Problem 15 - Thomas Calculus

Question

In Exercises \(13-16,\) find \(d s / d t\) $$ s=\frac{1+\csc t}{1-\csc t} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{ds}{dt} = \frac{-2\csc t \cdot \cot t}{(1-\csc t)^2} \).

Step 1: 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 \( \frac{ds}{dt} \).

Step 2: 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 \( v = 1 - \csc t \).

Step 3: Differentiate the Numerator and Denominator

Find \( \frac{du}{dt} \) and \( \frac{dv}{dt} \): - \( \frac{du}{dt} = \frac{d}{dt}(1 + \csc t) = -\csc t \cdot \cot t \)- \( \frac{dv}{dt} = \frac{d}{dt}(1 - \csc t) = \csc t \cdot \cot t \)

Step 4: 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.

Step 5: 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:\[\frac{ds}{dt} = \frac{-2\csc t \cdot \cot t}{(1-\csc t)^2}\]

Step 6: 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.

Key Concepts

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

By following these steps, you can solve problems involving derivatives of quotients. It's like a recipe where following the correct order and steps yields the correct derivative.

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 \( -\csc t \cdot \cot t \)

These derivatives are backbone knowledge for understanding and solving calculus problems involving trigonometric functions. By memorizing these, you set a strong foundation for tackling even more complex derivatives and calculus challenges.

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.

Mastering \( \csc t \) helps tackle derivative problems involved in calculus, especially when using the quotient rule to differentiate complex functions that include trigonometric components.