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Understanding the parameterization of curves is essential in multiple areas of calculus, particularly when dealing with line integrals. Parameterization is a method of expressing a curve using a parameter, usually denoted as t, that varies over an interval. In the case of the textbook problem, the curve C is a line segment connecting two points in the plane.

To parameterize a line segment from point A to point B, we consider a point moving from A to B as t ranges from 0 to 1. For our specific example, the line segment from (1,1) to (10,10) can be represented using the formula r(t) = (1-t)A + tB. This formula creates a blend between point A and point B depending on t.

The selection of the parameter t is crucial as it influences the simplicity of the integral's subsequent evaluation. By choosing a parameter that changes linearly, we could keep the parameterization, and therefore the computations, straightforward.

Derivatives play a pivotal role when dealing with parameterized curves, especially during the evaluation of line integrals. For scalar line integrals where the curve is parameterized as r(t), we need to calculate the derivative of this parameterized curve with respect to the parameter t, denoted by dr/dt.

In our exercise, to find dr/dt, we differentiate each component of the vector r(t) with respect to t. Since our curve's parameterization resulted in the expression r(t) = (1 + 9t, 1 + 9t), differentiating it gives us dr/dt = (9, 9). This derivative vector plays a critical role in determining the differential element ds, which represents an infinitesimal segment of the curve's length and is necessary for integrating along the curve.

The line integral evaluation involves summing up values of a function along a curve. In scalar line integrals, we're interested in integrating a function over a path with respect to arc length. The notation ∫_C f(x,y) ds represents the line integral of f(x,y) along C, where ds is the differential arc length.

To evaluate this, we must express both the function f(x,y) and ds in terms of the parameter t. Using our derived expressions for x, y, and ds, we can rewrite the original integral entirely in terms of t and bounds that correspond to the start and end of the curve. In the problem, integration was facilitated by expressing ds as the magnitude of dr/dt times dt, simplifying the process to a standard one-variable integral.

The substitution method is a powerful technique in integration, often used to simplify integrals into a more manageable form. In practice, it involves identifying a portion of the integrand that can be replaced with a new variable, reducing the complexity of the integral. This typically involves choosing a substitution that will cancel out problematic terms in the differential.

In our line integral example, the substitution u = 1+9t transforms the complex fraction into a simple power of u, and du = 9dt aligns with dt in the integral, further simplifying the integral's evaluation. Finally, we change the integration limits to match the new variable, ensuring the integral evaluates over the same portion of the curve. The substitution method thereby reduces the integral to a form where standard antiderivatives can be applied, leading to an efficient solution.