Question

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

Short answer

The graph is not smooth at \( t_0 = 1 \) and the limit is vertical, \( \lim _{t \rightarrow 1} \frac{dy}{dx} = \infty \).

Step-by-step solution

Compute Derivatives

To check where the graph is not smooth, we compute the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Calculate the derivative of \( x(t) = t^3 - 3t^2 + 3t - 1 \):\[ \frac{dx}{dt} = 3t^2 - 6t + 3 \]Calculate the derivative of \( y(t) = t^2 - 2t + 1 \):\[ \frac{dy}{dt} = 2t - 2 \].

Identify Condition for Non-smoothness

A parametric graph is not smooth where both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are zero simultaneously, as this indicates a cusp or critical point.

Solve for Smoothness Condition

Set both derivatives to zero and solve the system of equations:1. \( 3t^2 - 6t + 3 = 0 \) 2. \( 2t - 2 = 0 \) From the second equation, solve for \(t\):\[ 2t - 2 = 0 \Rightarrow t = 1 \].

Verify the Condition

Check if this \( t = 1 \) is a solution for both equations. Substitute \( t = 1 \) into the first equation:\[ 3(1)^2 - 6(1) + 3 = 0 \Rightarrow 0 = 0 \]Both conditions are satisfied, thus \( t_0 = 1 \) is where the graph is not smooth.

Compute the Limit of Derivative

Finally, calculate the limit \( \lim _{t \rightarrow 1} \frac{dy}{dx} \), which requires \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \).Since both derivatives are zero at \( t = 1 \), apply L'Hôpital's Rule:Differentiate \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) again with respect to \( t \):\( \frac{d^2y}{dt^2} = 2 \)\( \frac{d^2x}{dt^2} = 6t - 6 \).Now apply L'Hôpital's Rule:\( \lim _{t \rightarrow 1} \frac{2}{6t - 6} \rightarrow \frac{2}{0} \).This limit demonstrates a vertical tangent, suggesting the smooth transition in the context of size approaching infinity or precision configurability.

Key concepts

Derivatives

In calculus, derivatives are fundamental tools used to study how a function behaves as its input changes. When working with parametric equations—expressions where different variables are defined in terms of a common parameter—they help us analyze the slope and shape of a graph.

For the given problem, we determine the derivatives of the functions:

• For the function \(x(t) = t^3 - 3t^2 + 3t - 1\), the derivative is \(\frac{dx}{dt} = 3t^2 - 6t + 3\).
• For \(y(t) = t^2 - 2t + 1\), the derivative is \(\frac{dy}{dt} = 2t - 2\).

These derivatives describe how the coordinates x and y change concerning t.

By finding these derivatives, we gain insights about the rate of change along each axis and prepare to explore more about the nature of the curve they describe.

Smoothness of Graphs

A graph is considered smooth if it doesn't have any sharp corners or cusps. In the context of parametric equations, smoothness is typically checked by seeing whether both derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are zero at the same point.

In our exercise, we found:

• \(3t^2 - 6t + 3 = 0\)
• \(2t - 2 = 0\) which simplifies to \(t = 1\).

When both these derivatives are zero simultaneously, it indicates a potential point of non-smoothness like a cusp.

If t = 1 coincides for both conditions, the graph has a singular point where it's not smooth, which means it could either be a corner or a cusp on the curve.

L'Hôpital's Rule

L'Hôpital's Rule is a key technique in calculus used to find limits of indeterminate forms, where both the numerator and the denominator evaluate to zero or infinity. In our scenario, we reached a situation where the limit of \(\frac{dy}{dx}\) becomes \(\frac{0}{0}\) at t = 1.

To resolve this indeterminacy:

• Differentiate the derivatives: \(\frac{d^2y}{dt^2} = 2\) and \(\frac{d^2x}{dt^2} = 6t - 6\).
• Recalculate the limit with these second derivatives: \(\lim _{t \rightarrow 1} \frac{2}{6t - 6}\).

This application of L'Hôpital's Rule allows us to further analyze the behavior of the parametric curve, especially at points that might not be entirely straightforward.

Vertical Tangent

A vertical tangent line occurs when the slope of the tangent approaches infinity, indicating a sharp ascent or descent with respect to the axis. In parametric terms, this arises where the denominator of the slope \(\frac{dy}{dx}\), specifically \(\frac{dx}{dt}\), becomes zero while \(\frac{dy}{dt}\) is non-zero.

However, in our task, both are zero, leading us to apply L'Hôpital's Rule. In the end, we examined:

• \(\lim _{t \rightarrow 1} \frac{2}{6t - 6}\)

This solution simplifies, suggesting just such a vertical tangent where the graph's ascent or descent is undefined or infinite in typical coordinates. These tangents are significant as they often mark critical points on the graph indicative of sudden changes in direction.