Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(z=x y+x+y, x=r+s+t\), and \(y=r s t\), find \(\left.\frac{\partial z}{\partial s}\right|_{r=1, s=-1, t=2}\)

Step-by-Step Solution

Answer

The partial derivative \(\left.\frac{\partial z}{\partial s}\right|_{r=1, s=-1, t=2} = -1.\)

Step 1: Express z in terms of r, s, and t

Start by substituting the given expressions for \(x\) and \(y\) into \(z=x y + x + y\). Since \(x = r+s+t\) and \(y = rst\), substitute these into \(z\) to get: \[z = (r+s+t)(rst) + (r+s+t) + rst.\] Simplifying, we have \[z = rst(r+s+t) + r+s+t + rst.\]

Step 2: Simplify the expression for z

To simplify \(z = rst(r+s+t) + r+s+t + rst\), expand the first term: \[z = r^2st + rs^2t + rst^2 + r + s + t + rst.\] Combine like terms: \[z = r^2st + rs^2t + rst^2 + rst + r + s + t.\]

Step 3: Calculate the partial derivative of z with respect to s

Find \(\frac{\partial z}{\partial s}\) by differentiating each term of \(z\) with respect to \(s\): - \(\frac{\partial}{\partial s}(r^2st) = 2rst\),- \(\frac{\partial}{\partial s}(rs^2t) = 2rst\),- \(\frac{\partial}{\partial s}(rst^2) = rt^2\),- \(\frac{\partial}{\partial s}(rst) = rt\),- \(\frac{\partial}{\partial s}(r) = 0\),- \(\frac{\partial}{\partial s}(s) = 1\),- \(\frac{\partial}{\partial s}(t) = 0\).Therefore, \(\frac{\partial z}{\partial s} = 2rst + 2rst + rt^2 + rt + 1.\)

Step 4: Simplify the derivative

Combine like terms in the derivative expression: \(\frac{\partial z}{\partial s} = 4rst + rt^2 + rt + 1.\)

Step 5: Evaluate the derivative at given values

Substitute \(r=1, s=-1, t=2\) into the expression for \(\frac{\partial z}{\partial s}\): \[\frac{\partial z}{\partial s} = 4(1)(-1)(2) + (1)(2)^2 + (1)(2) + 1.\]Calculate each term:- \(4(1)(-1)(2) = -8\),- \(1(2)^2 = 4\),- \(1(2) = 2\),- Adding these gives: \(-8 + 4 + 2 + 1 = -1\).

Step 6: Final answer for the partial derivative

Conclude that after evaluating and simplifying, the partial derivative \(\frac{\partial z}{\partial s}\) at \(r=1, s=-1, t=2\) is \(-1\).

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with rates of change and the accumulation of quantities. It's divided into two main parts: differential calculus and integral calculus. Differential calculus focuses on the concept of the derivative. Derivatives are used to describe how a function changes as its input changes. This is crucial in various fields such as physics, engineering, and economics.

Derivatives answer questions like "How fast is something changing?" or "At what rate is something growing?" Think of it as looking at the steepness or slope of a curve. In single-variable calculus, we look at how changes in one variable affect the function. However, when dealing with functions of more than one variable, like in our given exercise, we dive into multivariable calculus.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions of more than one variable. In problems involving multiple variables, we often seek how one specific variable affects the entire system or function. This is where partial derivatives come into play. They help us understand how a function changes with respect to one variable while others are held constant.

In our exercise, the function is expressed in terms of three variables: \( r, s, \) and \( t \). We are tasked with finding the partial derivative of \( z \) with respect to \( s \). This means determining how \( z \) changes as \( s \) changes, while keeping \( r \) and \( t \) constant. The process involves differentiating the function with partial differentiation techniques and evaluating it at specific values, which we did to find the solution \( \frac{\partial z}{\partial s} = -1 \).

Partial derivatives are crucial in modeling real-world phenomena where changes occur in multiple dimensions simultaneously, such as weather systems, economics, and fluid dynamics.

Differentiation Rules

Differentiation rules are the tools that enable us to take derivatives effectively. In calculus, especially when dealing with multivariable functions, knowing differentiation rules is invaluable. Some key rules include the power rule, product rule, and chain rule.

• Power Rule: This rule helps us differentiate functions in the form of \(x^n\), where the derivative is \(nx^{n-1}\).
• Product Rule: This rule is used when differentiating the product of two functions. It states that for functions \(u\) and \(v\), the derivative is given by \((uv)' = u'v + uv'\).
• Chain Rule: This rule helps differentiate composite functions. If a function \(y = f(u)\) and \(u = g(x)\), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

In the exercise provided, these rules assist in calculating the partial derivative with respect to \(s\). Each term is addressed individually by applying these rules, allowing us to express how \(z\) changes as \(s\) changes. Understanding and applying these differentiation rules are crucial skills in tackling complex calculus problems efficiently.