Question

Find \(t=t_{0}\) where the graph of the given parametric equations is not smooth, then find \(\lim _{t \rightarrow t_{0}} \frac{d y}{d x}\). . \(x=\frac{1}{t^{2}+1}, \quad y=t^{3}\)

Short answer

The curve is not smooth at \( t = 0 \); the limit is 0.

Step-by-step solution

Understand the definition of 'not smooth'

A parametric curve is not smooth at a point where the derivative of the parametric equations is not defined or is discontinuous. This typically happens where the derivative is undefined or infinite.

Find where the derivative is undefined

To identify where the curve might not be smooth, find where the derivative \( \frac{dy}{dx} \) does not exist by first finding \( \frac{dx}{dt} \text{ and } \frac{dy}{dt} \).

Calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)

Differentiate each parametric equation with respect to \( t \):- For \( x = \frac{1}{t^2 + 1} \), \( \frac{dx}{dt} = \frac{-2t}{(t^2 + 1)^2} \).- For \( y = t^3 \), \( \frac{dy}{dt} = 3t^2 \).

Set the denominator to zero and solve for \( t \)

Since a possible point of non-smoothness occurs where \( \frac{dx}{dt} = 0 \), solve \( \frac{-2t}{(t^2 + 1)^2} = 0 \). This gives \( t = 0 \).

Evaluate the limit

Calculate the limit of \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \) as \( t \to 0 \). Since \( \frac{dy}{dt} = 3t^2 \) and \( \frac{dx}{dt} = \frac{-2t}{(t^2 + 1)^2} \), the expression simplifies to: \[ \lim_{t \to 0} \frac{3t^2}{\frac{-2t}{(t^2 + 1)^2}} = \lim_{t \to 0} \frac{-3t (t^2 + 1)^2}{2} \].

Evaluate the expression as \( t \to 0 \)

Simplify \( \frac{-3t (t^2 + 1)^2}{2} \) and find: \[ \lim_{t \to 0} \frac{-3t (t^2 + 1)^2}{2} = 0 \]. The limit is 0.

Key concepts

Smoothness of Curves

Curves defined by parametric equations are a vital area of study in calculus. A parametric curve is described as "smooth" when it flows continuously without any sharp turns or corners. Mathematically, a curve is not smooth where its derivative does not exist or becomes undefined or infinite.
This typically happens when the derivative of the x-component with respect to the parameter is zero. For example, in the given problem, we start by finding where \(\frac{dx}{dt}\) equals zero. A smoothly flowing curve is essential for reliable predictions about the motion of the curve at any given point. It ensures the continuity of the curve is logical and without sudden breaks. When \(t = 0\) in the problem, the curve is not smooth because \(\frac{dx}{dt}\) becomes zero, leading to an undefined derivative \(\frac{dy}{dx}\).
If the curve was not smooth at any point, it reflects a change in direction that isn't calculable, leading to potential errors in predicting behavior.

Derivatives in Parametric Form

In parametric equations, derivatives are computed slightly differently than in regular Cartesian equations. When you have an equation presented in parametric form, you need to calculate the derivatives of the parametric components separately.
For instance, given \(x = \frac{1}{t^2 + 1}\) and \(y = t^3\), you would separately find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). These derivatives represent how each variable x and y change concerning the parameter t. The importance of these derivatives comes into play when needing to calculate \(\frac{dy}{dx}\), the standard derivative showing the rate of change of y with respect to x.
It simplifies the process of finding \(dy/dx\) to \(\frac{dy/dt}{dx/dt}\), which is invaluable for determining the slope of the curve at a given point, helping understand how a variable y is changing directly as x changes.

Limits in Calculus

Limits are a fundamental concept in calculus, used to explore the behavior of functions as they approach specific values. In this problem, the task is to find the limit of \(\frac{dy}{dx}\) as the parameter \(t\) approaches a specific value.
By examining how the function behaves as \(t\) gets closer to zero, we can determine the value \(\frac{dy}{dx}\) approaches without necessarily reaching it. Here, limits are essential because they allow for the analysis of derivatives at points where regular inspection might lead to indeterminate forms, such as zero over zero or infinity. Calculating this limit involves simplifying the expression to a form that can be evaluated at the problematic point.
In the given exercise, simplifying \(\frac{-3t(t^2+1)^2}{2}\) enabled finding that as \(t\) approaches zero, the limit is zero, indicating a horizontal tangent at that point, contributing to understanding the behavior of the graph.

Parametric Differentiation

Parametric differentiation is the process of finding derivatives when a curve is expressed via parametric equations, using a third variable—often time—as a parameter.
The primary differentiating factor for parametric equations is using implicit differentiation through the parameter to find the slope of the curve. From the original example, obtaining \(\frac{dy}{dx}\) implies computing \(\frac{dy/dt}{dx/dt}\). This practice becomes especially useful in situations like physics or engineering, where an object's position changes in respect to time, making it easier to determine instantaneous velocity or acceleration.
Finding parametric derivatives is crucial to understanding the intricacies and behavior of a curve at a given point, as it delivers insights into how changes in one variable instigate changes in another, further revealing the dynamics of the motion or system represented by the curve.