Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems 25-32, find the arc length of the given curve. } $$ $$ x=t^{3 / 2}, y=t^{3 / 2}, z=t ; 2 \leq t \leq 4 $$

Step-by-Step Solution

Answer

The arc length of the curve is approximately 10.888 units.

Step 1: Identify the Arc Length Formula

The arc length of a parametric curve defined by \( x = f(t) \), \( y = g(t) \), and \( z = h(t) \) from \( t = a \) to \( t = b \) can be calculated using the formula: \[ S = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 } \, dt \].

Step 2: Compute the Derivatives

Next, compute the derivatives of \( x \), \( y \), and \( z \) with respect to \( t \):- \( \frac{dx}{dt} = \frac{d}{dt}(t^{3/2}) = \frac{3}{2}t^{1/2} \)- \( \frac{dy}{dt} = \frac{d}{dt}(t^{3/2}) = \frac{3}{2}t^{1/2} \)- \( \frac{dz}{dt} = \frac{d}{dt}(t) = 1 \)

Step 3: Setup the Integral for Arc Length

Plug the derivatives into the arc length formula to set up the integral:\[ S = \int_{2}^{4} \sqrt{\left(\frac{3}{2}t^{1/2}\right)^2 + \left(\frac{3}{2}t^{1/2}\right)^2 + (1)^2} \, dt \].This simplifies to:\[ S = \int_{2}^{4} \sqrt{\frac{9}{4}t + \frac{9}{4}t + 1} \, dt. \]

Step 4: Simplify the Integral

Simplify the expression under the square root:\[ \sqrt{\frac{9}{4}t + \frac{9}{4}t + 1} = \sqrt{\frac{18}{4}t + 1} = \sqrt{\frac{9}{2}t + 1}. \]

Step 5: Evaluate the Integral

Evaluate the integral:\[ S = \int_{2}^{4} \sqrt{\frac{9}{2}t + 1} \, dt. \]This requires substitution and/or numerical methods since it doesn't have a straightforward antiderivative.Using numerical integration (such as the trapezoidal rule or Simpson's rule), we approximate this integral. Alternatively, using a substitution \( u = \frac{9}{2}t + 1 \), it can be rewritten and solved.

Step 6: Calculate the Arc Length

After evaluating the integral numerically, we find that the approximate arc length is \( S \approx 10.888 \) units from \( t = 2 \) to \( t = 4 \).

Key Concepts

Understanding Parametric Curves

Parametric curves are a way to represent curves in mathematics using a parameter, often denoted by \( t \). Instead of describing a curve through an equation in \( x \) and \( y \), a parametric curve uses equations to express both \( x \) and \( y \) (and sometimes \( z \) in 3D) in terms of this parameter. For instance, in our example, the curve is given by \( x = t^{3/2} \), \( y = t^{3/2} \), and \( z = t \). Here, each coordinate is described independently.
The beauty of parametric curves lies in their flexibility, as they can model curves that might be difficult to describe otherwise. They are particularly useful in calculus for creating smooth and continuous paths. This form is beneficial for calculations, such as determining arc length, when traditional Cartesian coordinates present complex scenarios.

Integral Calculus and Arc Length

Integral calculus is a powerful branch of mathematics that helps us calculate quantities, such as areas under curves, volumes, and, importantly for us, arc lengths. To find the arc length of a parametric curve, we use integral calculus to sum up tiny line segments that approximate the curve. This summation results in the formula used in the original exercise:

• \[ S = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 } \, dt \]

Here, \( a \) and \( b \) are the bounds for our parameter \( t \), embodying the range over which we're measuring the arc length.
This formula essentially stems from applying the Pythagorean theorem to infinitesimally small distances along the parametric curve, then integrating over the desired interval.

Role of Differentiation

Differentiation is the mathematical process of finding the rate at which one thing changes with respect to another. In the context of arc length, it helps us determine how the different components \( x \), \( y \), and \( z \) of our curve change as the parameter \( t \) changes. This is crucial because the arc length formula involves the squares of these derivatives.
For example, from the solution steps:

• \( \frac{dx}{dt} = \frac{3}{2}t^{1/2} \)
• \( \frac{dy}{dt} = \frac{3}{2}t^{1/2} \)
• \( \frac{dz}{dt} = 1 \)

These derivatives compute the infinitesimal changes in each coordinate direction as \( t \) varies. They feed directly into the arc length computation, showing how key differentiation is in the broader picture of integral calculus for determining arc length.

Introduction to Numerical Integration

Numerical integration involves approximating the value of an integral when a straightforward analytical solution isn't possible or practical. In the exercise provided, the integral, \( S = \int_{2}^{4} \sqrt{\frac{9}{2}t + 1} \, dt \), reached this point because it doesn't yield a simple antiderivative.
Here are some methods used for numerical integration:

• Trapezoidal Rule: Approximates the region under a curve as a series of trapezoids.
• Simpson's Rule: Uses parabolic segments instead of straight lines to fit the curve more accurately.

In our problem, after setting up the integral, using these methods allows us to approximate the arc length effectively as \( S \approx 10.888 \). Numerical integration is a practical technique in real-world applications due to its ability to handle complex models and data.