91Ó°ÊÓ

A plane curve is a curve that lies entirely on a single two-dimensional plane, usually represented as the xy-plane. In our context, the curve is defined by the vector function \( \mathbf{r}(t) = t \mathbf{i} + 3t \mathbf{j} \). This equation tells us that as \( t \) changes, the point \( (t, 3t) \) traces out a specific path on the plane.

• Each component of the vector function represents a coordinate on the plane: \( x(t) = t \) and \( y(t) = 3t \).
• This particular plane curve is a line because both functions \( x(t) \) and \( y(t) \) are linear.

Visualizing these functions helps us understand that the curve is indeed a straight line with a slope of 3, passing directly through the origin. This line rises 3 units for every 1 unit it moves horizontally, showcasing a constant rate of change without any curvature.

Vector functions are crucial in describing the movement or trajectory of a point in space across dimensions. In the given exercise, \( \mathbf{r}(t) = t \mathbf{i} + 3t \mathbf{j} \) is a vector function.

• The function \( \mathbf{r}(t) \) combines both the horizontal and vertical components.
• The unit vectors \( \mathbf{i} \) and \( \mathbf{j} \) denote directions along the x-axis and y-axis, respectively.
• The scalar values multiplying these unit vectors (\( t \) and \( 3t \)) determine the magnitude and direction of the line.

Understanding vector functions like this provides a way to model paths or trajectories in a plane by combining multiple dimensions into a singular function.

In calculus, the derivative measures how a function changes as its input changes. For vector functions, each component can be differentiated separately. Here, to find the rate at which our curve's x and y coordinates change, we differentiate each component function in \( \mathbf{r}(t) \).

• The derivative of \( x(t) = t \) is \( dx/dt = 1 \).
• The derivative of \( y(t) = 3t \) is \( dy/dt = 3 \).

These derivatives tell us about the velocities in the x and y directions, respectively. Essentially, they're telling us how fast the particle represented by the vector function is moving along each axis at any point in time.
This constant rate of change for each component reflects the uniform motion along the line, a characteristic of linear functions, and provides the basis for further calculations, like finding curve lengths.

Integrals are essential for finding quantities like areas under curves or the length of a path. In this exercise, we use an integral to calculate the length of the curve defined by the vector function over a specific interval.

• The formula for curve length is \[ L = \int_a^b \sqrt{(dx/dt)^2 + (dy/dt)^2} \ dt. \]
• This formula effectively adds up tiny segments along the curve, represented by the square root expression under the integral sign.

In our example, substituting the derivatives \( dx/dt = 1 \) and \( dy/dt = 3 \) into this formula gives \[ L = \int_0^4 \sqrt{1^2 + 3^2} \ dt = \int_0^4 \sqrt{10} \ dt = 4\sqrt{10}. \]
Thus, the length of the curve over the interval from 0 to 4 is computed with the integral, demonstrating both the practical application of integration and the geometric property of measuring distance along a path.