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Magazine Chapter 3: Problem 15
In Exercises \(13-16,\) find \(d s / d t\) $$ s=\frac{1+\csc t}{1-\csc t} $$
Short Answer
Expert verified
The derivative is \( \frac{ds}{dt} = \frac{-2\csc t \cdot \cot t}{(1-\csc t)^2} \).
Step by step solution
01
Understand the Problem
We are tasked with finding the derivative of the function \( s(t) = \frac{1+\csc t}{1-\csc t} \) with respect to \( t \). This means we need to calculate \( \frac{ds}{dt} \).
02
Apply the Quotient Rule
The quotient rule states that for a function \( \frac{u}{v} \), the derivative \( \frac{d}{dt}(\frac{u}{v}) = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \). Let's identify \( u = 1 + \csc t \) and \( v = 1 - \csc t \).
03
Differentiate the Numerator and Denominator
Find \( \frac{du}{dt} \) and \( \frac{dv}{dt} \): - \( \frac{du}{dt} = \frac{d}{dt}(1 + \csc t) = -\csc t \cdot \cot t \)- \( \frac{dv}{dt} = \frac{d}{dt}(1 - \csc t) = \csc t \cdot \cot t \)
04
Substitute into the Quotient Rule Formula
Using the quotient rule: \[\frac{ds}{dt} = \frac{(1-\csc t)(-\csc t \cdot \cot t) - (1+\csc t)(\csc t \cdot \cot t)}{(1-\csc t)^2}\]Simplify the expression for further calculation.
05
Simplify the Expression
Simplifying the numerator: \[= -\csc t \cdot \cot t + \csc^2 t \cdot \cot t - \csc t \cdot \cot t - \csc^2 t \cdot \cot t \]Combine like terms:\[= -2\csc t \cdot \cot t \]Thus, the derivative is:\[\frac{ds}{dt} = \frac{-2\csc t \cdot \cot t}{(1-\csc t)^2}\]
06
Finalize the Result
The derivative \( \frac{ds}{dt} \) simplifies to \[ \frac{ds}{dt} = \frac{-2\csc t \cdot \cot t}{(1-\csc t)^2} \] This is our final result after simplifying the expression.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
The quotient rule in calculus is a method to find the derivative of a division of two functions. If you have a function given by \( \frac{u(t)}{v(t)} \), where both \( u \) and \( v \) are functions of the same variable \( t \), the quotient rule allows you to differentiate it effectively. The formula for the quotient rule is: \[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \]. This can seem a bit overwhelming, but breaking it down helps:
- Identify \( u \) and \( v \)
- Find the derivative of \( u \), denoted as \( \frac{du}{dt} \)
- Find the derivative of \( v \), denoted as \( \frac{dv}{dt} \)
- Substitute these into the quotient rule formula
Trigonometric Derivatives
Trigonometric functions have their specific derivatives which are crucial in calculus problems. When dealing with derivatives of functions involving trigonometric terms, the knowledge of these derivatives allows you to differentiate them easily. For example:
- The derivative of \( \sin t \) is \( \cos t \)
- The derivative of \( \cos t \) is \( -\sin t \)
- More complex ones include \( \csc t \), whose derivative is \( -\csc t \cdot \cot t \)
Cosecant Function
The cosecant function, abbreviated as \( \csc \), is one of the six fundamental trigonometric functions. It's the reciprocal of the sine function, defined as \( \csc t = \frac{1}{\sin t} \). Because the cosecant function is linked to the sine, its derivative has intriguing properties. The derivative of \( \csc t \) is \( -\csc t \cot t \).
- \( \csc t \) increases or decreases more rapidly than \( \sin t \), as it's infinite at some points where \( \sin t \) is zero.
- Understanding how \( \csc t \) behaves requires a solid grasp of the sine function since \( \csc t \) is undefined wherever \( \sin t \) is equal to zero.
