Problem 36 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the function. $$f(x, y, z, t)=\frac{x y^{2}}{t+2 z}$$

Step-by-Step Solution

Answer

The first partial derivatives are \( \frac{\partial f}{\partial x} = \frac{y^2}{t + 2z}, \frac{\partial f}{\partial y} = \frac{2xy}{t + 2z}, \frac{\partial f}{\partial z} = -\frac{2xy^2}{(t + 2z)^2}, \frac{\partial f}{\partial t} = -\frac{xy^2}{(t + 2z)^2}. \)

Step 1: Understand the Function

Given function is \( f(x, y, z, t) = \frac{x y^2}{t + 2z} \). It is a multivariable function with four variables: \( x, y, z, t \). We need to find the first partial derivatives with respect to each variable: \( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}, \frac{\partial f}{\partial t} \).

Step 2: Differentiate with Respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y, z, t \) as constants and differentiate \( f \) with respect to \( x \):\[ \frac{\partial f}{\partial x} = \frac{1 \cdot y^2}{t + 2z} = \frac{y^2}{t + 2z}. \]

Step 3: Differentiate with Respect to y

To find \( \frac{\partial f}{\partial y} \), treat \( x, z, t \) as constants and differentiate \( f \) with respect to \( y \):\[ \frac{\partial f}{\partial y} = x \cdot \frac{2y}{t + 2z} = \frac{2xy}{t + 2z}. \]

Step 4: Differentiate with Respect to z

To find \( \frac{\partial f}{\partial z} \), treat \( x, y, t \) as constants and differentiate \( f \) with respect to \( z \). Using the quotient rule, where \( u = xy^2 \) and \( v = t + 2z \), we have:\[ \frac{\partial f}{\partial z} = \frac{(t+2z)(0) - xy^2(2)}{(t+2z)^2} = -\frac{2xy^2}{(t+2z)^2}. \]

Step 5: Differentiate with Respect to t

To find \( \frac{\partial f}{\partial t} \), treat \( x, y, z \) as constants and differentiate \( f \) with respect to \( t \). Again using the quotient rule:\[ \frac{\partial f}{\partial t} = \frac{(t+2z)(0) - xy^2(1)}{(t+2z)^2} = -\frac{xy^2}{(t+2z)^2}. \]

Key Concepts

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions with more than one variable. It allows us to understand and analyze functions where outputs depend on several inputs, such as f(x, y, z, t). This type of calculus is powerful for modeling real-world scenarios.
For example, imagine predicting weather patterns, where temperature and humidity depend on multiple atmospheric variables. In multivariable calculus, we often work with partial derivatives, gradients, and multiple integrals. Understanding how each variable individually affects the function is crucial.
When we find partial derivatives, we're looking at how the function's value changes as just one variable changes, keeping the others constant. This insight is valuable for optimization and maximizing or minimizing outcomes depending on several factors. The exercise you solve helped highlight the process of finding partial derivatives in multivariable functions.

Differentiation

Differentiation focuses on finding how a change in one part of a mathematical system affects the whole system. In single-variable calculus, we compute derivatives to find the rate of change of a function with respect to its variable.In multivariable calculus, this extends to partial derivatives.
Partial derivatives are similar to regular derivatives; however, they focus on the rate of change concerning one variable while keeping other variables constant.For instance, in the given exercise, partial derivatives such as \( \frac{\partial f}{\partial x} \) or \( \frac{\partial f}{\partial y} \) show how the function \( f \) changes in response to changes in \( x \) or \( y \), respectively.
Differentiation is essential because it gives us insightful information about functions.It provides the slope of a function at any given point.This allows us to predict behaviors of systems, optimize solutions, and understand dynamic changes.

Quotient Rule

The quotient rule is a technique used for differentiating functions that are divided by one another.It is especially useful in multivariable calculus when dealing with ratios of functions.
The rule states that if you have a function \( \frac{u}{v} \), the derivative is \( \frac{v \cdot u' - u \cdot v'}{v^2} \).In simple terms, you take the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all over the square of the denominator.
In the exercise, it was necessary to use the quotient rule for calculating \( \frac{\partial f}{\partial z} \) and \( \frac{\partial f}{\partial t} \).This is because the function involves the fraction \( \frac{x y^2}{t+2z} \).The quotient rule ensures you correctly account for changes in both the numerator and the denominator when one variable alters.
Mastering the quotient rule in differentiation helps you tackle complex calculus problems with divided functions.