Question

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

Short answer

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

Step-by-step solution

Identify the Function and Variables

We start by identifying the main function and the given substitutions. The function is \( w = x^2 - y \ln x \), with substitutions \( x = \frac{s}{t} \) and \( y = s^2 t \).

Apply the Chain Rule for Partial Derivatives

To find \( \partial w / \partial t \), we need the partial derivatives of \( w \) in terms of \( x \) and \( y \) and apply the chain rule. According to the chain rule:\[\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial t}\]

Calculate Partial Derivatives of w with respect to x and y

The partial derivative \( \frac{\partial w}{\partial x} \, = \, 2x \, - \, \frac{y}{x}\). The partial derivative \( \frac{\partial w}{\partial y} \, = \, -\ln x\).

Calculate Partial Derivatives of x and y with respect to t

For the substitution \( x = \frac{s}{t} \), the derivative \( \frac{\partial x}{\partial t} \, = \, -\frac{s}{t^2}\). For \( y = s^2t \), the derivative \( \frac{\partial y}{\partial t} = s^2\).

Substitute and Simplify

Using the partial derivatives calculated:\[\frac{\partial w}{\partial t} \, = \, \left(2x - \frac{y}{x}\right) \cdot \left(-\frac{s}{t^2}\right) + \left(- \ln x\right) \cdot s^2\]Substitute \( x = \frac{s}{t} \) and \( y = s^2 t \) into the equation:\[= \left(2\frac{s}{t} - \frac{s^2t}{\frac{s}{t}}\right) \left(-\frac{s}{t^2}\right) - (\ln\left(\frac{s}{t}\right))s^2\] Simplify further to get the final derivative expression.

Final Simplified Expression

After simplification, the expression for \( \frac{\partial w}{\partial t} \) results in:\[\frac{\partial w}{\partial t} = -\frac{2s^2}{t^3} + \frac{s^3}{t^2} \ln t + \frac{s^3}{t^2} \ln s\]This result is the partial derivative of \( w \) with respect to \( t \).

Key concepts

Chain Rule in Calculus

The chain rule is a crucial technique in calculus, especially when dealing with derivatives of functions that are not directly dependent on the variable of differentiation. It allows us to differentiate composite functions. In our exercise, the function \( w = x^2 - y \ln x \) depends on variables \( x \) and \( y \), both of which are in turn dependent on \( t \). The chain rule helps us find how \( w \) changes as \( t \) changes by considering these dependencies.
To apply the chain rule for partial derivatives, we use the formula:\[\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial t}.\]
This formula calculates the total derivative of \( w \) with respect to \( t \) by summing the individual effects of changes in \( x \) and \( y \) with respect to \( t \). The chain rule ensures that no dependency on \( t \) is overlooked, allowing a thorough analysis of how changes in \( t \) influence \( w \).

Function of Several Variables

The function presented in the exercise is a function of several variables, specifically \( w(x, y) = x^2 - y \ln x \). A function of several variables means \( w \) depends directly on more than one independent variable, in this case, \( x \) and \( y \).
In multivariable calculus, understanding how these inputs independently and jointly affect the output \( w \) is fundamental. For example, in our exercise, \( x = \frac{s}{t} \) and \( y = s^2t \), which means both \( x \) and \( y \) involve \( t \) as a variable. Such dependencies indicate that small changes in \( t \) could result in changes in both \( x \) and \( y \), thus affecting \( w \).
This complexity of interaction between variables is why partial derivatives are so valuable. They allow us to focus on how the function changes with respect to one variable at a time, holding others constant.

Differentiation Techniques

Differentiating a function that includes several variables often requires a careful application of different differentiation techniques. One essential technique involves rewriting complex terms or substitutions. In the exercise, for instance, we have\( x = \frac{s}{t} \) and \( y = s^2t \), which we differentiate with respect to \( t \).
We calculate the partial derivatives \( \frac{\partial w}{\partial x} \) and \( \frac{\partial w}{\partial y} \) to apply the chain rule effectively. For \( \frac{\partial w}{\partial x} \), using the power and logarithmic rules, we get \( 2x - \frac{y}{x} \), and for \( \frac{\partial w}{\partial y} \), the derivative simplifies to \(-\ln x\).
Similarly, finding \( \frac{\partial x}{\partial t} \) and \( \frac{\partial y}{\partial t} \) involves recognizing each variable's structure; for \( x = \frac{s}{t} \), the derivative, treating \( s \) as a constant, is \(-\frac{s}{t^2} \), and for \( y = s^2t \), the derivative is \( s^2 \).
This blend of power rule, product rule, and quotient rule in differentiation are parts of the broader techniques you shroud apply confidently in calculus.

Multivariable Calculus

Multivariable calculus expands upon single-variable calculus concepts by dealing with functions that have multiple inputs. It enables us to model and solve real-world problems where dependencies cross more than one factor, making partial derivatives a centerpiece.
In our problem, we found the derivative \( \frac{\partial w}{\partial t} \), demonstrating how changes in variables \( x \) and \( y \) relate to \( t \). In real-world problems, such dependencies can describe physical phenomena like temperature changes over time across varying spatial coordinates.
Multivariable calculus also introduces the concept of higher-dimensional spaces, allowing for more complex modeling and analysis. In the function \( w(x, y, t) \), even though \( w \) only directly depends on \( x \) and \( y \), each is a function of other variables, expanding the dimensional space of the problem.
Therefore, understanding concepts like the chain rule and partial differentiation opens doors to solving problems requiring an appreciation of how multiple factors interact dynamically.