The image provides an explanation of parametric functions and their derivatives with respect to a parameter $t$, along with an example problem and a practice problem. Here's a breakdown of the content:

Key Concepts:

• Parametric functions involve a parameter $t$, and the goal is to find the derivative of $y$ with respect to $x$, denoted as $\frac{dy}{dx}$, by calculating the derivatives with respect to $t$.
• Formula: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Example Problem:

• Given: $x(t) = t^2 + 1 \quad \text{and} \quad y(t) = t^3$ Find $\frac{dy}{dx}$.

• Solution:

  • First, compute the derivatives with respect to $t$: $\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2$
  • Then, apply the formula: $\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$

Practice Problem:

• Given: $x(t) = \sin(t) \quad \text{and} \quad y(t) = \cos(t)$ Find $\frac{dy}{dx}$.

Let's solve the practice problem:

1. Compute the derivatives with respect to $t$:

   • $\frac{dx}{dt} = \cos(t)$
   • $\frac{dy}{dt} = -\sin(t)$

2. Apply the formula for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-\sin(t)}{\cos(t)} = -\tan(t)$

Thus, the solution to the practice problem is $\frac{dy}{dx} = -\tan(t)$.

Would you like more details on any part of the solution or further explanation? Here are some related questions:

1. How do you apply the chain rule when working with parametric equations?
2. Can parametric functions be expressed as implicit functions?
3. What happens when the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$ involve trigonometric functions?
4. How can parametric equations be used to describe curves in physics?
5. What are some applications of parametric differentiation in real-world problems?

Tip: Always ensure that $\frac{dx}{dt} \neq 0$ to avoid division by zero when calculating $\frac{dy}{dx}$.