Problem 24 - Thomas Calculus

Question

Find the line integral of \(f(x, y)=\sqrt{y} / x\) along the curve \(\mathbf{r}(t)=t^{3} \mathbf{i}+t^{4} \mathbf{j}, 1 / 2 \leq t \leq 1\)

Step-by-Step Solution

Answer

The line integral is \(\frac{125}{48} - \frac{13 \sqrt{13}}{48}\.\)

Step 1: Set Up the Line Integral

The line integral of the function \( f(x, y) = \frac{\sqrt{y}}{x} \) along the parametric curve \( \mathbf{r}(t) = t^3 \mathbf{i} + t^4 \mathbf{j} \) is given by \( \int_C f(x, y) \, ds \). Here \( ds \) is the differential arc length along the curve. We express the function in terms of \( t \): \( x(t) = t^3 \) and \( y(t) = t^4 \). So, \( f(x, y) = \frac{\sqrt{t^4}}{t^3} = \frac{t^2}{t^3} = \frac{1}{t} \).

Step 2: Find the Differential Arc Length

The differential arc length \( ds \) is obtained as \( ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \). Calculate \( \frac{dx}{dt} = 3t^2 \) and \( \frac{dy}{dt} = 4t^3 \). Substitute these into the formula for \( ds \): \[ ds = \sqrt{(3t^2)^2 + (4t^3)^2} \, dt = \sqrt{9t^4 + 16t^6} \, dt = \sqrt{t^4(9 + 16t^2)} \, dt = t^2\sqrt{9 + 16t^2} \, dt. \]

Step 3: Write the Integral Expression

The line integral with respect to arc length is \[ \int_{1/2}^{1} \frac{1}{t} t^2 \sqrt{9 + 16t^2} \, dt. \]Simplify the expression inside the integral: \[ \int_{1/2}^{1} t \sqrt{9 + 16t^2} \, dt. \]

Step 4: Evaluate the Integral

To evaluate \( \int_{1/2}^{1} t \sqrt{9 + 16t^2} \, dt \), use substitution: let \( u = 9 + 16t^2 \), then \( \frac{du}{dt} = 32t \) or \( t dt = \frac{1}{32} du \). Change limits: when \( t = 1/2 \), \( u = 9 + 16(1/2)^2 = 13 \); when \( t = 1 \), \( u = 25 \). The integral becomes:\[ \frac{1}{32} \int_{13}^{25} \sqrt{u} \, du. \]

Step 5: Compute the Integral Result

Compute the integral:\[ \frac{1}{32} \int_{13}^{25} \sqrt{u} \, du = \frac{1}{32} \left[ \frac{2}{3} u^{3/2} \right]_{13}^{25} = \frac{1}{48} \left[ (25)^{3/2} - (13)^{3/2} \right]. \]Find the values:- \( (25)^{3/2} = (5^2)^{3/2} = 5^3 = 125 \)- \( (13)^{3/2} = 13 \cdot \sqrt{13} \)Substitute:\[ \frac{1}{48} (125 - 13 \sqrt{13}). \]

Step 6: Solution Conclusion

Thus, the value of the line integral is \( \frac{125}{48} - \frac{13 \sqrt{13}}{48} \).

Key Concepts

Parametric Curves

When working with line integrals, we often deal with parametric curves. These curves are defined using parameters, typically denoted as functions of a single variable, usually \( t \). For example, a curve could be represented as \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \). This means that the \( x \) and \( y \) coordinates of the curve change as \( t \) varies.

• In our exercise, the curve is defined as \( \mathbf{r}(t) = t^3 \mathbf{i} + t^4 \mathbf{j} \).
• Here, \( t \) indicates how the curve progresses over the interval \( 1/2 \leq t \leq 1 \).

Understanding parametric curves is crucial because they allow us to express complex paths in a manageable mathematical form. This is especially useful for line integrals, where both the function and the path need to be considered together.

Differential Arc Length

The concept of differential arc length is central to evaluating line integrals. The arc length along a curve gives us a measure of distance that respects the curve's curvature. To find the differential arc length, \( ds \), we use the formula:

• \( ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \)

For our specific curve:

• We calculate \( \frac{dx}{dt} = 3t^2 \) and \( \frac{dy}{dt} = 4t^3 \).
• Substitute these into the formula to find \( ds = t^2 \sqrt{9 + 16t^2} \, dt \).

This expression represents an infinitesimally small segment of the curve, weighted by its true path length. It's vital to grasp this concept to set up and solve line integrals accurately, as it transforms our curve-integral problem into a more approachable mathematical expression.

Substitution Method

The substitution method is a powerful technique used to simplify integrals that are otherwise difficult to evaluate. During the evaluation of line integrals, substitution can often help to transform a complicated integral into a simpler form. In our exercise, we encountered an integral that required simplification:

• The integral \( \int_{1/2}^{1} t \sqrt{9 + 16t^2} \, dt \) was simplified using substitution.

Here's how the substitution process worked:

• We set \( u = 9 + 16t^2 \) and found \( \frac{du}{dt} = 32t \), so \( t \, dt = \frac{1}{32} \, du \).
• Substitute back in to change our variable of integration from \( t \) to \( u \): this transformed the limits to go from 13 to 25, resulting in the integral \( \frac{1}{32} \int_{13}^{25} \sqrt{u} \, du \).

Substitution not only simplifies computations but also enables us to integrate more complex functions by replacing variables with those for which the integral is simpler or already known. Mastering this technique can greatly extend the range of problems you can solve.