Problem 22 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

\( \frac{ds}{dt} = \frac{-1}{(1 - \cos t)^2} \)

Step 1: Identify the Function

We are given the function \( s = \frac{\sin t}{1 - \cos t} \). The goal is to find the derivative of \( s \) with respect to \( t \), denoted as \( \frac{ds}{dt} \).

Step 2: Differentiate Using the Quotient Rule

To differentiate a function given as a quotient \( \frac{u}{v} \), use the quotient rule: \( \frac{d}{dt} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \). Here, \( u = \sin t \) and \( v = 1 - \cos t \).

Step 3: Find \( u' \) and \( v' \)

Differentiate the functions \( u \) and \( v \) separately:1. \( u' = \cos t \) (the derivative of \( \sin t \)).2. \( v' = \sin t \) (the derivative of \( 1 - \cos t \) is \( 0 + \sin t \)).

Step 4: Apply the Quotient Rule

Plug \( u \), \( v \), \( u' \), and \( v' \) into the quotient rule formula:\[ \frac{ds}{dt} = \frac{(\cos t)(1 - \cos t) - (\sin t)(\sin t)}{(1 - \cos t)^2} \].

Step 5: Simplify the Expression

Simplify the expression by expanding and combining like terms:\[ \frac{ds}{dt} = \frac{\cos t - \cos^2 t - \sin^2 t}{(1 - \cos t)^2} \]. Use the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \).Thus, \( \cos^2 t + \sin^2 t = 1 \) implies \( \cos^2 t + 1 - \cos^2 t = 1 + \cos t \), simplifying further to:\[ \frac{ds}{dt} = \frac{-1}{(1 - \cos t)^2} \].

Key Concepts

Quotient Rule

The quotient rule is a method used in calculus to find the derivative of functions that are divided by each other. It is particularly useful when the function you need to differentiate is in the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of the same variable, like \( t \) in our exercise.
The quotient rule formula is:

• \( \frac{d}{dt} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \)

Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \), respectively. To apply the rule, you first find these derivatives separately. Then, plug them back into the formula to arrive at the derivative of the entire expression.
For instance, in the given function \( s = \frac{\sin t}{1-\cos t} \), you identify \( u = \sin t \) and \( v = 1 - \cos t \). You then compute the derivatives \( u' = \cos t \) and \( v' = \sin t \). Finally, substitute everything into the quotient rule formula to find \( \frac{ds}{dt} \).
This rule is powerful because it systematically provides a way to handle more mathematically complex divisions involving variable-dependent functions.

Trigonometric Functions

Trigonometric functions are a fundamental part of calculus and are frequently encountered in derivative problems. Common trigonometric functions include \( \sin t \), \( \cos t \), and \( \tan t \).
In our exercise, we deal with the functions \( \sin t \) and \( \cos t \). Knowing how to differentiate these functions is essential:

• The derivative of \( \sin t \) is \( \cos t \).
• The derivative of \( \cos t \) is \( -\sin t \).

These identities are vital because they allow us to compute the derivatives necessary for applying the quotient rule.
Moreover, trigonometric identities such as \( \sin^2 t + \cos^2 t = 1 \) come in handy when simplifying expressions. In this problem, by substituting \( \sin^2 t \) and \( \cos^2 t \) with their equivalents from the identity, we streamline the differentiation process considerably. This simplification is crucial to solving trigonometric differentiation problems more easily and effectively.

Simplifying Expressions

Simplifying expressions in calculus often requires using algebraic identities and properties to make the derivative more manageable. After applying the quotient rule, you may obtain a complex expression that needs further refinement.
In our specific exercise, we encountered the expression:

• \( \frac{ds}{dt} = \frac{\cos t - \cos^2 t - \sin^2 t}{(1 - \cos t)^2} \)

Utilizing the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \), we can reformulate the expression for simplification. By substituting \( \sin^2 t \) with \( 1 - \cos^2 t \), we can combine and reduce terms:

• \( \cos^2 t + \sin^2 t = 1 \)

Through this identity, the expression simplifies to:

• \( \frac{ds}{dt} = \frac{-1}{(1 - \cos t)^2} \)

Such simplification is necessary because it reveals the underlying simpler structure of the problem, making it easier to understand and solve. By reducing complex terms, you help ensure that the final derivative form is both correct and simplified.