Problem 21 - Calculus Early Transcendentals

Question

Find the following derivatives. \(z_{s}\) and \(z_{t},\) where \(z=x y-x^{2} y, x=s+t,\) and \(y=s-t\)

Step-by-Step Solution

Answer

Question: Find the partial derivatives of the function \(z(x,y) = xy - x^2y\) with respect to \(s\) and \(t\) given that \(x = s + t\) and \(y = s - t\). Answer: Using the chain rule, we found the partial derivatives of the function \(z(x,y)\) with respect to \(s\) and \(t\) to be: \(z_s = y - 2xy + x - x^2 = s - t - 2(s+t)(s-t) + s+ t - (s + t)^2\) \(z_t = y - 2xy - x + x^2 = s - t - 2(s+t)(s-t) - s - t + (s + t)^2\)

Step 1: Find the partial derivatives of \(z\) with respect to \(x\) and \(y\).

We begin by calculating \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). To find these derivatives, differentiate the given expression for \(z\) with respect to \(x\) and \(y\) respectively while treating the other variable as constant: \(\frac{\partial z}{\partial x} = y - 2xy\) \(\frac{\partial z}{\partial y} = x - x^2\)

Step 2: Find the partial derivatives of \(x\) and \(y\) with respect to \(s\) and \(t\).

Now, we differentiate the given expressions for \(x(s,t)\) and \(y(s,t)\) with respect to \(s\) and \(t\): \(\frac{\partial x}{\partial s} = 1\) \(\frac{\partial x}{\partial t} = 1\) \(\frac{\partial y}{\partial s} = 1\) \(\frac{\partial y}{\partial t} = -1\)

Step 3: Apply the chain rule.

Using the chain rule, we can now find the partial derivatives of \(z\) with respect to \(s\) and \(t\). The chain rule states that \(\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}\) and \(\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}\). Using our results from Steps 1 and 2: \(z_s = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s} = (y - 2xy)(1) + (x - x^2)(1) = y - 2xy + x - x^2\) \(z_t = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} = (y - 2xy)(1) + (x - x^2)(-1) = y - 2xy - x + x^2\) Now we have found the derivatives \(z_s\) and \(z_t\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with functions of multiple variables. When we want to calculate the derivative of a composite function, we use the chain rule to relate the rate of change of one variable with the rate of change of another through their dependencies. In the context of partial derivatives, the chain rule is applied to functions like our example, where you have multiple interrelated variables.

For the problem at hand, the function \( z \) is dependent on \( x \) and \( y \), and these, in turn, depend on \( s \) and \( t \). To find the partial derivatives \( z_s \) and \( z_t \), one must use the chain rule to express changes in \( z \) with respect to \( s \) and \( t \) based on the dependencies through \( x \) and \( y \):

• \( \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \)
• \( \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \)

To apply the chain rule effectively, it's essential to first identify all dependencies and calculate each partial derivative separately.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of several variables. Unlike single-variable calculus, where a function depends on just one variable, multivariable calculus handles scenarios where functions depend on two or more variables. This field is crucial for understanding systems that are influenced by multiple factors in real-world applications such as physics, engineering, and economics.

In our example, the function \( z = xy - x^2y \) is a representation involving variables \( x \) and \( y \). Each of these components further depends on \( s \) and \( t \). The purpose of evaluating \( z \) in terms of \( s \) and \( t \) highlights why multivariable calculus is essential.

• Multivariable calculus allows for understanding the complex interplay between multiple changing variables.
• It provides tools like partial derivatives to measure how each variable individually affects the function.

Through these tools, altering one or both of \( s \) and \( t \) helps in analyzing how \( z \) changes.

Derivative Computation

Derivative computation involves finding the rate at which a function changes. For multivariable functions, this means taking partial derivatives, which measure how a function changes as one variable is varied, while others are held constant.

In this exercise, the first step is to compute the partial derivatives of \( z \) with respect to \( x \) and \( y \). This involves treating each of \( x \) and \( y \) as an independent variable, while considering the other as constant.

• \( \frac{\partial z}{\partial x} = y - 2xy \)
• \( \frac{\partial z}{\partial y} = x - x^2 \)

Once these are calculated, the next step involves computing the derivatives of \( x \) and \( y \) with respect to \( s \) and \( t \). This helps establish how basic variables like \( x \) and \( y \) vary with \( s \) and \( t \):

• \( \frac{\partial x}{\partial s} = 1 \)
• \( \frac{\partial x}{\partial t} = 1 \)
• \( \frac{\partial y}{\partial s} = 1 \)
• \( \frac{\partial y}{\partial t} = -1 \)

With these derivatives, you can determine how the original function \( z \) changes due to variations in \( s \) and \( t \), using the chain rule to tie everything together.