Question

Differentials with more than two variables Write the differential dw in terms of the differentials of the independent variables. $$w=f(p, q, r, s)=\frac{p q}{r s}$$

Short answer

Question: Given the function \(w = \frac{pq}{rs}\), find the differential of w (dw) in terms of the differentials of the independent variables p, q, r, and s. Answer: \(dw = (\frac{q}{rs}) dp + (\frac{p}{rs}) dq + (-\frac{pq}{r^2 s}) dr + (-\frac{pq}{rs^2}) ds\)

Step-by-step solution

Find the partial derivatives of w with respect to p, q, r, and s

First, we will find the partial derivatives of w with respect to each of the independent variables p, q, r, and s. The partial derivative of w with respect to p will give us how much w changes with respect to a small change in p, while keeping the other variables q, r, and s constant. The same applies to the other partial derivatives. Using our given function, \(w = \frac{pq}{rs}\), \(\frac{\partial w}{\partial p} = \frac{q}{rs}\) \(\frac{\partial w}{\partial q} = \frac{p}{rs}\) \(\frac{\partial w}{\partial r} = -\frac{p q}{r^2 s}\) \(\frac{\partial w}{\partial s} = -\frac{pq}{r s^2}\)

Write the differential dw as the sum of the product of the partial derivatives and the differentials of the independent variables

Now that we have found the partial derivatives of w with respect to each of the independent variables, we can write the differential of w (dw) as the sum of the product of the partial derivatives and the differentials of the independent variables: \(dw = \frac{\partial w}{\partial p} dp + \frac{\partial w}{\partial q} dq + \frac{\partial w}{\partial r} dr + \frac{\partial w}{\partial s} ds\)

Substitute the partial derivatives into the expression for dw

We will now substitute the partial derivatives we found in Step 1 into the expression for dw: \(dw = (\frac{q}{rs}) dp + (\frac{p}{rs}) dq + (-\frac{p q}{r^2 s}) dr + (-\frac{pq}{rs^2}) ds\)

Simplify the expression for dw (if possible)

In this case, it is not possible to simplify the expression for dw any further. Thus, the final expression for the differential of w in terms of the differentials of the independent variables is: \(dw = (\frac{q}{rs}) dp + (\frac{p}{rs}) dq + (-\frac{pq}{r^2 s}) dr + (-\frac{pq}{rs^2}) ds\) This is the differential of the function w in terms of the differentials of the independent variables p, q, r, and s.

Key concepts

Partial Derivatives

Partial derivatives are a foundational concept in multivariable calculus, representing how a function changes as one of its variables is varied, while the other variables are held constant. To understand this, imagine gently pushing a rubber sheet stretched tight—you're affecting only a small part of the sheet while the rest remains still.

In the provided exercise, we explored the function w in relation to four independent variables: p, q, r, and s. We asked, 'How does w change as we tweak each variable slightly?' Calculating the partial derivatives \( \frac{\partial w}{\partial p} \), \( \frac{\partial w}{\partial q} \), \( \frac{\partial w}{\partial r} \), and \( \frac{\partial w}{\partial s} \) provided us with those rates of change. Each partial derivative is like a GPS direction for w, telling us how to navigate the change in the function's value steadily, one variable at a time.

Multivariable Calculus

Multivariable calculus extends the concepts of single variable calculus to functions with more than one input. Think of it like moving from tracking a single path in the woods to exploring an entire map with many trails. In the given problem, w depends on four trails—or variables: p, q, r, and s.

The principle behind differentials within multivariable calculus is about measuring small stretches or movements along these trails. We express the tiny stretch in w, denoted as dw, as a combination of stretches along each variable's path, represented as dp, dq, dr, and ds. The differentials tell us the total change in w given the joint effect of minor changes in all of the variables it depends on.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives, and they play a monumental role in modeling real-world phenomena such as population growth or motion of objects. The equation represents a relationship: one side tells us about changes or differentials, and the other side tells us about the function itself.

While our exercise doesn't solve a differential equation per se, it illustrates a foundational skill in forming such equations: understanding how differentials relate to each other. In constructing the differential dw, we established a relationship not just between w and its partial derivatives but also the independent variables' differentials. It's like piecing together a puzzle where each piece's shape is defined by how it changes, creating a larger, more complex picture of the whole.

Independent Variables

Independent variables are the inputs or parameters that we believe cause some effect on the output of a function. They are free to vary, and their selection is crucial in the study of functions, especially in cases where those functions model complex, real-life situations.

In our exercise, p, q, r, and s are the independent variables of the function w. When we calculate the differential dw, we consider the effect of infinitesimally small changes in these variables—one at a time—on the outcome. Think of them as knobs on a control panel; as you adjust each knob, the display (our function w) reacts. Together, they paint a dynamic picture of how w behaves as its environment shifts.