Problem 39 - Calculus Early Transcendentals

Question

Write the differential \(d w\) in terms of the differentials of the independent variables. $$w=f(x, y, z)=x y^{2}+z x^{2}+y z^{2}$$

Step-by-Step Solution

Answer

Question: Given the function \(w = f(x, y, z) = xy^2 + zx^2 + yz^2\), find the differential \(dw\). Answer: For the function \(w = f(x, y, z) = xy^2 + zx^2 + yz^2\), the differential \(dw\) is: $$dw = (y^2 + 2zx) dx + (2xy + z^2) dy + (x^2 + 2yz) dz$$

Step 1: Find the partial derivatives with respect to x, y, and z

First, we need to find the partial derivatives of the function with respect to the independent variables: \(\frac{\partial w}{\partial x}\), \(\frac{\partial w}{\partial y}\), and \(\frac{\partial w}{\partial z}\). Recall that to find a partial derivative \(\frac{\partial w}{\partial x}\), we differentiate \(w\) with respect to \(x\) while treating all other variables as constants. The same applies to the other two partial derivatives. $$\frac{\partial w}{\partial x} = \frac{\partial (xy^2+zx^2+yz^2)}{\partial x} = y^2 + 2zx$$ $$\frac{\partial w}{\partial y} = \frac{\partial (xy^2+zx^2+yz^2)}{\partial y} = 2xy + z^2$$ $$\frac{\partial w}{\partial z} = \frac{\partial (xy^2+zx^2+yz^2)}{\partial z} = x^2 + 2yz$$

Step 2: Write the differential of w in terms of differentials of independent variables

Now we need to express the differential \(dw\) in terms of the differentials of the independent variables \(dx\), \(dy\), and \(dz\). Recall the general formula for the total differential: $$dw = \frac{\partial w}{\partial x} dx + \frac{\partial w}{\partial y} dy + \frac{\partial w}{\partial z} dz$$ Applying the formula to our problem with the partial derivatives found in Step 1: $$dw = (y^2 + 2zx) dx + (2xy + z^2) dy + (x^2 + 2yz) dz$$ So, the differential \(dw\) in terms of the differentials of the independent variables is: $$dw = (y^2 + 2zx) dx + (2xy + z^2) dy + (x^2 + 2yz) dz$$

Key Concepts

Partial Derivatives

Partial derivatives are a crucial part of multivariable calculus and are used to understand how a function changes when one of its variables changes, while keeping the others constant. Imagine you have a function, let's call it \(w\), that is dependent on three variables \(x\), \(y\), and \(z\). If you only want to know how \(w\) changes as \(x\) changes, you look at the partial derivative of \(w\) with respect to \(x\). This is represented as \(\frac{\partial w}{\partial x}\). The process involves treating \(y\) and \(z\) as constants and only differentiating with respect to \(x\).
For example, in the function given as \(w = xy^2 + zx^2 + yz^2\), the partial derivative with respect to \(x\) would be \(y^2 + 2zx\).

• The partial derivative with respect to \(y\) considers \(x\) and \(z\) as constants, resulting in \(2xy + z^2\).
• For \(z\), treating \(x\) and \(y\) as constants, it gives \(x^2 + 2yz\).

Getting used to this method helps in analyzing functions with multiple variables.

Total Differential

The total differential is a way to approximate the change in a function when all its variables change by small amounts. In our function \(w = xy^2 + zx^2 + yz^2\), the total differential \(dw\) helps to express how \(w\) changes concerning tiny changes in \(x\), \(y\), and \(z\).
The total differential formula combines all partial derivatives. It looks like this: \[dw = \frac{\partial w}{\partial x}dx + \frac{\partial w}{\partial y}dy + \frac{\partial w}{\partial z}dz\]

• This expression allows you to plug in the changes \(dx\), \(dy\), and \(dz\) to estimate \(dw\).
• The result is a linear approximation of the change in \(w\).
• In our specific case, using the partial derivatives from earlier, the total differential is:\[dw = (y^2 + 2zx)dx + (2xy + z^2)dy + (x^2 + 2yz)dz\]

Understanding the total differential is vital for situations when exact solutions are difficult, and approximations are needed.

Independent Variables

Independent variables are the backbone of any function like \(w = f(x, y, z)\). They are parameters that we can change independently, without one affecting the others. In this context, \(x\), \(y\), and \(z\) serve as our independent variables.

• When dealing with multivariable functions, each independent variable affects the output of the function distinctively.
• This is why partial derivatives are calculated for each independent variable – each represents a different way the function \(w\) can change.
• In the example, when you change \(x\), only the terms that involve \(x\) will contribute to changing \(w\).

Grasping the concept of independent variables is key when working with functions of several variables, as it allows you to explore the function's behavior by manipulating any of these variables separately.