Welcome back to Professor Baker's Math Class! Today, we are exploring Chapter 10, Sections 1 and 2, where we shift our perspective from standard $y=f(x)$ functions to Parametric Equations. This is a crucial leap in calculus that allows us to describe curves that fail the vertical line test and model the motion of particles over time.

1. What are Parametric Equations?

Usually, we define graphs by relating $x$ and $y$ directly. However, in parametric equations, both $x$ and $y$ are defined as functions of a third variable, called the parameter (often denoted as $t$).

  • $x = f(t)$
  • $y = g(t)$

As $t$ varies, the point $(x, y)$ traces out a curve $C$. While $t$ often represents time in physics problems, it can represent an angle (like $\theta$) in geometric problems.

2. Tangents and Slopes

Just because we are using a parameter doesn't mean we stop caring about the slope! To find the tangent line at a specific point on the curve, we use the Chain Rule. The slope $\frac{dy}{dx}$ is calculated as:

$$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \quad \text{provided } \frac{dx}{dt} \neq 0 $$

Key takeaways from the notes:

  • Horizontal Tangents: Occur where $\frac{dy}{dt} = 0$ (and $\frac{dx}{dt} \neq 0$).
  • Vertical Tangents: Occur where $\frac{dx}{dt} = 0$ (and $\frac{dy}{dt} \neq 0$).

3. Areas under Parametric Curves

Finding the area under a curve remains a fundamental skill. If a curve is defined parametrically, we can adapt our standard integral $A = \int y \, dx$ using the Substitution Rule:

$$ A = \int_{\alpha}^{\beta} g(t) f'(t) \, dt $$

In our class notes, we looked at the fascinating example of the Cycloid—the curve traced by a point on a rolling circle. Using the formula above, we proved that the area under one arch of a cycloid is exactly $3\pi r^2$!

4. Arc Length

How long is a curve? When working with Cartesian coordinates, the formula can be messy. However, parametric equations offer a very symmetric and Pythagorean-looking formula for Arc Length ($L$):

$$ L = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt $$

We practiced this by finding the length of a loop defined by cubic and quadratic components, resulting in an approximation of $15.21$.

5. Surface Area of Revolution

Finally, we extended our reach to 3D geometry. If we rotate a parametric curve around the x-axis, the surface area ($S$) is found by:

$$ S = \int_{\alpha}^{\beta} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt $$

A classic application of this is proving the surface area of a sphere. By defining a semi-circle parametrically ($x=r\cos\theta, y=r\sin\theta$) and rotating it, the integral simplifies beautifully to the famous formula $S = 4\pi r^2$.

Make sure to review the attached PDF for the step-by-step derivation of the Cycloid area and the specific algebra used in the Arc Length examples. Keep practicing, and happy calculating!