91Ó°ÊÓ

Parametric equations allow us to define a curve in the plane using two separate equations for each coordinate of the points on the curve, typically in terms of a third variable, called the parameter. In this exercise, the parameter is denoted as \( t \). Instead of expressing \( y \) as a direct function of \( x \), we use:

• \( x = \cos t \)
• \( y = 1 + \sin t \)

These equations specify a curve through the trajectory of points \((x, y)\) as \( t \) varies. When \( t = \frac{\pi}{2} \), substituting \( t \) into these equations gives the specific point \((0, 2)\) on the curve. This method of parameterizing via \( t \) is useful in tracing more complex paths and is especially helpful in dealing with objects moving over time.

The chain rule is a fundamental principle in calculus used to differentiate composite functions. When dealing with parametric equations, we often need to find \( \frac{dy}{dx} \), the derivative of \( y \) with respect to \( x \), while we begin with \( y \) and \( x \) defined in terms of \( t \). To achieve this, we differentiate each equation with respect to \( t \):

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

We can then relate these to find \( \frac{dy}{dx} \) using the formula:\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]This gives us \( \frac{dy}{dx} = -\cot t \). At \( t = \frac{\pi}{2} \), substituting this value results in \( \frac{dy}{dx} = 0 \). Hence, the slope of the tangent line at this point is 0, indicating a horizontal tangent. This chain rule insight is crucial when dealing with parametric curves and their slopes.

The second derivative \( \frac{d^2y}{dx^2} \) of a parametric curve provides information about the curve's concavity and inflection points. To find this, we initially differentiate \( \frac{dy}{dx} \) with respect to \( t \), given:\[\frac{dy}{dx} = -\cot t\]Using the derivative \( \frac{d}{dt} \left(-\cot t\right) = \csc^2 t \), we then apply the chain rule again, multiplying by \( \frac{dt}{dx} \):\[\frac{d^2y}{dx^2} = \csc^2 t \cdot \frac{1}{\frac{dx}{dt}}\]Since \( \frac{dx}{dt} = -\sin t \), we substitute it to yield:\[\frac{d^2y}{dx^2} = \csc^2 t \cdot \frac{1}{-\sin t} = \frac{1}{-\sin t} \times \frac{1}{\sin t} = \frac{1}{-\sin^2 t}\]At \( t = \frac{\pi}{2} \), the value simplifies to -1, indicating that at this point, the curve is concave down. This demonstrates how parametric calculus can offer insights into a curve’s behavior beyond merely finding tangent lines.