Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the arc length of the given curve. \(x=t^{2}, y=(4 / 3) t^{3 / 2}, z=t ; 0 \leq t \leq 8\)

Step-by-Step Solution

Answer

The arc length of the curve is 72.

Step 1: Identify the Formula for Arc Length

To find the arc length of a 3D parametric curve defined by \(x = f(t)\), \(y = g(t)\), and \(z = h(t)\), the formula is: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt \] where \(a\) and \(b\) are the limits of \(t\). In this problem, \(a = 0\) and \(b = 8\).

Step 2: Compute the Derivatives

Calculate the derivatives for each component with respect to \(t\). - For \(x = t^2\), \(\frac{dx}{dt} = 2t\).- For \(y = \frac{4}{3}t^{3/2}\), \(\frac{dy}{dt} = 2t^{1/2}\).- For \(z = t\), \(\frac{dz}{dt} = 1\).

Step 3: Substitute the Derivatives into the Arc Length Formula

Substitute the derivatives calculated in Step 2 into the arc length formula:\[ L = \int_{0}^{8} \sqrt{(2t)^2 + (2t^{1/2})^2 + (1)^2} \, dt \]This simplifies to:\[ L = \int_{0}^{8} \sqrt{4t^2 + 4t + 1} \, dt \]

Step 4: Simplify the Expression Under the Square Root

Notice the expression under the square root: \(4t^2 + 4t + 1\) is a perfect square.It can be rewritten as \((2t + 1)^2\).

Step 5: Integrate the Simplified Expression

Use the simplified expression to evaluate the integral:\[ L = \int_{0}^{8} |2t + 1| \, dt \]Since \(2t + 1\) is always positive over the interval \([0, 8]\), we can drop the absolute value:\[ L = \int_{0}^{8} (2t + 1) \, dt \]. Evaluate this integral.

Step 6: Evaluate the Integral

Compute the integral:\[ L = \int_{0}^{8} (2t + 1) \, dt = \left[ t^2 + t \right]_{0}^{8} \].Calculate the definite integral:\[ L = (64 + 8) - (0 + 0) = 72 \].

Key Concepts

3D Parametric Curves

In mathematics, a 3D parametric curve is a way to represent a curve in three-dimensional space using a set of parametric equations. These equations express the coordinates of the points on the curve as functions of a parameter, commonly denoted as \( t \). For the given problem, the parametric equations are \( x = t^2 \), \( y = \frac{4}{3}t^{3/2} \), and \( z = t \). These equations describe how the position of a point on the curve changes as \( t \) varies. 3D parametric curves are essential in fields such as computer graphics and physics, where they are used to model trajectories and shapes in space. They provide a dynamic way to describe objects and paths that are not easily captured by traditional equations.

Integral Calculus

Integral Calculus is a branch of mathematics focusing on integrals and their properties. It is concerned with finding the whole from the part, meaning it helps in determining quantities like areas, volumes, and arc lengths from small incremental data. In the context of finding the arc length of a parametric curve, Integral Calculus plays a crucial role. We use integrals to sum up infinitely small pieces of a curve's length to find the total length. This is done through a definite integral, signified by the integral sign \( \int \) along with limits of integration, which precisely define the interval over which you are summing. The use of Integral Calculus allows us to solve complex geometry problems involving smooth curves and surfaces.

Derivatives

A derivative represents the rate at which a quantity changes. In calculus, it is a fundamental tool for understanding how functions behave. For parametric curves, derivatives with respect to the parameter \( t \) help us determine the components of velocity in the direction of each axis in three-dimensional space. In this particular exercise, we calculate derivatives such as \( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt} \). These represent how fast the \( x \), \( y \), and \( z \) coordinates are changing. Differentiating each component of the curve is essential for finding the velocity vector along the curve and further using it to calculate arc lengths using the arc length formula. It essentially empowers us to manage dynamic changes in mathematical models.

Definite Integrals

Definite Integrals are used to calculate the accumulation of quantities and are central to Integral Calculus. They are expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are limits of integration and \( f(x) \) is the function being integrated. In the process of finding arc lengths of curves, definite integrals help us sum up the precise length of curved sections between two points \( a \) and \( b \). In the exercise, the integral \( \int_{0}^{8} (2t + 1) \, dt \) was evaluated to find the arc length over the interval \([0, 8]\). Evaluating a definite integral involves finding the antiderivative of the function, then computing the difference between its values at \( b \) and \( a \), giving us a solid understanding of the end-to-end measure of the quantity.