Problem 30 - Calculus

Question

Write down a ratio that approaches \(d y / d x\) at \(x=\pi .\) For \(y=\sin x\) and \(\Delta x=.01\) compute that ratio.

Step-by-Step Solution

Answer

The ratio is approximately -1 at \(x = \pi\).

Step 1: Understand the Derivative Notation

The expression \( \frac{dy}{dx} \) represents the derivative of \( y \) with respect to \( x \). It is the instantaneous rate of change of \( y \) with respect to \( x \). In this problem, \( y = \sin x \) and the task is to approximate the derivative at \( x = \pi \) using a numerical approach with a small interval \( \Delta x \).

Step 2: Set Up the Difference Quotient

The difference quotient is used to approximate the derivative: \( \frac{\Delta y}{\Delta x} \). For a small change \( \Delta x \), \( \Delta y = \sin(x + \Delta x) - \sin x \). Hence, the difference quotient becomes:\( \frac{\sin(x + \Delta x) - \sin x}{\Delta x} \)

Step 3: Substitute Values

Substitute \( x = \pi \) and \( \Delta x = 0.01 \) into the difference quotient expression:\[ \frac{\sin(\pi + 0.01) - \sin \pi}{0.01} \]Since \( \sin \pi = 0 \), the expression simplifies to:\[ \frac{\sin(\pi + 0.01)}{0.01} \]

Step 4: Calculate \( \sin(\pi + 0.01) \)

Recall that \( \sin(\pi + \theta) = -\sin(\theta) \). Therefore, \( \sin(\pi + 0.01) = -\sin(0.01) \). Since \( \sin(0.01) \approx 0.01 \) when \( \theta \) is in radians and very small, we have \( \sin(\pi + 0.01) \approx -0.01 \).

Step 5: Compute the Ratio

Substitute \( \sin(\pi + 0.01) \approx -0.01 \) into the expression from Step 3:\[ \frac{-0.01}{0.01} = -1\]Thus, the value of the ratio that approximates \( \frac{dy}{dx} \) at \( x = \pi \) is \(-1\).

Key Concepts

Derivative of Sine Function

The derivative of the sine function is a fundamental concept in calculus. Mathematically, the derivative of \( \sin x \) with respect to \( x \) is \( \cos x \). This means that at any point \( x \), the rate at which \( \sin x \) is changing is equal to the value of \( \cos x \). For instance, at \( x = \pi \), the derivative \( \cos(\pi) = -1 \). This derivative tells us how the sine function is behaving at any given point. It's essential to understand that this derivative provides the "instantaneous" rate of change of \( \sin x \) at \( x \). In practical terms, it helps us understand how a wave oscillates as represented by \( \sin x \). The sinusoidal nature of the function is reflected in its derivative, showcasing peaks and troughs at predictable intervals.

Difference Quotient

The difference quotient is a crucial tool used to approximate the derivative of a function. It represents the average rate of change of the function over a small interval. The general form of a difference quotient for a function \( f(x) \) is given by:

• \[ \frac{f(x + \Delta x) - f(x)}{\Delta x} \]

This expression calculates the slope of the secant line passing through the points \( (x, f(x)) \) and \( (x + \Delta x, f(x + \Delta x)) \). As \( \Delta x \) approaches zero, the difference quotient converges to the derivative of the function at \( x \). In the context of the exercise, using the difference quotient with \( f(x) = \sin x \) allows us to find an approximation for the derivative \( \frac{dy}{dx} \) at \( x = \pi \). By setting \( \Delta x = 0.01 \), we apply the difference quotient to the sine function and find that it closely approximates the derivative at that point.

Instantaneous Rate of Change

The instantaneous rate of change is a concept that describes how a function changes at a specific point, and it is synonymous with the derivative. For example, in physics, it can represent the velocity of an object at a particular moment. In mathematics, when we talk about the instantaneous rate of change of a function \( y = f(x) \), we're referring to the derivative \( \frac{dy}{dx} \). At any specific value \( x \), the derivative tells us the rate at which \( y \) is changing with respect to \( x \). In our exercise, the instantaneous rate of change at \( x = \pi \) for \( y = \sin x \) gives us the slope of the tangent line to the curve at that point. This is equivalent to evaluating the derivative at \( x = \pi \), yielding \( -1 \). Thus, the rate at which the sine function changes instantaneously at \( x = \pi \) is \(-1\), indicating it is decreasing at that point.