Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems 25-32, find the arc length of the given curve. } $$ $$ x=t / 4, y=t / 3, z=t / 2 ; 1 \leq t \leq 3 $$

Step-by-Step Solution

Answer

The arc length is \(\frac{\sqrt{61}}{6}\).

Step 1: Understand the Formula for Arc Length

The arc length \( L \) of a curve described parametrically by \( (x(t), y(t), z(t)) \) is given by the integral: \(L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 \space } dt\). We have \( x = \frac{t}{4}, \ y = \frac{t}{3}, \ z = \frac{t}{2} \), for \ 1 \leq t \leq 3 \.

Step 2: Compute Derivatives

Calculate the derivatives with respect to \( t \): - \( \frac{dx}{dt} = \frac{1}{4} \)- \( \frac{dy}{dt} = \frac{1}{3} \)- \( \frac{dz}{dt} = \frac{1}{2} \)These derivatives give the rates of change of each coordinate of the curve with \( t \).

Step 3: Write the Arc Length Formula

Substitute the derivatives calculated into the arc length formula:\( L = \int_{1}^{3} \sqrt{\left( \frac{1}{4} \right)^2 + \left( \frac{1}{3} \right)^2 + \left( \frac{1}{2} \right)^2 \space } dt \).

Step 4: Simplify Inside the Integral

Simplify the expression under the square root: - Calculate: \( \left( \frac{1}{4} \right)^2 = \frac{1}{16}, \ \left( \frac{1}{3} \right)^2 = \frac{1}{9}, \ \left( \frac{1}{2} \right)^2 = \frac{1}{4} \)- Combine these: \( \frac{1}{16} + \frac{1}{9} + \frac{1}{4} = \frac{9}{144} + \frac{16}{144} + \frac{36}{144} = \frac{61}{144} \)Thus, \( L = \int_{1}^{3} \sqrt{ \frac{61}{144} } dt = \int_{1}^{3} \frac{\sqrt{61}}{12} dt \).

Step 5: Evaluate the Integral

Integrate over \( t \) from 1 to 3:- The integral is \( \int_{1}^{3} \frac{\sqrt{61}}{12} dt = \frac{\sqrt{61}}{12} \left[ t \right]_{1}^{3} \)- Plugging the limits of integration: \( \frac{\sqrt{61}}{12} (3 - 1) = \frac{\sqrt{61}}{12} \times 2 = \frac{\sqrt{61}}{6} \).

Step 6: Final Answer

The arc length of the curve from \( t=1 \) to \( t=3 \) is \( \frac{\sqrt{61}}{6} \). This value represents the total length of the curve in three-dimensional space over the specified interval.

Key Concepts

Parametric Equations

Parametric equations are a powerful way to describe curves. Instead of expressing one variable in terms of another, like \( y = f(x) \), parametric equations define both \( x \) and \( y \) (and possibly \( z \)) in terms of a third variable, usually \( t \), known as the parameter. This offers a more intuitive way to trace complex curves as it allocates separate equations for each coordinate.

In the exercise, the parametric equations are \( x = \frac{t}{4} \), \( y = \frac{t}{3} \), and \( z = \frac{t}{2} \). As the parameter \( t \) varies between 1 and 3, these equations define a specific curve in three-dimensional space.

• By setting limits on \( t \), you define the segment of the curve to be analyzed or measured.
• These equations let you see how positions change uniformly with time or other indices, offering a dynamic perspective on curve motion or path.

This is particularly useful when dealing with paths and motions in physics, where multiple dimensions come into play.

Integral Calculus

Integral calculus is essential to calculate arc lengths, among other things, because it sums infinitely small sections together to find a total quantity. When determining the arc length of a parametric curve, we use definite integrals to measure the distance over the interval of the parameter \( t \).

The formula for arc length given by parametric equations \((x(t), y(t), z(t))\) from \( t=a \) to \( t=b \) is:
\[ L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 } \, dt \]

• The expression under the integral sign accounts for the rate of change (derivatives) of each coordinate with respect to \( t \).
• A definite integral adds up the infinitesimal segments along the curve, providing a cumulative measure of total length.

This technique elegantly marries geometry with calculus. It helps understand not just lengths, but also areas, volumes, and related concepts through summation and limits.

Three-Dimensional Curves

Three-dimensional curves add complexity beyond the two-dimensional case as they involve an additional coordinate, \( z \). This brings a new layer of depth and challenges our spatial understanding.

The given curve in the exercise is defined by parametric equations that specify positions in space by \( x = \frac{t}{4} \), \( y = \frac{t}{3} \), and \( z = \frac{t}{2} \). Here, each axis component of the curve develops independently. This means:

• Both curvature and torsion need to be considered, which makes analyzing 3D curves complex yet fascinating.
• The arc length formula integrates these perspectives of constant change (along \( x, y, \) and \( z \) directions) to find a true spatial length measure.

Applications of 3D curves range from computer graphics to physics, where real-world paths and shapes often span three dimensions. Understanding their arc lengths helps evaluate trajectories and pathways effectively.