Question

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

Short answer

The arc length of the curve is approximately 7.56 units.

Step-by-step solution

Understand the Formula for Arc Length

The formula for the arc length of a space curve defined by the parametric equations \(x(t), y(t), z(t)\) is given by \[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt\]where \(t\) ranges from \(a\) to \(b\). Our task is to apply this formula.

Compute First-Order Derivatives

Compute \(\frac{dx}{dt}\), \(\frac{dy}{dt}\), and \(\frac{dz}{dt}\).- For \( x = t^{3/2} \), \( \frac{dx}{dt} = \frac{3}{2}t^{1/2} \).- For \( y = t^{3/2} \), \( \frac{dy}{dt} = \frac{3}{2}t^{1/2} \).- For \( z = t \), \( \frac{dz}{dt} = 1 \).

Set Up the Arc Length Integral

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

Solve the Integral

The integral to solve is:\[L = \int_{2}^{4} \sqrt{\frac{9}{2}t + 1} \, dt\].Let \(u = \frac{9}{2}t + 1\), then \(\frac{du}{dt} = \frac{9}{2}\) or \(dt = \frac{2}{9} du\).When \(t = 2\), \(u = 10\) and when \(t = 4\), \(u = 19\).The integral becomes:\[L = \int_{10}^{19} \sqrt{u} \cdot \frac{2}{9} \, du\].This simplifies and evaluates to:\[L = \frac{2}{9} \cdot \left[ \frac{2}{3}u^{3/2} \right]_{10}^{19} = \frac{4}{27} \left[(19)^{3/2} - (10)^{3/2}\right]\].

Calculate the Final Result

Evaluate the numerical expression:- \((19)^{3/2} = \sqrt{19^3} = \sqrt{6859} \approx 82.64\).- \((10)^{3/2} = \sqrt{10^3} = \sqrt{1000} = 31.62\).So the length \(L\) is:\[L = \frac{4}{27} \left[ 82.64 - 31.62 \right] = \frac{4}{27} \times 51.02 \approx 7.56 \].

Key concepts

Parametric Equations

Parametric equations are essential for describing the position of points along a curve using parameters. In this context, a parameter, often represented as \( t \), varies over a certain interval and uniquely among x, y, and z coordinates. For this exercise, the curves are defined by \( x = t^{3/2} \), \( y = t^{3/2} \), and \( z = t \) with \( t \) ranging from 2 to 4.

This simple representation allows us to understand the behavior of each coordinate independently. Such equations are pivotal in generating complex three-dimensional paths where the relation between coordinates can't be easily expressed in conventional Cartesian equations. Each parameter value provides a point in the 3D space for a smooth and continuous depiction of curves.

Integral Calculus

Integral calculus is an area of calculus focused on the summation of parts to find the whole. In this exercise, integral calculus helps calculate the arc length of the curve. We apply the arc length formula, an integral, to estimate the total length of the path created in a three-dimensional space.

The formula used here 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 integration evaluates from \( t = 2 \) to \( t = 4 \).

• This integral sums infinitely small line segments between these limits, taking into account differentials of each coordinate.

Understanding this principle is vital in calculating distances over curves that are not straight.

Derivatives

Derivatives are a crucial tool to analyze the rate of change of a function concerning a variable. Here, derivatives of the parametric functions \( x(t) \), \( y(t) \), and \( z(t) \) are necessary to substitute into the arc length formula.

For instance, for \( x = t^{3/2} \), we derive: \( \frac{dx}{dt} = \frac{3}{2}t^{1/2} \),and the same for \( y \). For \( z = t \), the derivative is simply 1 as it changes linearly with respect to \( t \).

• The derivatives reflect how quickly each coordinate changes with respect to \( t \) and are used to calculate the velocity of each respective coordinate in the space curve.

This is vital to predict the precise shape and length of the curve.

Mathematical Solutions

Mathematical solutions often culminate in calculating a precise result after navigating through a series of mathematical operations. In this case, after substituting the derivatives back into the arc length formula, simplifying, and solving the integral, the final arc length of the curve turns into a numerical estimation.

The problem concludes by evaluating \[L = \frac{4}{27} \times 51.02 \] which approximately equals 7.56, representing the length of the curve between \( t = 2 \) and \( t = 4 \).

• Ensuring accuracy through each transformation and computation step is fundamental to yield an exact and correct final answer.
• Following through with logical steps ensures that the solution remains structured and understandable, assisting in learning how to apply these concepts effectively in similar problems.

Understanding each component of the solution aids in mastering such complex integrals.