Problem 57 - Calculus and its applications

Question

Find \(f_{x}\) and \(f_{t}\). $$f(x, t)=\frac{x^{2}+t^{2}}{x^{2}-t^{2}}$$

Step-by-Step Solution

Answer

\(f_x = \frac{-4xt^2}{(x^2 - t^2)^2}\) and \(f_t = \frac{4tx^2}{(x^2 - t^2)^2}\).

Step 1: Understand the Problem

We need to find the partial derivatives of the function \(f(x, t) = \frac{x^2 + t^2}{x^2 - t^2}\). This involves calculating \(f_x\) (the partial derivative with respect to \(x\)) and \(f_t\) (the partial derivative with respect to \(t\)).

Step 2: Differentiate with Respect to x

To find \(f_x\), differentiate the function \(f(x, t)\) with respect to \(x\) while treating \(t\) as a constant. Apply the quotient rule: if \(u(x) = x^2 + t^2\) and \(v(x) = x^2 - t^2\), then \(f_x = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\).

Step 3: Apply Quotient Rule to Find f_x

Calculate as follows: \(u'(x) = 2x\) and \(v'(x) = 2x\), so:\[ f_x = \frac{(x^2 - t^2)(2x) - (x^2 + t^2)(2x)}{(x^2 - t^2)^2} = \frac{2x(x^2 - t^2 - x^2 - t^2)}{(x^2 - t^2)^2} = \frac{-4xt^2}{(x^2 - t^2)^2} \].

Step 4: Differentiate with Respect to t

To find \(f_t\), differentiate the function \(f(x, t)\) with respect to \(t\) while treating \(x\) as a constant. Again, use the quotient rule with \(u(t) = x^2 + t^2\) and \(v(t) = x^2 - t^2\).

Step 5: Apply Quotient Rule to Find f_t

Calculate as follows: \(u'(t) = 2t\) and \(v'(t) = -2t\), so:\[ f_t = \frac{(x^2 - t^2)(2t) - (x^2 + t^2)(-2t)}{(x^2 - t^2)^2} = \frac{2t(x^2 - t^2 + x^2 + t^2)}{(x^2 - t^2)^2} = \frac{4tx^2}{(x^2 - t^2)^2} \].

Key Concepts

Quotient Rule

When working with the function \(f(x, t) = \frac{x^2 + t^2}{x^2 - t^2}\), one powerful tool we use is the quotient rule. This is an essential rule in calculus used for differentiating functions that are ratios of two differentiable functions.
To apply the quotient rule, you need to differentiate both the numerator and the denominator separately and then insert them into this specific formula:

• You take the derivative of the numerator and multiply it by the denominator.
• Subtract from it the product of the numerator and the derivative of the denominator.
• All over the square of the denominator.

So for our function:

• Let \(u(x) = x^2 + t^2\) and \(v(x) = x^2 - t^2\).
• The derivatives are \(u'(x)\) and \(v'(x)\), which when substituted into the rule gives: \(f_x = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\).

This systematic approach helps us handle even complex expressions smoothly.

Differentiation

Differentiation is a fundamental concept in calculus. It enables us to find rates at which things change. When we differentiate, we are looking for the derivative, which represents the rate of change of a function.
In contexts like this exercise, where we want partial derivatives, it's slightly different. We focus only on one variable at a time, treating other variables as constants.
For example, to find \(f_x\), we differentiate the function \(f(x, t) = \frac{x^2 + t^2}{x^2 - t^2}\) with respect to \(x\), treating \(t\) as if it doesn’t change. It provides information about how \(f\) changes as \(x\) changes, helping us understand the role each variable plays in the function.

• This focus allows for a deep dive into each component of the function.
• It simplifies the problem by breaking it down into more manageable pieces.

Differentiation, especially partial, is a versatile tool used in many scientific and engineering fields.

Multivariable Calculus

Multivariable calculus extends concepts from single-variable calculus to functions of several variables, like \(f(x, t)\) we are dealing with here. This branch explores how functions change when more than one variable is in play.
Partial differentiation, a core subject in this field, allows us to see how a function behaves individually with respect to each variable.
Some important aspects include:

• Partial derivatives provide insights into the slope of the function in the direction of each variable.
• The function's behavior in a multidimensional space is decoded piece by piece.

For our function, by finding \(f_x\) and \(f_t\), we gain a detailed understanding of how each influences the outcome.
This layered approach helps build a comprehensive picture of the function's dynamics and can highlight how changing one variable might affect the overall system, critical in optimization and modeling scenarios.