Chapter 4: Problem 195
Complete each task. Let \(z=f(x, y)=x e^{y}\). Compute \(\Delta z\) from \(P(1,2)\) to \(Q(1.05,2.1)\) and then find the approximate change in \(z\) from point \(P\) to point Q. \(\quad\) Recall \(\Delta z=f(x+\Delta x, y+\Delta y)-f(x, y), \quad\) and \(d z\) and \(\Delta z\) are approximately equal.
Short Answer
Step by step solution
Identify Given Points and Function
Calculate Delta x and Delta y
Compute z at Point P
Compute z at Point Q
Compute the Exact Change in z, Δz
Approximate dz using Differentials
Compare Δz and dz
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
To calculate these, we treat the chosen variable as a constant and differentiate only with respect to the variable of interest. In our example, for the function \(z = x e^y\),
- The partial derivative with respect to \(x\) is \(\frac{\partial z}{\partial x} = e^y\), which shows how \(z\) changes with \(x\) when \(y\) is fixed.
- Similarly, the partial derivative with respect to \(y\) is \(\frac{\partial z}{\partial y} = x e^y\), illustrating how \(z\) shifts with changes in \(y\).
Differentials
For a function \(z = f(x, y)\), the differential \(dz\) represents an approximation of change as \(x\) and \(y\) change slightly by \(\Delta x\) and \(\Delta y\). It is calculated as:
\[ dz = \frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y \]
In our example, the approximate change \(dz\) is calculated using the partial derivatives found at point \(P(1,2)\). After substituting \(\partial z / \partial x\) and \(\partial z / \partial y\) into the formula and multiplying by \(\Delta x\) and \(\Delta y\), we obtain a quick estimation of how \(z\) changes from \(P\) to \(Q\). Using differentials is like zooming into the function's surface to get a local view of the slope for a slight shift in variables.
Function Approximation
When moving from point \(P\) to point \(Q\), instead of finding the exact change in a rather complex way, we use differentials. This simplifies the computation process. By calculating \(dz\) with partial derivatives, we form a linear approximation for the actual change in \(z\), referred to as \(\Delta z\).
This approximation assumes small changes, making differentials effective and straightforward for estimating transformations in functions. Although it is just an estimate, it's remarkably close to the value obtained through precise calculation. Approximations often make analysis manageable and provide a valuable first look at how a multivariable function evolves.
Change in Functions
To determine \(\Delta z\), we compute the function's value at the new point and subtract the initial value. This gives the exact change in function value when moving from point \(P\) to point \(Q\).
However, this calculation can be tedious and is not always feasible without sufficient information. Instead, employing differentials provides a practical alternative. By using the linear approximation \(dz\), based on partial derivatives, we get an efficient estimate of how the function morphs with small changes. Thus, while \(\Delta z\) offers precision, differentials allow for swift and efficient evaluations of functional change.