Question

Approximate function change Use differentials to approximate the change in z for the given changes in the independent variables. \(z=\ln (1+x+y)\) when \((x, y)\) changes from (0,0) to (-0.1,0.03)

Short answer

Question: Use differentials to approximate the change in z for the function \(z = \ln(1 + x + y)\) when \((x, y)\) changes from (0, 0) to (-0.1, 0.03). Answer: The approximated change in z is -0.07.

Step-by-step solution

Calculate partial derivatives of z

To find the partial derivatives of the function \(z = \ln(1 + x + y)\), calculate the derivatives with respect to x and y. Partial derivative with respect to x: \(\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \ln(1 + x + y) = \frac{1}{1+x+y}\) Partial derivative with respect to y: \(\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} \ln(1 + x + y) = \frac{1}{1+x+y}\)

Evaluate partial derivatives at the initial point (0, 0)

Now we need to evaluate the partial derivatives at the initial values of x and y, (0, 0). At \((x, y) = (0, 0)\): \(\frac{\partial z}{\partial x} = \frac{1}{1+0+0} = 1\) \(\frac{\partial z}{\partial y} = \frac{1}{1+0+0} = 1\)

Calculate the differentials

We are given that \((x, y)\) changes from (0, 0) to (-0.1, 0.03), so the differentials dx and dy are: dx = -0.1 dy = 0.03

Approximate the change in z using differentials

Now we will use the differentials and the partial derivatives at the initial point to approximate the change in z. Δz ≈ \(\frac{\partial z}{\partial x} \cdot dx + \frac{\partial z}{\partial y} \cdot dy\) Δz ≈ \(1 \cdot (-0.1) + 1 \cdot (0.03)\) Δz ≈ \(-0.1 + 0.03\) Δz ≈ \(-0.07\) So, the approximated change in z is -0.07 when \((x, y)\) changes from (0, 0) to (-0.1, 0.03), using differentials.

Key concepts

Partial Derivatives

Partial derivatives are an extension of the concept of derivatives to functions of multiple variables. In single-variable calculus, a derivative represents the rate at which a function changes as its input changes. For a function with more than one variable, such as \(z = f(x, y)\), the partial derivative with respect to one variable gives the rate of change in the function while keeping the other variables constant.

In our exercise, we are given the function \(z = \ln(1 + x + y)\). To start solving the problem, we first need to find the partial derivatives of \(z\) with respect to both \(x\) and \(y\). These partial derivatives are crucial because they tell us how \(z\) changes with small changes in \(x\) and \(y\). For \(z\), both partial derivatives turn out to be \(\frac{1}{1+x+y}\), indicating that changing either \(x\) or \(y\) affects \(z\) in the same proportion, given the function’s symmetrical nature with respect to \(x\) and \(y\).

Function Approximation

Approximation of functions is a core concept in calculus, especially when evaluating expressions analytically might be complex or impossible. Using differentials is one efficient way to approximate the change in a function’s value.

In our context, we are dealing with small changes in \(x\) and \(y\) from (0,0) to (-0.1,0.03). By leveraging the linear approximation provided by partial derivatives at the initial point, we can estimate the change in \(z\). This is done by multiplying the obtained partial derivatives with their respective small changes, or differentials, \(dx\) and \(dy\). The sum of these products gives us the approximate change, \(\Delta z\), in \(z\). This methodology simplifies many real-world problems where exact calculations are cumbersome, allowing for faster and often sufficiently accurate solutions.

Change in Variables

Change in variables is a fundamental concept that allows us to understand how a function behaves when its input variables are modified. To practically apply this in the current problem, we see that \((x, y)\) shifts from (0,0) to (-0.1,0.03).

The differentials \(dx = -0.1\) and \(dy = 0.03\) represent these small changes in \(x\) and \(y\), respectively. Understanding this shift is essential, as these small changes are what we input into our differential approximation technique. Through partial derivatives, the changes in each independent variable are accounted for separately so that their contribution to the change in \(z\), the dependent variable, can be computed. In calculus, this principle is widely used not only in theoretical problems but also when modeling real-world processes where multiple variables interact.

Natural Logarithm

The natural logarithm, denoted \(\ln\), is a special mathematical function. It is the logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. The natural logarithm is extensively used in calculus due to its unique properties that simplify differentiation and integration.

In the given exercise, the function \(z = \ln(1 + x + y)\) utilizes the natural logarithm. One property that proves useful is the derivative of \(\ln(u)\), which is \(\frac{1}{u}\cdot\frac{du}{dx}\). This makes calculating partial derivatives of \(z\) relatively straightforward, as you only need to account for the change inside the logarithm.

This function illustrates how a small change in \(x\) or \(y\) results in a logarithmic scale change in \(z\), a useful characteristic in modeling phenomena such as exponential growth and decay, pH in chemistry, and other such applications where ratios are more meaningful than differences.