Problem 405 - Calculus Volume 3

Question

For the following exercises, find all first partial derivatives. \(u(x, y)=x^{4}-3 x y+1, x=2 t, y=t^{3}\)

Step-by-Step Solution

Answer

The partial derivatives are \(\frac{\partial u}{\partial x} = 4x^3 - 3y\) and \(\frac{\partial u}{\partial y} = -3x\).

Step 1: Understand the Function

We are given the function \(u(x, y) = x^4 - 3xy + 1\), where \(x\) and \(y\) are in terms of \(t\) such that \(x = 2t\) and \(y = t^3\). The goal is to find the first partial derivatives \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\).

Step 2: Find \(\frac{\partial u}{\partial x}\)

To find \(\frac{\partial u}{\partial x}\), treat \(y\) as a constant and differentiate \(u(x, y) = x^4 - 3xy + 1\) with respect to \(x\). This results in: \[\frac{\partial u}{\partial x} = 4x^3 - 3y.\]

Step 3: Find \(\frac{\partial u}{\partial y}\)

To find \(\frac{\partial u}{\partial y}\), treat \(x\) as a constant and differentiate \(u(x, y) = x^4 - 3xy + 1\) with respect to \(y\). This results in: \[\frac{\partial u}{\partial y} = -3x.\]

Step 4: Evaluate at Specific Parametric Forms (Optional)

If you need to find the partial derivatives at specific \(x\) and \(y\), substitute the parameterization \(x = 2t\) and \(y = t^3\) into the partial derivatives found in Steps 2 and 3: Substitute into \(\frac{\partial u}{\partial x} = 4x^3 - 3y\), we get:\[\frac{\partial u}{\partial x} = 4(2t)^3 - 3(t^3)\] Substitute into \(\frac{\partial u}{\partial y} = -3x\), we get:\[\frac{\partial u}{\partial y} = -3(2t)\]

Step 5: Simplify the Derivatives (Optional)

Simplify the derivatives: For \(\frac{\partial u}{\partial x} = 4(2t)^3 - 3(t^3)\): \[\frac{\partial u}{\partial x} = 32t^3 - 3t^3 = 29t^3.\] For \(\frac{\partial u}{\partial y} = -3(2t)\): \[\frac{\partial u}{\partial y} = -6t.\]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus focused on finding the rate at which a function changes at any given point. In simpler terms, it's about determining how a small change in one variable affects another. When dealing with functions of several variables, partial differentiation is used. This involves calculating the derivative of a function with respect to one variable, keeping other variables constant.

For the function given in our exercise, partial differentiation lets us explore how changing just one of the variables \(x\) or \(y\) influences the output \(u(x, y)\).

- To find \(\frac{\partial u}{\partial x}\), treat \(y\) as constant, and differentiate the function as if it's only affected by \(x\). This reflects the direct influence of \(x\) on \(u\) without considering \(y\)'s impact. - Similarly, finding \(\frac{\partial u}{\partial y}\) involves treating \(x\) as constant and determining how changes in \(y\) exclusively affect \(u\). Ultimately, differentiation provides tools for analyzing dynamic changes across variables, significantly benefitting studies involving multi-variable systems.

Parametric Equations

Parametric equations are useful for expressing a set of quantities as explicit functions of independent parameters. They describe curves, surfaces, or more complex structures using equations that rely on a third variable, often denoted as \(t\) for time. In contexts like our exercise, parametric equations provide a framework to express \(x\) and \(y\) in terms of \(t\), where \(x = 2t\) and \(y = t^3\).

The impact of parametric equations cannot be understated:

• They allow us to explore how changes in \(t\) manifest in both \(x\) and \(y\), paving the way to understand the resulting effects on the function \(u(x, y)\).
• Substituting these into partial derivatives enriches the analysis, showing the derivative's dependence on \(t\).
• Assigning specific values to \(t\) offers practical ways to evaluate the function at given points or trajectories.

Parametric equations bridge the gap between traditional Cartesian coordinate expressions and dynamic expressions influenced by an independent variable.

Calculus

Calculus, a branch of mathematics, heavily infuses real-world applications to model change and motion. It comprises two main topics: differentiation (discussed earlier) and integration. While differentiation deals with the rates of change, integration focuses on the accumulation of quantities.

In our given exercise, calculus underpins the methods to derive conclusions by differentiating a multivariable function partially. When applying calculus:

• We gain insights into how small alterations in variables impact the overall function.
• It allows the evaluation of complex problems by breaking them into manageable pieces.
• Analyzing functions in terms of parametric variations shines light on behavior that isn't intuitive in a standard coordinate system.

Mastery of calculus emphasizes understanding the rules and processes, empowering students to forge connections between theory and practical occurrences in physics, engineering, economics, and beyond.