91Ó°ÊÓ

A vector-valued function is a mathematical function that outputs vectors rather than scalar quantities. In this exercise, we consider a vector-valued function \( \mathbf{r}(t) = \frac{1}{3}t^3 \mathbf{i} + \frac{1}{2}t^2 \mathbf{j} \). Here, \( t \) is a parameter, and \( \mathbf{i}, \mathbf{j} \) are unit vectors along the coordinate axes.
Vector-valued functions describe curves in the plane or space by effectively assigning a position vector to every point in the curve as the parameter \( t \) varies. As \( t \) increases, \( \mathbf{r}(t) \) traces out a path or trajectory in two-dimensional space.
This way of representing curves helps us to explore their geometry and kinematics using calculus.

The derivative of a vector-valued function \( \mathbf{r}(t) \), denoted \( \mathbf{r}'(t) \), determines the rate at which the curve's position changes as the parameter \( t \) varies. For our specific function, \( \mathbf{r}(t) = \frac{1}{3}t^3 \mathbf{i} + \frac{1}{2}t^2 \mathbf{j} \), the derivative is:
\[ \mathbf{r}'(t) = t^2 \mathbf{i} + t \mathbf{j}. \]
This derivative vector is tangent to the curve at each point \( t \) and indicates the direction of movement along the curve. It also provides the velocity of a particle traveling along the curve, since the derivative measures how the position changes over time.

• Magnitude of \( \mathbf{r}'(t) \): It represents the speed of movement along the curve and is given by \( ||\mathbf{r}'(t)|| = t\sqrt{t^2 + 1} \).

This concept is fundamental in understanding both the local and global behavior of curves in space.

Integral calculus comes into play when we want to compute the arc length of a curve described by a vector-valued function. The arc length from a starting parameter value to another is given by the integral of the magnitude of the derivative vector.
In our exercise, the arc length \( s(t) \) is:
\[ s(t) = \int_0^t u\sqrt{u^2 + 1} \, du. \]
Using integral calculus, we solve this integral to find the total distance traveled along the curve from \( t = 0 \) to any point \( t \). The process involves substitution to simplify and compute the integral.

• This step allows expressing the parameter \( t \) in terms of the arc length \( s \), crucial for re-parametrizing the curve based on distance traveled.

Understanding these integrals helps in deriving an arc length parameterization that offers a consistent way to describe the curve's geometry.

Parametric curves are curves that are expressed as a set of equations where each coordinate is a function of one or more parameters. Here, the curve is parametrized by \( t \), and represented by \( \mathbf{r}(t) = \frac{1}{3}t^3 \mathbf{i} + \frac{1}{2}t^2 \mathbf{j} \).
Parametric equations offer flexibility in describing complex curves that are not easily expressed in Cartesian form. By using an additional parameter \( t \), parametric curves can model motion, where \( t \) often represents time, allowing us to track how a point moves along the curve.
Reparametrizing the curve using arc length, \( s \), results in an expression \( \mathbf{r}(s) \), where each part of the curve is related directly to the distance traveled along it:
\[ \mathbf{r}(s) = \frac{1}{3} \left( \sqrt{(3s + 1)^{2/3} - 1} \right)^3 \mathbf{i} + \frac{1}{2} \left( \sqrt{(3s + 1)^{2/3} - 1} \right)^2 \mathbf{j}. \]
This expression modifies the parameterization such that equal distances on the curve correspond to equal differences in the parameter \( s \), simplifying the analysis of the curve's physical properties.