91Ó°ÊÓ

Parametric equations allow us to describe a curve in the coordinate plane by defining both the x and y coordinates in terms of a third variable, typically denoted as t. Instead of having a direct relationship between x and y, parametric equations express each as a function of t.
In the given exercise, we have:

• The x-component as a function of t: \( x(t) = t^3 \)
• The y-component as a function of t: \( y(t) = \frac{3t^2}{2} \)

The parameter t varies over an interval that gives us a range of points \((x,y)\) that form the curve. In our example, t ranges from 0 to \(\sqrt{3}\). By using parametric equations, we can analyze and measure properties of curves that may not be easily accessible through straightforward Cartesian coordinates.

A definite integral represents the accumulation of quantities, such as area under a curve, over a specified interval. In the context of finding the length of a parametric curve, we use a definite integral to sum up tiny increments of length along the curve from one endpoint to the other.
In this exercise, the length \( L \) of the curve is given by the integral:\[L = \int_{0}^{\sqrt{3}} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt\]This integral captures the entire length of the curve between \( t = 0 \) and \( t = \sqrt{3} \) by adding up infinitely small linear segments. The part inside the square root represents the Pythagorean addition of changes in the x and y components, effectively providing us with the hypotenuse of the triangles formed along the curve.

Differentiation is the process of finding the derivative of a function, which represents the rate of change of the function with respect to one of its variables. When dealing with parametric curves, we compute derivatives to understand how x and y change as t changes.
For our parametric functions:

• The derivative of \( x(t) = t^3 \) with respect to t is \( \frac{dx}{dt} = 3t^2 \).
• The derivative of \( y(t) = \frac{3t^2}{2} \) with respect to t is \( \frac{dy}{dt} = 3t \).

These derivatives are crucial for determining the slope of the curve at any given point and are essential for integrating to find the curve's length. Differentiation allows us to break down these changes into simple, calculable pieces that fit into our integral for length.

The substitution method is a useful technique for evaluating integrals that may seem complex at first glance. By making a substitution, we can simplify parts of the integral to a more manageable form. In our exercise, this method plays a critical role in simplifying the calculation of the parametric curve's length.
When evaluating the length integral:

• We set \( u = t^2 + 1 \)
• Therefore, \( du = 2t \, dt \) implies \( dt = \frac{du}{2t} \)

Substituting back into our integral transforms it into a form that is easier to solve:\[\frac{3}{2} \int_{1}^{4} \sqrt{u} \, du\]This simplification allows us to compute the integral and ultimately find that the length of the curve is 7 units. The substitution method effectively transforms our integral into a form that uses basic calculus techniques, making difficult problems more approachable.