Problem 15 - The Calculus with Analytic Geometry

Question

In Exercises 15 through 18 , find the total derivative \(d u / d t\) by using the chain rule; do not express \(u\) as a function of \(t\) before, differentiating. $$ u=\tan ^{-1}\left(\frac{y}{x}\right) ; x=\ln t ; y=e^{t} $$

Step-by-Step Solution

Answer

Using the chain rule, \[ \frac{du}{dt} = \left( \frac{-e^t}{ ( \ln t )^2 + ( e^t )^2 } \right) \left( \frac{1}{t} \right) + \left( \frac{\ln t}{ ( \ln t )^2 + ( e^t )^2 } \right) \left( e^t \right) \]

Step 1: Define the given functions

Given: \[ u = \tan^{-1}\left( \frac{y}{x} \right) \]with\[ x = \ln t \]and\[ y = e^t \]

Step 2: Apply the Chain Rule for \( \frac{du}{dt} \)

Using the chain rule, the total derivative \( \frac{du}{dt} \) is given by:\[ \frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} \]

Step 3: Determine \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \)

For \( u = \tan^{-1}\left( \frac{y}{x} \right) \), we use the partial derivatives:\[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \tan^{-1}\left( \frac{y}{x} \right) = \frac{ \left( - y / x^2 \right) }{ 1 + \left( y / x \right)^2 } = \frac{-y}{x^2 + y^2} \]\[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} \tan^{-1}\left( \frac{y}{x} \right) = \frac{ \left( 1 / x \right) }{ 1 + \left( y / x \right) ^2 } = \frac{x}{x^2 + y^2} \]

Step 4: Find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)

We need to determine the derivatives of \( x \) and \( y \) with respect to \( t \):\[ \frac{dx}{dt} = \frac{d}{dt} ( \ln t ) = \frac{1}{t} \]\[ \frac{dy}{dt} = \frac{d}{dt} ( e^t ) = e^t \]

Step 5: Substitute and simplify

Substitute \( \frac{\partial u}{\partial x} \), \( \frac{\partial u}{\partial y} \), \( \frac{dx}{dt} \), and \( \frac{dy}{dt} \) into the chain rule formula:\[ \frac{du}{dt} = \left( \frac{-y}{x^2 + y^2} \right) \left( \frac{1}{t} \right) + \left( \frac{x}{x^2 + y^2} \right) \left( e^t \right) \]Now substitute \( x = \ln t \) and \( y = e^t \):\[ \frac{du}{dt} = \left( \frac{-e^t}{ ( \ln t )^2 + ( e^t )^2 } \right) \left( \frac{1}{t} \right) + \left( \frac{\ln t}{ ( \ln t )^2 + ( e^t )^2 } \right) \left( e^t \right) \]

Key Concepts

Chain Rule

The chain rule is a fundamental theorem in calculus used to find the derivative of a function based on its composition of other functions. It is especially useful when a variable depends on multiple other variables.
In essence, the chain rule helps us break down complex derivatives into simpler parts that are easier to handle. When you have a composite function, say, where a variable \( u \) depends on \( x \) and \( y \), which in turn depend on \( t \), you can use the chain rule to find \( \frac{du}{dt} \). The formula for the total derivative using the chain rule is:

• \( \frac{du}{dt} = \frac{\text{\textpartial} u}{\text{\textpartial} x} \frac{dx}{dt} + \frac{\text{\textpartial} u}{\text{\textpartial} y} \frac{dy}{dt} \)

In the exercise provided, we have a function \( u = \tan^{-1}\big(\frac{y}{x}\big) \) with \( x = \text{ln} \thinspace t \) and \( y = e^t \).
Using the chain rule helps us systematically compute \( \frac{du}{dt} \) without having to explicitly express \( u \) in terms of \( t \), which can be complex and cumbersome.

Partial Derivatives

Partial derivatives are a core concept in multivariable calculus. They measure how a function changes as one variable changes while keeping other variables constant.
For a function like \( u = \tan^{-1}\big(\frac{y}{x}\big) \), we need to find its partial derivatives with respect to \( x \) and \( y \). These are the first steps in applying the chain rule.
In the given exercise, we calculate the partial derivatives as:

• \( \frac{\text{\textpartial} u}{\text{\textpartial} x} = \frac{\text{\textpartial}}{\text{\textpartial} x} \tan^{-1}\big(\frac{y}{x}\big) = \frac{-y}{x^2 + y^2} \)
• \( \frac{\text{\textpartial} u}{\text{\textpartial} y} = \frac{\text{\textpartial}}{\text{\textpartial} y} \tan^{-1}\big(\frac{y}{x}\big) = \frac{x}{x^2 + y^2} \)

Calculating partial derivatives precisely is crucial for correctly applying the chain rule, as they provide the rate of change of \( u \) concerning \( x \) and \( y \).

Differentiation

Differentiation is the process of finding the rate at which a function changes at any given point. It is one of the most important concepts in calculus.
For this exercise, we need to differentiate the functions \( x = \text{ln} \thinspace t \) and \( y = e^t \) concerning \( t \). This will give us the necessary derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):

• \( \frac{dx}{dt} = \frac{d}{dt}( \text{\textln} \thinspace t ) = \frac{1}{t} \)
• \( \frac{dy}{dt} = \frac{d}{dt}( e^t ) = e^t \)

Now, substituting these derivatives back into our chain rule formula allows us to find the total derivative \( \frac{du}{dt} \):

• Substitute \( \frac{\text{\textpartial} u}{\text{\textpartial} x}, \frac{\text{\textpartial} u}{\text{\textpartial} y}, \frac{dx}{dt} \), and \( \frac{dy}{dt} \) into \( \frac{du}{dt} = \frac{\text{\textpartial} u}{\text{\textpartial} x} \frac{dx}{dt} + \frac{\text{\textpartial} u}{\text{\textpartial} y} \frac{dy}{dt} \)
• Plug \( x = \text{\textln} \thinspace t \) and \( y = e^t \) to get the final expression

By carefully following the differentiation steps, we simplify the complex function into something manageable and solvable.