Problem 26 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the function. $$f(x, t)=\arctan (x \sqrt{t})$$

Step-by-Step Solution

Answer

\(\frac{\partial f}{\partial x} = \frac{\sqrt{t}}{1+x^2 t}, \quad \frac{\partial f}{\partial t} = \frac{x}{2\sqrt{t}(1+x^2 t)}\)

Step 1: Understand the Concept of Partial Derivatives

Partial derivatives involve differentiating a function with respect to one variable while keeping other variables constant. For a function of two variables, such as \(f(x, t)=\arctan(x \sqrt{t})\), partial derivatives \(\frac{{\partial f}}{{\partial x}}\) and \(\frac{{\partial f}}{{\partial t}}\) must be calculated.

Step 2: Calculate the Partial Derivative with Respect to x

To find \(\frac{{\partial f}}{{\partial x}}\), treat \(t\) as a constant. The derivative of \(\arctan(u)\) with respect to \(u\) is \(\frac{1}{{1+u^2}}\). So, using chain rule:\[\frac{{\partial f}}{{\partial x}} = \frac{1}{{1+(x\sqrt{t})^2}} \cdot \frac{{d}}{{dx}}(x\sqrt{t}) = \frac{1}{{1+x^2 t}} \cdot \sqrt{t} = \frac{\sqrt{t}}{{1+x^2 t}}\]

Step 3: Calculate the Partial Derivative with Respect to t

To find \(\frac{{\partial f}}{{\partial t}}\), treat \(x\) as a constant. First apply the chain rule. With the same outer derivative \(\frac{1}{{1+u^2}}\) for \(\arctan(u)\):\[\frac{{\partial f}}{{\partial t}} = \frac{1}{{1+(x\sqrt{t})^2}} \cdot \frac{d}{{dt}}(x\sqrt{t}) = \frac{1}{{1+x^2 t}} \cdot \frac{x}{2\sqrt{t}} = \frac{x}{2\sqrt{t}(1+x^2 t)}\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. This rule helps in breaking down the differentiation process into manageable parts by considering the function's inner and outer derivatives. For instance, in the function \(f(x, t) = \arctan(x\sqrt{t})\), we encounter two layers:

• The outer function is \(\arctan(u)\), where the chain rule provides its derivative as \(\frac{1}{1+u^2}\).
• The inner function involves the expression \(x\sqrt{t}\), where its derivative with respect to one variable must be considered.

When applying the chain rule, you begin by differentiating the outer function while keeping the inner function unchanged, then multiply by the derivative of the inner function. This process helps in finding partial derivatives, especially in functions of more than one variable, by treating one variable as constant at a time.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function's output value changes concerning its input values. When dealing with functions of two variables, like \(f(x, t) = \arctan(x\sqrt{t})\), partial differentiation is used.Here's what you need to know:

• Consider the role of each variable separately while performing partial differentiation.
• Keep all other variables constant except for the one you are differentiating with respect to.

In partial derivatives, we focus on ambiguous parts of a function by isolating the effect that each variable has. Therefore, the notation \(\frac{\partial f}{\partial x}\) is used to describe differentiation with respect to \(x\), while treating \(t\) as a fixed number, allowing us to solve for changes in one dimension of the function's landscape.

Functions of Two Variables

Functions of two variables can be visualized as surfaces in a three-dimensional space. These functions rely on two independent variables, and their output can vary accordingly. In the context of \(f(x, t) = \arctan(x\sqrt{t})\), each variable—\(x\) and \(t\)—contributes to determining the function's behavior.Important aspects to keep in mind include:

• The surface created by the function changes shape as the values of \(x\) and \(t\) change.
• Finding the first partial derivatives will give you the slopes of tangent lines parallel to the \(x\) or \(t\) axes at any specific point \((x,t)\).

Understanding how each variable affects the output individually, and together, enables a deeper grasp of how surfaces behave and changes are perceived.

Arctan Function

The \(\arctan\) function, or the inverse tangent function, is the angle whose tangent is \(x\). It maps the outputs from the real number line to the interval \([-\pi/2, \pi/2]\). In calculus, \(\arctan\)'s derivative provides insights into changes and transitions across its domain.Points to note about the \(\arctan\) function:

• Its derivative is \(\frac{1}{1+u^2}\), which results from the property of the inverse tangent.
• This formula becomes crucial when differentiating composite functions involving \(\arctan\) through the chain rule.

Within the function \(f(x, t) = \arctan(x\sqrt{t})\), the \(\arctan\) part defines how outputs are scaled transformation-wise, modifying the intensity of variation as compared to a linear model, thus having unique properties worth analyzing in-depth.