Question

Find \(t=t_{0}\) where the graph of the given parametric equations is not smooth, then find \(\lim _{t \rightarrow t_{0}} \frac{d y}{d x}\). \(x=\cos ^{2} t, \quad y=1-\sin ^{2} t\)

Short answer

The curve is not smooth at \( t = \frac{\pi}{2}, \pi, \frac{3\pi}{2} \). The limit is \( -1 \).

Step-by-step solution

Understand the Parametric Equations

The given parametric equations are \( x = \cos^2 t \) and \( y = 1 - \sin^2 t \). Rewrite \( y \): since \( \sin^2 t + \cos^2 t = 1 \), it follows that \( y = \cos^2 t \). Thus, both \( x \) and \( y \) represent the same expression: \( \cos^2 t \).

Identify Points of Non-smoothness

A parametric curve \( (x(t), y(t)) \) is not smooth at \( t = t_0 \) if the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are both zero at \( t = t_0 \). Compute \( \frac{dx}{dt} = -2\cos t \sin t \) and \( \frac{dy}{dt} = 2\cos t \sin t \). Both derivatives are zero when \( \cos t \sin t = 0 \).

Solve for Non-smoothness Condition

Using \( \cos t \sin t = 0 \), this happens when either \( \cos t = 0 \) or \( \sin t = 0 \). Therefore, \( t = \frac{\pi}{2}, \frac{3\pi}{2}, 0, \pi, 2\pi, \ldots \). Check which falls within the fundamental period \([0, 2\pi]\).

Calculate the Limit

To evaluate \( \lim_{t \to t_0} \frac{dy}{dx} \), note that \( \frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{ 2\cos t \sin t }{ -2\cos t \sin t } = -1 \). As both \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) approach zero as \( t \to t_0 \), the derivative evaluation simplifies to \( -1 \) whenever both derivatives approach zero limitlessly. Thus, \( \lim_{t \to t_0} \frac{dy}{dx} = -1 \).

Key concepts

Understanding Parametric Equations

Parametric equations provide a way to define a set of related quantities expressed as functions of an independent parameter, often denoted by \( t \). In the original exercise, the equations are given as \( x = \cos^2 t \) and \( y = 1 - \sin^2 t \). The parametric nature of these equations allows for the description of curves in a coordinate system, with \( x \) and \( y \) depending on \( t \).
Let's illustrate it:

• Both \( x(t) \) and \( y(t) \) describe movements along a curve formed in the \( xy \)-plane.
• The independent variable \( t \) can be thought of as time, controlling how a point moves along the curve.
• Solving for \( y \) using trigonometric identity \( \sin^2 t + \cos^2 t = 1 \) simplifies \( y = \cos^2 t \). Both expressions for \( x \) and \( y \) turn out to be the same here: \( \cos^2 t \).

This redundancy implies a specific path along the curve, particularly simplifying the analysis of smoothness and direction of the tangent at different points.

Limits and Continuity in Parametric Curves

Understanding where and how a parametric curve is not smooth involves exploring the limits and behavior of its derivatives. A curve is smooth if, intuitively, you can draw it without lifting your pen, and this is mathematically linked to the continuity of derivatives.
Parametric curves are not smooth at a point \( t = t_0 \) where both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are equal to zero. Calculating the derivatives for the provided parametric equations:

• \( \frac{dx}{dt} = -2\cos t \sin t \)
• \( \frac{dy}{dt} = 2\cos t \sin t \)

At values of \( t \) where \( \cos t \sin t = 0 \), both derivatives are zero, causing potential discontinuities. Specifically, this happens at \( t = \frac{\pi}{2}, \frac{3\pi}{2}, 0, \pi, 2\pi, \ldots \). Within one cycle of \([0, 2\pi]\), non-smoothness occurs at \( t = \frac{\pi}{2} \) and \( t = \frac{3\pi}{2} \). Such values need special consideration when discussing the curve's behavior.

Determining the Slope of Parametric Curves

Finding the slope of a parametric curve involves determining \( \frac{dy}{dx} \) at a given point \( t \). This is crucial in understanding the direction in which the curve moves within the plane at that point. When both numerator and denominator of \( \frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } \) are zero, a special analysis is required.
In our case:

• \( \frac{dy}{dt} = 2\cos t \sin t \)
• \( \frac{dx}{dt} = -2\cos t \sin t \)

The ratio \( \frac{dy}{dx} \) simplifies to \(-1\) because the terms \(2\cos t \sin t\) cancel each other. This constant value reflects a vital understanding: as \( t \) approaches the critical points of non-smoothness, the slope of the curve approaches \(-1\). Understanding such limits helps better visualize and sketch the behavior of the curve.