91Ó°ÊÓ

Parametric equations are a powerful way to describe curves and lines using parameters like time. They tell us where a point on the curve is at any given time or t-value. Unlike regular equations, which directly relate x and y, parametric equations let us express x and y as different functions of one or more parameters. This is especially useful for describing complex motions, like the path of a particle in space. For lines, parametric equations often look like:

• For x: \[ x = x_0 + at \]
• For y: \[ y = y_0 + bt \]

where \( (x_0, y_0) \) is a point on the line, and a and b are the directions of the line, or slopes in terms of parametrics. These equations provide flexibility by letting us encapsulate both linear and non-linear paths as functions of t.

A vector-valued function, like \( \mathbf{r}(t) = e^{2t} \mathbf{i} - 2\cos(3t) \mathbf{j} \), outputs vectors instead of just numbers. The derivative of a vector-valued function works like taking derivatives of each of its component functions. The component functions are like small pieces—each is function moving in either the i or j direction. To find the derivative \( \mathbf{r}'(t) \), differentiate each piece separately:

• The derivative of \( e^{2t} \mathbf{i} \) is \[ 2e^{2t} \mathbf{i} \]
• Using the chain rule, \( -2\cos(3t) \mathbf{j} \) becomes \[ 6\sin(3t) \mathbf{j} \]

Combining these gives the full derivative \( \mathbf{r}'(t) = 2e^{2t} \mathbf{i} + 6\sin(3t) \mathbf{j} \). This function gives us vectors pointing in the direction of the tangent to the curve at any t.

The chain rule is a fundamental tool in calculus used when differentiating composite functions. It tells us how to take the derivative of a function that is made up of other functions. Think of it as peeling apart layers of an onion: each layer is a function inside another.Consider the function \( -2\cos(3t) \), made up of an inner and outer function:

• Outer: \( \cos(u) \)
• Inner: \( u = 3t \)

The chain rule states: First, take the derivative of the outer function \( \cos(u) \), which is \( -\sin(u) \), then multiply it by the derivative of the inner function \( 3t \), which is 3. Putting it together with the constant factor -2, we get the derivative:\[ -2\cos(3t) \rightarrow 6\sin(3t) \]This operation is why we could express part of the derivative of \( \mathbf{r}(t) \) in terms of \( \sin \), demonstrating the inner change in t affects the outer cosine function's rate of change.