Problem 6 - Active Calculus Multivariable

Question

Let \(z=g(u, v)\) and \(u(r, s), v(r, s)\). How many terms are there in the expression for \(\partial z / \partial r ?\) __________ terms

Step-by-Step Solution

Answer

The expression for \(\frac{\partial z}{\partial r}\) has a total of __2__ terms.

Step 1: Find \(\frac{\partial u}{\partial r}\), \(\frac{\partial u}{\partial s}\), \(\frac{\partial v}{\partial r}\), and \(\frac{\partial v}{\partial s}\).

First, compute the partial derivatives of \(u\) with respect to both \(r\) and \(s\). Next, compute the partial derivatives of \(v\) with respect to both \(r\) and \(s\). #Step 2: Compute the chain rule terms#

Step 2: Find \(\frac{\partial z}{\partial u}\frac{\partial u}{\partial r}\), \(\frac{\partial z}{\partial u}\frac{\partial u}{\partial s}\), \(\frac{\partial z}{\partial v}\frac{\partial v}{\partial r}\), and \(\frac{\partial z}{\partial v}\frac{\partial v}{\partial s}\).

Apply the chain rule to find the four terms that will be summed to compute the partial derivative of\(z\) with respect to \(r\). Multiply the partial derivatives in pairs as shown above to get the chain rule terms. #Step 3: Sum the terms to find the partial derivative of z with respect to r#

Step 3: Compute \(\frac{\partial z}{\partial r} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial r} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial r}\).

Sum the relevant chain rule terms to find the expression for the partial derivative of \(z\) with respect to \(r\). The expression for \(\frac{\partial z}{\partial r}\) has a total of __2__ terms.

Key Concepts

Multivariable Calculus

When we step into the world of multivariable calculus, we enter a realm where functions are no longer limited to a single variable. Instead, they depend on two or more variables, such as in the expression \( z=g(u, v) \) from our problem.

Imagine standing on a mountain, where your position is determined by two coordinates: east-west, and north-south. If the temperature at your location varies based on these two directions, then we would need multivariable calculus to describe how the temperature changes as you move. This concept is similar to what we do in multivariable calculus: we explore how changes in variables like \( u \) and \( v \) affect a function like \( z \).

In our exercise, we're asked to relate the changes in \( z \) not directly to \( u \) and \( v \), but through another set of variables \( r \) and \( s \). This is where we begin to apply the machinery of multivariable calculus, including partial derivatives and the chain rule, to uncover these relationships.

Partial Derivatives

Partial derivatives are at the heart of multivariable calculus. They reveal how a multivariable function changes as one variable varies while others are held constant. To anchor this concept, think about how a gardener watches over a sunflower. The gardener might wonder how the height of the sunflower increases with more water or sunlight. But when considering water, they want to ignore changes in sunlight. This is akin to taking a partial derivative.

In our textbook exercise, the partial derivatives \( \frac{\partial u}{\partial r} \), \( \frac{\partial u}{\partial s} \), \( \frac{\partial v}{\partial r} \), and \( \frac{\partial v}{\partial s} \) are pieces of this puzzle. Each one tells us how the respective variable (\( u \) or \( v \)) changes as we adjust either \( r \) or \( s \) alone. Notice that we don't consider how \( r \) and \( s \) might simultaneously influence \( u \) or \( v \); that's the essence of a partial derivative—focusing on one variable at a time.

Chain Rule Application

The chain rule is a powerful tool in both single-variable and multivariable calculus. It provides a method to compute the derivative of a composite function. Imagine a clock with gears: turning one gear causes another to turn. The chain rule helps us understand how much the second gear turns when we turn the first one.

In our exercise, the chain rule describes how changes in \( r \) (and \( s \)) impact \( z \), through their effect on \( u \) and \( v \). Specifically, it tells us that to find the partial derivative of \( z \) with respect to \( r \), we consider both how \( r \) affects \( u \) (and thus \( z \) through \( u \)) and how \( r \) affects \( v \) (and thus \( z \) through \( v \)).

To apply the chain rule, we multiply each partial derivative of \( z \) with respect to \( u \) and \( v \) by the partial derivatives of \( u \) and \( v \) with respect to \( r \), respectively. Then, summing these products gives us the total effect of \( r \) on \( z \), which, in our case, results in two terms, highlighting the interconnectedness of these variables.