Problem 20 - Thomas Calculus

Question

Evaluate \(\int _ { C } \sqrt { x + 2 y } d s ,\) where \(C\) is a. the straight-line segment \(x = t , y = 4 t ,\) from \(( 0,0 )\) to \(( 1,4 )\) . b. \(C _ { 1 } \cup C _ { 2 } ; C _ { 1 }\) is the line segment from \(( 0,0 )\) to \(( 1,0 )\) and \(C _ { 2 }\) is the line segment from \(( 1,0 )\) to \(( 1,2 )\) .

Step-by-Step Solution

Answer

a. \(2\sqrt{17}\); b. \(\frac{13}{3}\).

Step 1: Determine the Parameterization for Part a

For the line segment from \((0,0)\) to \((1,4)\), given that \(x = t\) and \(y = 4t\) for \(0 \leq t \leq 1\), the position vector is \(\vec{r}(t) = (t, 4t)\).

Step 2: Calculate the Differential Arc Length for Part a

The differential arc length is computed as \(ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). For \(x = t\) and \(y = 4t\), this becomes \(ds = \sqrt{1^2 + 4^2} \, dt = \sqrt{17} \, dt\).

Step 3: Evaluate the Integral for Part a

Substitute \(x = t\) and \(y = 4t\) into the integrand to get \(\sqrt{x + 2y} = \sqrt{t + 8t} = \sqrt{9t}\). Therefore, the integral becomes \(\int_{0}^{1} \sqrt{9t} \sqrt{17} \, dt = \sqrt{17} \int_{0}^{1} 3\sqrt{t} \, dt\).

Step 4: Integrate for Part a

The integral \(\int 3\sqrt{t} \, dt\) can be computed by substituting \(\sqrt{t} = t^{1/2}\). So, \(\int_{0}^{1} 3t^{1/2} \, dt = 3 \left[ \frac{2}{3}t^{3/2} \right]_{0}^{1} = [2t^{3/2}]_{0}^{1} = 2\). Multiply by \(\sqrt{17}\) to get \(2\sqrt{17}\).

Step 5: Determine the Parameterization for Part b Component C1

For \(C_1\), the segment from \((0,0)\) to \((1,0)\), the parameterization is \(x = t\), \(y = 0\) for \(0 \leq t \leq 1\).

Step 6: Calculate the Differential Arc Length for Component C1

For \(C_1\), \(ds = \sqrt{1^2 + 0^2} \, dt = \, dt\).

Step 7: Evaluate C1 Integral

The integrand becomes \(\sqrt{x + 2y} = \sqrt{t}\). The integral is \(\int_{0}^{1} \sqrt{t} \, dt = \frac{2}{3}[t^{3/2}]_{0}^{1} = \frac{2}{3}\).

Step 8: Determine the Parameterization for Part b Component C2

For \(C_2\), the segment from \((1,0)\) to \((1,2)\), \(x = 1\), \(y = t\) for \(0 \leq t \leq 2\).

Step 9: Calculate the Differential Arc Length for Component C2

For \(C_2\), \(ds = \sqrt{0^2 + 1^2} \, dt = \, dt\).

Step 10: Evaluate C2 Integral

The integrand becomes \(\sqrt{x + 2y} = \sqrt{1 + 2t}\). The integral is \(\int_{0}^{2} \sqrt{1 + 2t} \, dt\). Performing a substitution, let \(u = 1 + 2t\); hence, \(du = 2\,dt\). The integral becomes \(\frac{1}{2}\int_{1}^{5} \sqrt{u} \, du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{5} = \frac{1}{3} \left(5^{3/2} - 1^{3/2}\right)\).

Step 11: Compute Total for Part b

Add the results from \(C_1\) and \(C_2\) to find the total integral for part b: \(\frac{2}{3} + \frac{1}{3} \left(5^{3/2} - 1\right) = \frac{2}{3} + \frac{1}{3} \times 11 = \frac{2}{3} + \frac{11}{3} = \frac{13}{3}\).

Key Concepts

Parameterization of Curves

One important step in solving line integrals is parameterizing the curves along which the integral is computed. Imagine the curve as a path or line you are tracking. We express the points on this path using parameters, which are typically denoted by variables like \(t\). This is essentially setting up a function that will describe the curve in terms of these parameters. For example, the line segment from \((0,0)\) to \((1,4)\) can be parameterized as \(x = t\) and \(y = 4t\) for \(0 \leq t \leq 1\). This makes calculations easier, as it transforms the problem from being about arbitrary points to being about specific points described by equations.
To summarize, parameterization transforms a curve into a simpler form:

• Describe the curve with equations in terms of a parameter \(t\).
• Simplifies integration by providing a concrete path to follow.
• Converts multi-variable relationships into single-variable expressions.

Differential Arc Length

Another key concept in line integrals is the differential arc length, \(ds\). It represents a tiny piece of the path's length, and it's crucial because it helps weight the contribution of each segment to the overall integral. Mathematically, it's expressed as \(\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). This formula originates from the Pythagorean theorem, adapting it to accommodate curves rather than straight lines.
Using \(x = t\) and \(y = 4t\), the formula simplifies the calculation of arc length as:

• \(\sqrt{1^2 + 4^2} \, dt = \sqrt{17} \, dt\), making it easier to proceed with the integration.

The differential arc length is crucial because:

• It accounts for the changes in both \(x\) and \(y\) components of a curve.
• Combines the curve's trajectory into manageable pieces for summation/integration.
• Results in a precise measure of the segment's contribution to the integral.

Integration by Substitution

Integration by substitution is a powerful tool for solving complex integrals by turning them into simpler forms. It resembles reversing the chain rule: given a complicated expression, we cleverly change variables to make the integral easier. For a function \(\sqrt{1 + 2t}\) for example, using \(u = 1 + 2t\) as a substitution simplifies the integration process significantly.
As you change variables:

• Pay attention to the new limits of integration which also change based on substitution.
• Calculate the differential \(du\), remembering it relates to \(dt\) as \(du = 2 \, dt\).
• The integral transforms, reducing to simpler forms like \([\frac{2}{3} u^{3/2}]\) which are straightforward to calculate

This technique simplifies complicated changes:

• Makes intricate integrals manageable and solvable.
• Cycles back to familiar integral forms which can be computed with basic rules.

Piecewise Curve Integration

Piecewise curve integration is a technique where a complex path or curve is split into simpler segments, each evaluated separately, and then recombined. Consider a path from \((0,0)\) to \((1,0)\) to \((1,2)\)—this can be split into two segments: \(C_1\) and \(C_2\). Each segment is individually parameterized, and separate integrals are evaluated before being summed.
This approach is vital for:

• Managing complex paths by dividing them into simple, manageable sections.
• Applying different parameterizations for each section, based on its unique path.
• Ensuring each section's contribution is considered and calculated independently.

For piecewise integration:

• Carefully account for each curve's parameterization and differential arc length.
• Calculate individual integrals, carefully ensuring their parameters match actual sections.
• Combine these results to find the total integral of the piecewise curve.

This method leverages simplicity in parts to tackle the complexity as a whole, breaking down a difficult problem into digestible sections.