91Ó°ÊÓ

When you're working with parametric equations, the slope of the curve is represented by the derivative \( \frac{dy}{dx} \). This describes how steep the curve is at any given point. In a standard form, you would think about the slope as the change in \( y \) over the change in \( x \), but with parametric equations, we use a different approach.

First, compute the derivatives for each of the parametric equations: \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). In essence, these derivatives give us the rate of change of \( x \) and \( y \) with respect to the parameter \( t \).

The slope \( \frac{dy}{dx} \) for the parametric curve is found by dividing \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \). Thus, this method allows us to explore how the curve behaves in terms of its direction and steepness at any particular point.

Finding the derivative \( \frac{dy}{dx} \) for parametric equations is a crucial step to understanding how the curve behaves. In parametric form, the derivative is derived from the separate derivatives of \( x \) and \( y \) with respect to \( t \).

Here’s how you do it:

• Compute \( \frac{dx}{dt} \), which tells you how \( x \) changes as \( t \) changes.
• Compute \( \frac{dy}{dt} \), which tells you how \( y \) changes as \( t \) changes.
• Divide \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \) to find \( \frac{dy}{dx} \).

In this exercise, we start with \( \frac{dy}{dt} = -4 \) and \( \frac{dx}{dt} = \frac{1}{2\sqrt{t}} \). The derivative \( \frac{dy}{dx} \) is then calculated as \( -8\sqrt{t} \). This particular derivative tells us how steep the curve is at each point as the parameter \( t \) increases.

Solving for the parameter \( t \) is often the key to understanding specific points on a parametric curve, especially when given a specific slope value.

In this case, to find when the slope is exactly 0, we set our expression for \( \frac{dy}{dx} \) equal to 0 and solve for \( t \). For this exercise, we determine that \( -8\sqrt{t} = 0 \).

This equation simplifies to give us \( t = 0 \). When you reach this solution, it indicates that the parameter \( t \) at which the curve has a slope of 0 is precisely at \( t = 0 \). This solution will later help find the actual point on the curve.

Once you've solved for the parameter \( t \), the next step is to find the actual coordinates on the curve. This involves substituting the value of \( t \) back into the given parametric equations.

For the example exercise, when you substitute \( t = 0 \) into \( x = 2 + \sqrt{t} \) and \( y = 2 - 4t \), you’ll get:

• \( x = 2 + \sqrt{0} = 2 \)
• \( y = 2 - 4 \times 0 = 2 \)

Therefore, the point on the curve where the slope is 0 is \((2, 2)\).

This process of finding the exact coordinates ensures that you accurately locate where the specific parameter conditions—like a given slope—occur on the parametric curve.