Problem 40 - Calculus Early Transcendentals

Question

Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places. $$x=\ln t, \quad y=\sqrt{t+1}, \quad 1 \leq t \leq 5$$

Step-by-Step Solution

Answer

The length of the curve is approximately 3.1424.

Step 1: Parameterize the curve

The curve is given by the functions \( x = \ln t \) and \( y = \sqrt{t+1} \) with parameter \( t \). We will use this parameterization to set up the integral for the arc length of the curve.

Step 2: Find derivatives

Calculate the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x = \ln t \), \( \frac{dx}{dt} = \frac{1}{t} \). For \( y = \sqrt{t+1} \), \( \frac{dy}{dt} = \frac{1}{2\sqrt{t+1}} \).

Step 3: Apply the arc length formula

The formula for the length of a curve given by parameter \( t \) is \( L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \). Using our derivatives, we have:\[ L = \int_1^5 \sqrt{\left( \frac{1}{t} \right)^2 + \left( \frac{1}{2\sqrt{t+1}} \right)^2} \, dt \]

Step 4: Simplify the integral expression

Combine the terms inside the square root:\[ L = \int_1^5 \sqrt{\frac{1}{t^2} + \frac{1}{4(t+1)}} \, dt \]

Step 5: Use a calculator to evaluate the integral

Input the integral expression into a calculator capable of solving definite integrals:\[ L = \int_1^5 \sqrt{\frac{1}{t^2} + \frac{1}{4(t+1)}} \, dt \]The calculator will give a numerical value for the length of the curve.

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that deals with finding the integral of functions. In the context of arc length, it helps determine the distance along a curve. The integral collects tiny "pieces" of length along the curve through summation. For curves parameterized in terms of a variable, the arc length is found by:

• Computing the derivatives of the functions for the x and y coordinates.
• Substituting these derivatives into the arc length formula which involves integrating over the specified interval.

Understanding integrals is crucial as they form the basis to calculate total quantities, such as length, area, and volume.
In this problem, integral calculus allows us to compute the length of a curve defined by the parameterization given in the exercise.

Parameterization

Parameterization is a technique used to express a curve using a parameter. Typically, this parameter is denoted as 't.' For curves in a plane, two equations define the x and y coordinates of each point on the curve.
Parameterizing a curve transforms a complex problem into manageable components. In this exercise, the functions:

• \( x = \ln t \)
• \( y = \sqrt{t+1} \)

are parameterized using 't', with 't' encompassing the interval from 1 to 5. This standardizes the method by which the position on the curve is expressed and is the starting point for later calculus operations.

Derivatives

Derivatives represent how a function is changing at any given point and are pivotal in calculating the arc length. Since arc length relies on finding how steeply x and y change with 't,' differentiating these functions provides this rate of change.
In the problem:

• The derivative of \( x = \ln t \) with respect to \( t \) is \( \frac{1}{t} \).
• The derivative of \( y = \sqrt{t+1} \) with respect to \( t \) is \( \frac{1}{2\sqrt{t+1}} \).

These derivatives are substituted into the arc length formula to assess how the component directions contribute to the total curve length in space.

Numerical Integration

Numerical integration involves approximating the value of an integral, especially when finding the exact integral analytically is difficult. Many functions, similar to the arc length integral here, may not have a simple antiderivative.
By using numerical methods, such as the trapezoidal rule or advanced algorithms in calculators, these integrals can be computed to great accuracy. In this exercise, once the integral is set up, a calculator simplifies what would otherwise be a demanding calculation:

• The arc length integral, \( L = \int_1^5 \sqrt{\frac{1}{t^2} + \frac{1}{4(t+1)}} \, dt \) is evaluated numerically.

This approach is valuable for ensuring precision and efficiency in deriving solutions for integrals without closed-form.