Problem 21 - University Calculus: Early Transcendentals

Question

Find \(d s / d t\). $$s=\frac{1+\csc t}{1-\csc t}$$

Step-by-Step Solution

Answer

\( \frac{d s}{d t} = \frac{-2\csc t \cot t}{(1 - \csc t)^2} \)

Step 1: Identify the Derivative Rule

We need to find the derivative \( \frac{d s}{d t} \) of the function \( s = \frac{1 + \csc t}{1 - \csc t} \). Since this is a quotient, we will use the Quotient Rule: \( \frac{d}{d t} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \). Here, let \( u = 1 + \csc t \) and \( v = 1 - \csc t \).

Step 2: Differentiate the Numerator

Differentiate the numerator: \( u = 1 + \csc t \). The derivative of \( 1 \) is \( 0 \), and the derivative of \( \csc t \) is \( -\csc t \cot t \). Therefore, \( u' = -\csc t \cot t \).

Step 3: Differentiate the Denominator

Differentiate the denominator: \( v = 1 - \csc t \). The derivative of \( 1 \) is \( 0 \), and the derivative of \( -\csc t \) is \( \csc t \cot t \). Thus, \( v' = \csc t \cot t \).

Step 4: Apply the Quotient Rule

Substitute \( u' \), \( u \), \( v' \), and \( v \) into the quotient rule formula: \[ \frac{d s}{d t} = \frac{-\csc t \cot t (1 - \csc t) - (1 + \csc t) \csc t \cot t}{(1 - \csc t)^2} \].

Step 5: Simplify the Expression

Simplify the expression for \( \frac{d s}{d t} \):\[ \frac{d s}{d t} = \frac{-\csc t \cot t + \csc^2 t \cot t - \csc t \cot t - \csc^2 t \cot t }{(1 - \csc t)^2} \] which simplifies to \[ \frac{d s}{d t} = \frac{-2\csc t \cot t}{(1 - \csc t)^2} \].

Step 6: State the Final Result

The derivative \( \frac{d s}{d t} \) is \( \frac{-2\csc t \cot t}{(1 - \csc t)^2} \).

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental technique in calculus for finding the derivative of a function that is the ratio of two differentiable functions. It's particularly useful when dealing with quotients in the form \(\frac{u}{v}\) where both \(u\) and \(v\) are functions of a variable, in this case \(t\). The formula for the Quotient Rule is: \[ \frac{d}{dt} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \] Where:

• \(u\) is the numerator function
• \(v\) is the denominator function
• \(u'\) is the derivative of \(u\)
• \(v'\) is the derivative of \(v\)

It’s important to apply this carefully to maintain the order of operations as given in the formula. Substituting correctly organized derivatives helps ensure the accuracy of the solution. Using the correct numerator and denominator derivatives is crucial for seamless computation. This rule helps simplify the expression for the derivative, which can otherwise become complex due to the multiplication and subtraction involved.

Trigonometric Derivatives

Trigonometric derivatives are derivatives of the six fundamental trigonometric functions: sine, cosine, tangent, cosecant, secant, and cotangent. These derivatives are essential tools in calculus for solving problems involving trigonometric expressions. Here are some fundamental rules:

• The derivative of \(\sin t\) is \(\cos t\).
• The derivative of \(\cos t\) is \(-\sin t\).
• The derivative of \(\tan t\) is \(\sec^2 t\).
• The derivative of \(\csc t\) (cosecant) is \(-\csc t \cot t\).
• The derivative of \(\sec t\) is \(\sec t \tan t\).
• The derivative of \(\cot t\) is \(-\csc^2 t\).

In the given problem, we are primarily concerned with the function \(\csc t\) and its derivative. Understanding how these derivatives are derived and applied is crucial for managing more complex trigonometric functions, especially when they appear in the context of calculus operations like differentiation.

Cosecant Function

The cosecant function, often abbreviated as "csc," is one of the six primary trigonometric functions. It is the reciprocal of the sine function, defined as \(\csc t = \frac{1}{\sin t}\). It is used to represent the ratio of the hypotenuse to the side opposite a given angle in a right triangle. Understanding the behavior of the cosecant function is essential when dealing with expressions involving reciprocal trigonometric functions. Notably, the derivative of the cosecant function \(\csc t\) plays a significant role in calculus: - The derivative of \(\csc t\) is \(-\csc t \cot t\). This derivative is particularly useful when applied in the context of the Quotient Rule, as seen in this exercise. Recognizing when and how to apply this derivative helps in accurately simplifying expressions where cosecant appears in more complicated terms, allowing calculations and analysis to proceed smoothly.