Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 7-12, find \(\partial w / \partial t\) by using the Chain Rule. Express your final answer in terms of \(s\) and \(t\). $$ w=x^{2}-y \ln x ; x=s / t, y=s^{2} t $$

Step-by-Step Solution

Answer

\( \frac{\partial w}{\partial t} = \frac{s^2 (t^2 - 2)}{t^3} \)

Step 1: Identify the Variables and Given Function

We are given the function \( w = x^2 - y \ln x \) where \( x = \frac{s}{t} \) and \( y = s^2 t \). The goal is to find \( \frac{\partial w}{\partial t} \).

Step 2: Express w in Terms of s and t

Substitute \( x = \frac{s}{t} \) and \( y = s^2 t \) into the function \( w \), resulting in:\[w = \left(\frac{s}{t}\right)^2 - s^2 t \ln\left(\frac{s}{t}\right)\]

Step 3: Simplify the Expression for w

Simplify the expression:\[w = \frac{s^2}{t^2} - s^2 t (\ln s - \ln t)\]

Step 4: Differentiate w with Respect to t

Use the formula for partial derivatives and chain rule on both terms separately:Derivative of \( \frac{s^2}{t^2} \) with respect to \( t \):\[\frac{\partial}{\partial t}\left(\frac{s^2}{t^2}\right) = -\frac{2s^2}{t^3}\]Derivative of \( - s^2 t (\ln s - \ln t) \) with respect to \( t \):\[- s^2 (\ln s - \ln t) - \frac{s^2 t}{t} = s^2\]

Step 5: Combine the Derivatives

Add the derivatives from Step 4:\[\frac{\partial w}{\partial t} = -\frac{2s^2}{t^3} + s^2\]Express the second term in terms of \( t \):\[= -\frac{2s^2}{t^3} + \frac{s^2 t^2}{t^3}\]Combine using a common denominator to get:\[= \frac{s^2 t^2 - 2s^2}{t^3}\]

Step 6: Final Expression for the Partial Derivative

The final expression for the partial derivative in terms of \( s \) and \( t \) is:\[\frac{\partial w}{\partial t} = \frac{s^2 (t^2 - 2)}{t^3}\]

Key Concepts

Partial Derivatives

Partial derivatives are used in calculus to measure how much a multivariable function changes as one of the input variables is varied, while keeping the others constant. Imagine a surface in a 3D space; anything you change affects how that surface behaves.

Calculating a partial derivative involves selecting one variable to change at a time and computing how much the function changes in response to this isolated change. It's like focusing on one direction of a surface at a time.

• Notation: The partial derivative of a function \(f\) with respect to \(x\) is denoted by \(\frac{\partial f}{\partial x}\).
• Interpretation: It's the slope or rate of change of the function in the direction of the selected variable.
• Application: Use in physics, engineering, economics, etc., where changes in multiple variables affect outcomes.

Multivariable Calculus

Multivariable calculus is an extension of single-variable calculus to functions of multiple variables. It involves a variety of new concepts such as partial derivatives, multiple integrals, and vector calculus.

In multivariable calculus, you deal with functions that have more than one input, and you investigate how these inputs affect one another and the overall function. It's as if you're controlling multiple dials on a machine and checking how each dial influences the device's performance.

• Multiple Dimensions: Functions now have inputs like \( (x, y) \) or \( (x, y, z) \), leading to graphs that are planes, surfaces, or even higher-dimensional objects.
• Real-world Problems: Useful in weather prediction, economics, engineering designs, etc.
• Visualization: Often involves contour maps, 3D plots, or vector fields to represent function changes visually.

Function Differentiation

Function differentiation, especially in the context of multivariable calculus, extends the idea of differentiating individual variables to functions with multiple variables. Differentiation finds out how things change; multivariable differentiation examines how these changes occur in multiple dimensions.

When talking about differentiating functions in a multivariable context, you're frequently utilizing the chain rule, which helps to manage how changes in input variables impact an outcome variable.

• The Chain Rule: A method for calculating the derivative of a composite function. It involves nested functions, carrying out differentiation step-by-step.
• Example with Multiple Variables: If you're given a function \( w(x, y) \), and you know \( x = f(s, t) \) and \( y = g(s, t) \), you can find \( \frac{\partial w}{\partial t} \) using the chain rule. The trick is accounting for how both \(x\) and \(y\) influence \(w\) as \(t\) changes.
• Purpose: Essential for finding gradients, optimization problems, and modeling in natural and social sciences.