Defining and Differentiating Parametric Equations Chapter Notes | Calculus BC - Grade 9 PDF Download

In Unit 9 of AP Calculus BC, we explore three fascinating topics: parametric equations, polar coordinates, and vector-valued functions. This post focuses on parametric equations, which are incredibly useful for modeling real-world phenomena like projectile motion or circular paths. Let’s dive into what makes parametric equations unique and how to differentiate them!

What Are Parametric Equations?

Parametric equations describe a relationship where the x- and y-coordinates of a point are defined independently in terms of a third variable, often called the parameter (typically t for time). Unlike the Cartesian system, where we assume steady movement along the x-axis, parametric equations offer flexibility to model complex motions, such as curves or trajectories, by allowing both coordinates to vary with t.
For example, consider the parametric equations:
x(t) = t² - 1, y(t) = 3t
Here, the x-coordinate depends on t² - 1, and the y-coordinate depends on 3t. At t = 1, you’d plot the point (0, 3) on the xy-plane. The parameter t doesn’t appear on the graph itself—it’s a tool to define the relationship between x and y independently.
There are several methods for calculating derivatives of real-valued functions, such as the limit definition, the power rule, the product rule, and the quotient rule. These methods can be extended to parametric functions, which are functions that depend on both a real variable and one or more parameters.
"To summarize, methods for calculating derivatives of real-valued functions can be extended to parametric functions."
One way to extend these methods to parametric functions is to treat the parameters as constants and use the usual rules for differentiation. For example, if a parametric function is given by f(x, p) = px², where x is the real variable and p is the parameter, then the derivative with respect to x can be calculated using the power rule as df/dx=2p∗x.
However, our method for computing derivatives will actually be much simpler than the method above.

Differentiating Parametric Equations

Since parametric equations are graphed on the xy-plane, finding the slope of a tangent line to a parametric curve still requires calculating dy/dx. However, because x and y are both functions of t, we use a special approach. The slope of the tangent line is given by:
dy/dx = (dy/dt) / (dx/dt)
This means you take the derivative of y(t) with respect to t (denoted dy/dt) and divide it by the derivative of x(t) with respect to t (denoted dx/dt). Note that dx/dt must not equal zero, as this could indicate a vertical tangent, where the slope is undefined.
This method, known as the parametric derivative, is unique to parametric equations and allows us to find the instantaneous rate of change at any point on the curve. It’s a powerful tool for analyzing motion or other dynamic systems.

Extending Differentiation to Parametric Functions

Still confused about the theory? Let's go into further detail on what this idea really means physically and mathematically.
This idea is known as the "parametric derivative" in calculus, often used to find the instantaneous rate of change of a parametric curve at a specific point. This method can only be used for parametric equations, where the curve is defined using a set of parametric equations in terms of a parameter, such as t.
To find the slope of the tangent line at a point on the curve, we first find the derivative of the x-coordinate function with respect to the parameter and the derivative of the y-coordinate function with respect to the parameter. These derivatives are denoted as dx/dt and  dy/dt respectively.
Then, at a specific point on the curve (x, y) the slope of the tangent line is found by taking the ratio of the derivative of the y-coordinate function with respect to the parameter (dy/dt) to the derivative of the x-coordinate function with respect to the parameter (dx/dt). This is the equation dy/dx = dy/dt /dx/dt we saw earlier! It’s important to note that the above equation is only valid if dx/dt is not equal to zero at the point of interest, as a vertical tangent line would not have a well-defined slope.

Practice Examples

Let’s solidify our understanding with two examples of finding the slope of a tangent line for parametric equations.