with a multivariable function, such as f(x, y), the partial derivative with respect to one variable, say x, is represented as \( \frac{\partial f}{\partial x} \). It measures the rate at which the function changes as the variable x varies, while all other variables are held fixed.
Crucial to understanding exact differentials, partial derivatives allow us to break down complex multivariable functions into simpler slices, reflecting how changes in one direction affect the function's value. In the context of exact differentials, we look for a relationship between the partial derivatives to verify the existence of a potential function whose differential embodies the original multivariable function.
Crucial to understanding exact differentials, partial derivatives allow us to break down complex multivariable functions into simpler slices, reflecting how changes in one direction affect the function's value. In the context of exact differentials, we look for a relationship between the partial derivatives to verify the existence of a potential function whose differential embodies the original multivariable function.
Differential Calculus
Differential calculus focuses on the concept of the derivative, which is a central idea in calculus for understanding how functions change. Fundamentally, differential calculus deals with the calculation of derivatives and their applications, such as rates of change and slopes of curves.
When a function f is differentiable, we can find its differential df, which is an expression approximating the change in f for a small change in its variables. For a function with a single variable, y = f(x), its differential is dy = f'(x)dx. However, with multivariable functions, the total differential involves a sum of terms like df = \( F_x dx + F_y dy \), reflecting contributions from each direction. An exact differential, therefore, is one where this precise form captures the variation in the function across its entire domain.
When a function f is differentiable, we can find its differential df, which is an expression approximating the change in f for a small change in its variables. For a function with a single variable, y = f(x), its differential is dy = f'(x)dx. However, with multivariable functions, the total differential involves a sum of terms like df = \( F_x dx + F_y dy \), reflecting contributions from each direction. An exact differential, therefore, is one where this precise form captures the variation in the function across its entire domain.
Multivariable Calculus
Multivariable calculus extends the principles of single-variable calculus to functions with multiple inputs. It teaches us how to handle functions of two or more variables, such as f(x, y), and how to work with their derivatives and integrals.
In relation to exact differentials, multivariable calculus introduces the concept of a potential function, an antecedent whose gradients (or partial derivatives) give rise to the original function. The test for exactness of a differential, as mentioned in the exercise, arises within this framework. If we can show that the mixed partial derivatives of a function's potential counterpart are equal (meaning they are symmetric with respect to interchange of differentiation order: \( \frac{\partial^2 F}{\partial x \partial y} = \frac{\partial^2 F}{\partial y \partial x} \)), we confirm that an exact differential exists.
This principle ensures that our function behaves nicely under integration and that the path taken during integration doesn't alter the outcome—critical for fields like physics and engineering, where such functions often represent physical quantities.
In relation to exact differentials, multivariable calculus introduces the concept of a potential function, an antecedent whose gradients (or partial derivatives) give rise to the original function. The test for exactness of a differential, as mentioned in the exercise, arises within this framework. If we can show that the mixed partial derivatives of a function's potential counterpart are equal (meaning they are symmetric with respect to interchange of differentiation order: \( \frac{\partial^2 F}{\partial x \partial y} = \frac{\partial^2 F}{\partial y \partial x} \)), we confirm that an exact differential exists.
This principle ensures that our function behaves nicely under integration and that the path taken during integration doesn't alter the outcome—critical for fields like physics and engineering, where such functions often represent physical quantities.
