Chapter 11: Problem 26
$$ \text { In Problems 25-32, find the arc length of the given curve. } $$ $$ x=t / 4, y=t / 3, z=t / 2 ; 1 \leq t \leq 3 $$
Short Answer
Expert verified
The arc length is \(\frac{\sqrt{61}}{6}\).
Step by step solution
01
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 \.
02
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 \( t \).
03
Write the Arc Length Formula
Substitute the derivatives calculated into the arc length formula:\( L = \int_{1}^{3} \sqrt{\left( \frac{1}{4} \right)^2 + \left( \frac{1}{3} \right)^2 + \left( \frac{1}{2} \right)^2 \space } dt \).
04
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, \( L = \int_{1}^{3} \sqrt{ \frac{61}{144} } dt = \int_{1}^{3} \frac{\sqrt{61}}{12} dt \).
05
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: \( \frac{\sqrt{61}}{12} (3 - 1) = \frac{\sqrt{61}}{12} \times 2 = \frac{\sqrt{61}}{6} \).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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 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.
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:
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.