Chapter 9: Problem 39
Given \(u=\ln (x+y+z)\), where \(x=a \cos t, y=b \sin t\), and \(z=c t\), find \(\frac{d u}{d t}\).
Short Answer
Expert verified
The derivative is \( \frac{-a \sin t + b \cos t + c}{a \cos t + b \sin t + c t} \).
Step by step solution
01
Represent the Variables with Respect to t
Given the expressions for the variables, substitute them into the function. We have:\[ u = \ln(x + y + z) = \ln(a \cos t + b \sin t + c t) \] This represents the function in terms of \(t\).
02
Derive the Chain Rule
To find \( \frac{d u}{d t} \), apply the chain rule for derivatives. The chain rule states that the derivative \( \frac{d}{dt}[\ln(f(t))] = \frac{1}{f(t)} \cdot \frac{df(t)}{dt} \).
03
Differentiate the Inner Function
Find \( \frac{d}{dt}(f(t)) \) where \( f(t) = a \cos t + b \sin t + c t \).Differentiate each term:- The derivative of \( a \cos t \) is \( -a \sin t \).- The derivative of \( b \sin t \) is \( b \cos t \).- The derivative of \( c t \) is \( c \).Hence, \[ \frac{d}{dt}(f(t)) = -a \sin t + b \cos t + c \].
04
Apply the Chain Rule
Using \( \frac{d}{dt}[\ln(f(t))] = \frac{1}{f(t)} \cdot \frac{df(t)}{dt} \), substitute the expression from Step 3:\[ \frac{d u}{d t} = \frac{1}{a \cos t + b \sin t + c t} \cdot (-a \sin t + b \cos t + c) \].
05
Simplify the Expression
The expression for \( \frac{d u}{d t} \) is already in a simplified form since it contains no further complex components to reduce.The final derivative is:\[ \frac{d u}{d t} = \frac{-a \sin t + b \cos t + c}{a \cos t + b \sin t + c t} \].
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental concept in calculus used for finding the derivative of composite functions. A composite function is essentially a function inside another function. In our exercise, we have the natural logarithm function, \( u = \ln(f(t)) \), where \( f(t) = a \cos t + b \sin t + ct \).
To apply the chain rule, we differentiate the outer function and multiply it by the derivative of the inner function. This is given by the formula:
To apply the chain rule, we differentiate the outer function and multiply it by the derivative of the inner function. This is given by the formula:
- \( \frac{d}{dt}[\ln(f(t))] = \frac{1}{f(t)} \cdot \frac{df(t)}{dt} \)
Differentiation
Differentiation is the process of finding how a function changes at any given point, which is mathematically represented by a derivative. In the context of this exercise, we aim to determine how \( u \), a function of \( t \), changes as \( t \) changes.
To differentiate \( f(t) = a \cos t + b \sin t + ct \) with respect to \( t \), we use basic differentiation rules:
Differentiation helps us convert an expression into its rate of change format, offering insights into how different variables in a function independently or collectively impact its behavior.
To differentiate \( f(t) = a \cos t + b \sin t + ct \) with respect to \( t \), we use basic differentiation rules:
- The derivative of \( \cos t \) is \( -\sin t \).
- The derivative of \( \sin t \) is \( \cos t \).
- The derivative of \( ct \) is simply \( c \), because \( c \) is a constant.
Differentiation helps us convert an expression into its rate of change format, offering insights into how different variables in a function independently or collectively impact its behavior.
Trigonometric Functions
Trigonometric functions like sine and cosine are periodic functions that play a crucial role in various fields including calculus. In our exercise, \( x = a \cos t \) and \( y = b \sin t \) are functions of \( t \) which describe oscillatory behavior.
Here are some key points to remember:
Here are some key points to remember:
- \( \cos t \) oscillates between -1 and 1, repeating every \( 2\pi \) interval.
- \( \sin t \) also oscillates between -1 and 1, similarly repeating every \( 2\pi \).
- Derivatives involving \( \cos t \) and \( \sin t \) maintain the trigonometric nature of the function.