Question

Find the length of the curve \(y=e^{x}\) over \([0, \ln (2)]\).

Short answer

The length of the curve is approximately 1.53 units.

Step-by-step solution

Understand the problem

We want to find the length of the curve given by the function \(y = e^x\) over the interval \([0, \ln(2)]\). This means we need to calculate the arc length of the function's graph on this interval.

Recall the formula for arc length

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

Differentiate the function

Start by differentiating the function \(y = e^x\). The derivative \(\frac{dy}{dx}\) is \(e^x\).

Set up the integral for arc length

Substitute \(\frac{dy}{dx} = e^x\) into the arc length formula. The integral becomes: \[ L = \int_{0}^{ ext{ln}(2)} \sqrt{1 + (e^x)^2} \, dx = \int_{0}^{ ext{ln}(2)} \sqrt{1 + e^{2x}} \, dx \].

Solve the integral

We need to compute \(L = \int_{0}^{ ext{ln}(2)} \sqrt{1 + e^{2x}} \, dx \). This integral can be solved using substitution. Set \(u = e^x\), then \(du = e^x \, dx\), making \(dx = \frac{du}{u}\). When \(x = 0\), \(u = 1\) and when \(x = \ln(2)\), \(u = 2\). The integral becomes: \[ L = \int_{1}^{2} \frac{\sqrt{1 + u^2}}{u} \, du \].

Finalize the integral limits

Evaluate the integral \(L = \int_{1}^{2} \frac{\sqrt{1 + u^2}}{u} \, du \). This integral might require numerical estimation or further techniques such as trigonometric substitution if no standard solution method seems viable directly. However, acknowledge that solving by standard calculus techniques might give complex forms and direct evaluation or numerical approximation might be employed in practical scenarios.

Key concepts

Integration Techniques

Integration is a fundamental concept in calculus used to find areas, volumes, and arc lengths. In our exercise, the goal is to calculate the arc length of the curve described by the function \(y = e^x\) over a particular interval, from 0 to \(\ln(2)\). To do this, we use a specific type of integral called an arc length integral. The formula for the arc length \( L \) of a curve \( y = f(x) \) from \( x = a \) to \( x = b \) is:\[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2 } \, dx \]To solve such an integral, various integration techniques might be used, depending on the complexity of the function. For example, substitution is one technique often employed when the integral is not straightforward. In our exercise, after differentiating the function, we use substitution by letting \( u = e^x \), turning our original integral into a potentially simpler form. Additionally, in some cases, trigonometric substitution or numerical methods may be necessary, especially when dealing with integrals that do not resolve easily into basic antiderivative forms. Being familiar with a range of integration techniques expands the ability to tackle diverse calculus problems more effectively.

Differentiation

Differentiation is the process of finding the derivative of a function. It helps to determine how a function changes at any given point and is essential in the calculation of rates and tangents. In this exercise, we need to differentiate the exponential function \( y = e^x \) to progress to calculating its arc length.The derivative \( \frac{dy}{dx} \) of \( y = e^x \) is straightforward since the derivative of \( e^x \) with respect to \( x \) is itself \( e^x \). Differentiation of exponential functions, particularly \( e^x \), is often simpler compared to polynomial or trigonometric functions because of this property where the function is equal to its derivative.Understanding differentiation is crucial, as it is the first step in transforming the arc length formula into an integral that can be further manipulated. By obtaining the derivative of \( e^x \), we gain the necessary component to feed into our arc length formula, \( \sqrt{1 + (e^x)^2} \).Differentiation provides both direct practical value in finding slopes and fundamental theoretical insights necessary for solving complex calculus problems.

Exponential Functions

Exponential functions are characterized by their constant rate of growth or decay, making them essential in modeling various real-world phenomena, ranging from population growth to radioactive decay. The function \( y = e^x \) is a classic example where the base of the exponential is Euler's number, \( e \). In the context of our exercise, \( y = e^x \) represents a curve whose steepness grows exponentially as \( x \) increases. This property plays a significant role when calculating arc lengths, as steeper curves typically yield longer arc lengths over equivalent intervals compared to less steep curves.Understanding the behavior of exponential functions is crucial in calculus, especially given their prevalence. They appear frequently in integration and differentiation problems where their unique properties often simplify computations. Specifically, \( y = e^x \) maintains the distinctive feature where its rate of change is equal to itself, which simplifies finding derivatives and working through complex calculus problems.Grasping these properties is fundamental to solving a variety of mathematical challenges and leveraging exponential functions for practical applications.