Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the arc length of the given curve. \(x=t^{2}, y=\frac{4 \sqrt{3}}{3} t^{3 / 2}, z=3 t ; 1 \leq t \leq 4\)

Step-by-Step Solution

Answer

The arc length of the curve is 24.

Step 1: Find Derivatives of Parametric Equations

The parametric equations are given as \( x = t^2 \), \( y = \frac{4\sqrt{3}}{3} t^{3/2} \), and \( z = 3t \). Compute the derivatives:- \( \frac{dx}{dt} = 2t \)- \( \frac{dy}{dt} = 2\sqrt{3} t^{1/2} \)- \( \frac{dz}{dt} = 3 \)

Step 2: Apply Arc Length Formula for Parametric Curves

The arc length \( L \) for a parametric curve from \( t=a \) to \( t=b \) is given by the integral:\[ L = \int_a^b \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 } \, dt \]Substituting the derivatives, the arc length becomes:\[ L = \int_1^4 \sqrt{(2t)^2 + (2\sqrt{3} t^{1/2})^2 + 3^2} \, dt \]

Step 3: Simplify the Integrand

Simplify the expression under the square root:\[ (2t)^2 = 4t^2, \quad (2\sqrt{3} t^{1/2})^2 = 12t, \quad 3^2 = 9 \]Adding these gives:\[ 4t^2 + 12t + 9 \]The integrand becomes:\[ \sqrt{4t^2 + 12t + 9} \]

Step 4: Factor and Integrate

Notice that \( 4t^2 + 12t + 9 = (2t + 3)^2 \), hence:\[ \sqrt{4t^2 + 12t + 9} = |2t+3| \]For the interval \( 1 \leq t \leq 4 \), \( 2t+3 \) is always positive, so:\[ \sqrt{4t^2 + 12t + 9} = 2t + 3 \]The definite integral becomes:\[ L = \int_1^4 (2t + 3) \, dt \]

Step 5: Evaluate the Integral

Evaluate the integral:\[ L = \left[ t^2 + 3t \right]_1^4 \]Calculate:- When \( t = 4 \), \( t^2 + 3t = 16 + 12 = 28 \)- When \( t = 1 \), \( t^2 + 3t = 1 + 3 = 4 \)Thus, the arc length is:\[ L = 28 - 4 = 24 \]

Step 6: State the Final Answer

The arc length of the curve over the interval \( 1 \leq t \leq 4 \) is 24.

Key Concepts

Parametric Equations

Parametric equations express the coordinates of points that make up a geometric object, such as a curve, using one or more parameters. In this exercise, the parameter is denoted by \( t \), and the coordinates \( (x, y, z) \) of the curve are defined by the following expressions: \( x = t^2 \), \( y = \frac{4\sqrt{3}}{3} t^{3/2} \), and \( z = 3t \).

Unlike traditional Cartesian coordinates, which directly express \( y \) as a function of \( x \), parametric equations allow each dimension to be expressed independently in terms of \( t \). This flexibility can be particularly advantageous when dealing with curves in three dimensions or more complex shapes.

Parametric equations can describe a wide variety of curves, such as circles and ellipses, and even more complicated shapes that are difficult to describe using standard equations. In this problem, parametric equations help us analyze and find the arc length of the given 3D curve.

Derivatives

In calculus, derivatives represent the rate at which a quantity changes. When dealing with parametric equations, calculating derivatives with respect to the parameter \( t \) is essential for understanding how the curve behaves.

For the curve defined by \( x = t^2 \), \( y = \frac{4\sqrt{3}}{3} t^{3/2} \), and \( z = 3t \), we differentiate each equation with respect to \( t \) to find:

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

These derivatives describe the velocity of the curve along the \( x \), \( y \), and \( z \) directions.

Understanding these derivatives is crucial as they are used to construct the integrand needed for calculating the arc length of the curve. Essentially, they help us see how fast each component of the curve is changing as the parameter \( t \) changes.

Definite Integral

A definite integral gives us the accumulated quantity of a function over a specific interval. In the context of this problem, the definite integral is used to calculate the total arc length of the curve from \( t = 1 \) to \( t = 4 \).

Once we have the derivatives \( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt} \), we use these to form an expression that can be integrated. The arc length is calculated using the formula:\[L = \int_1^4 \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt\]

This integral takes into account the rate of change in each dimension of the curve and sums up these little bits of change to find the entire length. It is crucial to evaluate this integral over specified bounds, ensuring that we only measure the desired section of the curve.

Arc Length Formula

The arc length formula for parametric curves helps us determine the total length of a curve described by parametric equations over a certain interval. This formula is founded on the idea of taking the square root of the sum of the squares of the derivatives of the parametric equations.

In this exercise, the formula is applied as follows:\[L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt\]

For the curve \( x = t^2 \), \( y = \frac{4\sqrt{3}}{3} t^{3/2} \), and \( z = 3t \), the formula evaluates to:

• The integrand becomes \( \sqrt{4t^2 + 12t + 9} \) as a result of combining squares of the derivatives.
• Further simplification shows that \( \sqrt{4t^2 + 12t + 9} = |2t+3| \), which equals \( 2t+3 \) in this range.
• The resulting integral \( \int_1^4 (2t + 3) \, dt \) is straightforward and evaluates to 24.

The arc length formula elegantly combines differential and integral calculus to measure the geometric property of curves. Understanding it aids greatly in geometrical analysis and problem-solving.