Problem 60 - Calculus for Biology and Medicine

Question

Set up, but do not evaluate, the integrals for the lengths of the following curves: \(y=\sin x, 0 \leq x \leq \frac{\pi}{2}\)

Step-by-Step Solution

Answer

Set up the integral: \[L = \int_{0}^{\frac{\pi}{2}} \sqrt{1 + \cos^2 x} \, dx\]

Step 1: Identify the formula for arc length

The arc length of a curve described by the function \(y = f(x)\) from \(x = a\) to \(x = b\) is given by the formula: \\[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]

Step 2: Determine the derivative of the function

Calculate the derivative of \(y = \sin x\) with respect to \(x\). The derivative \(\frac{dy}{dx} = \cos x\).

Step 3: Substitute the derivative into the arc length formula

Substitute \(\frac{dy}{dx} = \cos x\) into the arc length formula: \\[L = \int_{0}^{\frac{\pi}{2}} \sqrt{1 + (\cos x)^2} \, dx\]

Step 4: Set up the integral

The integral for the length of the curve \(y = \sin x\) from \(x = 0\) to \(x = \frac{\pi}{2}\) is: \\[L = \int_{0}^{\frac{\pi}{2}} \sqrt{1 + \cos^2 x} \, dx\]

Key Concepts

Derivative of a Function

When we talk about the derivative of a function, we are referring to a fundamental concept in calculus. The derivative provides a way to understand how a function changes at any given point. In simpler terms, it tells us the rate of change or the slope of the function at a particular point.
To find the derivative of a function, we use rules like the power rule, product rule, and chain rule, depending on the function type. For example, the derivative of the function \( y = \sin x \) is \( \cos x \).
This is because the derivative of sine, according to standard differentiation rules, is cosine. Understanding how to calculate derivatives is essential for solving various problems in calculus, such as finding arc lengths, optimizing functions, and solving differential equations.

Integral Calculus

Integral calculus is another crucial part of calculus. It deals with the concept of integration, which is the reverse process of differentiation. While differentiation breaks down functions into rates of change, integration sums up these changes to find areas, volumes, and total accumulations.
The integral we set up for finding arc length is a common application of integral calculus. Typically, to find the arc length of a curve from point \(a\) to \(b\), we use the formula: \[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]
In the given exercise, we applied this by substituting the derivative of \( y = \sin x \) (which is \( \cos x \)) into the formula, resulting in \[ L = \int_{0}^{\frac{\pi}{2}} \sqrt{1 + \cos^2 x} \, dx \]
Integration helps us transform and solve the equation, even though in this case, we only needed to set it up without evaluating it. By setting up such integrals, we prepare to solve for values that represent real-world lengths, areas, or other accumulated quantities.

Trigonometric Functions

Trigonometric functions, such as \( \sin x \), \( \cos x \), and \( \tan x \), are functions that relate the angles of a triangle to the lengths of its sides. They are fundamental in elementary math and are applied extensively in various fields including physics, engineering, and mathematics.
In calculus, these functions are important because they are differentiable and integrable. For example, the function \( y = \sin x \) is one of these trigonometric functions, and its properties make it interesting to study in the context of calculus.
We use their derivatives in many problems. The derivative of \( \sin x \) is \( \cos x \), and similarly, \( \cos x \) differentiates to \(-\sin x\). These functions oscillate between -1 and 1, which makes calculations like the arc length integral both achievable and practical. Understanding trigonometric functions deeply enhances our ability to model periodic phenomena, such as sound waves, light waves, and alternating current electricity.