Problem 89 - University Calculus: Early Transcendentals

Question

Find \(d s / d t\) when \(\theta=3 \pi / 2\) if \(s=\cos \theta\) and \(d \theta / d t=5\).

Step-by-Step Solution

Answer

\(\frac{ds}{dt} = 5\) when \(\theta = \frac{3\pi}{2}\).

Step 1: Understand the Problem

We need to find the derivative \( \frac{ds}{dt} \) when \( \theta = \frac{3\pi}{2} \), given that \( s = \cos \theta \) and the rate of change of \( \theta \) with respect to time, \( \frac{d\theta}{dt} = 5 \).

Step 2: Use the Chain Rule

We need to find \( \frac{ds}{dt} \), which involves differentiating \( s = \cos \theta \) with respect to \( t \). Using the chain rule, we have:\[ \frac{ds}{dt} = \frac{ds}{d\theta} \cdot \frac{d\theta}{dt} \]

Step 3: Differentiate with Respect to \( \theta \)

Find \( \frac{ds}{d\theta} \) by differentiating \( s = \cos \theta \) with respect to \( \theta \):\[ \frac{ds}{d\theta} = -\sin \theta \]

Step 4: Substitute \( \theta = \frac{3\pi}{2} \) and \( \frac{d\theta}{dt} = 5 \)

Substitute \( \theta = \frac{3\pi}{2} \) into \( \frac{ds}{d\theta} \):\[ \frac{ds}{d\theta} = -\sin \left( \frac{3\pi}{2} \right) = -(-1) = 1 \]Substitute \( \frac{d\theta}{dt} = 5 \) and \( \frac{ds}{d\theta} = 1 \) into the chain rule expression:\[ \frac{ds}{dt} = 1 \cdot 5 = 5 \]

Step 5: Final Calculation

Therefore, the rate of change of \( s \) with respect to time \( t \), \( \frac{ds}{dt} \), is \( 5 \) when \( \theta = \frac{3\pi}{2} \).

Key Concepts

Chain rule

The chain rule is a fundamental concept in calculus used to compute the derivative of a function that is composed of other functions. In simpler terms, if you have a function inside another function, the chain rule helps you to differentiate the outer function with respect to the inner function.
For example, consider the function given by two variables — say, if we have a function defined as \( s = \cos \theta \), and \( \theta \) itself changes with time \( t \). Here, \( s \) changes as \( \theta \) changes, which in turn changes over time. The connection of all these changes can be computed using the chain rule.

• Given functions: \( s = \cos \theta \)
• Rate of change of \( \theta \) with respect to time is \( \frac{d\theta}{dt} = 5 \)

To find how \( s \) changes with time, mathematically expressed as \( \frac{ds}{dt} \), we apply the chain rule:\[ \frac{ds}{dt} = \frac{ds}{d\theta} \cdot \frac{d\theta}{dt} \]This formula captures the idea that the rate of change of \( s \) over time depends on how \( s \) changes with respect to \( \theta \) and how \( \theta \) changes with respect to \( t \). It's a multiplication of these smaller, simpler derivatives.

Trigonometric derivatives

To solve problems involving trigonometric functions and their derivatives, it's essential to understand how to differentiate basic trigonometric functions. Here, we focus on the cosine function.
A critical step to finding the rate of change when working with trigonometric functions is to first determine the derivative of the trigonometric function in question.

• The derivative of \( \cos \theta \) with respect to \( \theta \) is \( -\sin \theta \).

Let's apply this to our exercise:Given \( s = \cos \theta \), we differentiate \( s \) with respect to \( \theta \):\[ \frac{ds}{d\theta} = -\sin \theta \]This expression provides the rate at which \( s \) changes as \( \theta \) varies. We then need this intermediate step to apply the chain rule effectively. For instance, when \( \theta = \frac{3\pi}{2} \), substituting in gives:\[ \frac{ds}{d\theta} = -\sin \left( \frac{3\pi}{2} \right) = 1 \]This tells us that at \( \theta = \frac{3\pi}{2} \), the sine function reaches its peak value in terms of the rate of change in the negative direction, which simplifies to 1 when negated.

Rate of change

The concept of rate of change is crucial in calculus and is key to understanding the behavior of functions over time. It can be thought of as how one quantity changes in relation to another. In this exercise, the primary focus is on how the function \( s = \cos \theta \) changes with respect to time.
Specifically, we are asked to find \( \frac{ds}{dt} \), which tells us how the cosine of an angle changes as the angle itself changes over time when \( \theta \) is at a specific angle.

• At \( \theta = \frac{3\pi}{2} \), we needed an algebraic solution to know \( \sin \left( \frac{3\pi}{2} \right) = -1 \) so that \( \frac{ds}{d\theta} = 1 \).
• Given \( \frac{d\theta}{dt} = 5 \), our final rate of change can be computed as follows:\[ \frac{ds}{dt} = \frac{ds}{d\theta} \cdot \frac{d\theta}{dt} = 1 \times 5 = 5 \]

The rate \( 5 \) demonstrates how quickly \( s \) is changing with respect to time at the moment when \( \theta \) equals \( \frac{3\pi}{2} \). This is essential in various fields such as physics, engineering, and economics, where knowing how parameters change over time can make predictions and influence decisions.