To find where the tangent line to the curve is horizontal or vertical, we first need to compute the derivatives of $x(t)$ and $y(t)$ with respect to $t$, and then use these derivatives to find the slope of the tangent line.

Step 1: Find the derivatives

Given: $x(t) = t^2 - 5$ $y(t) = t^3 - 2t$

We need $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

1. Calculate $\frac{dx}{dt}$: $\frac{dx}{dt} = 2t$

2. Calculate $\frac{dy}{dt}$: $\frac{dy}{dt} = 3t^2 - 2$

Step 2: Determine when the tangent line is horizontal

A tangent line is horizontal when the slope of the curve, given by $\frac{dy}{dx}$, is zero. This occurs when $\frac{dy}{dt} = 0$ (provided $\frac{dx}{dt} \neq 0$).

Set $\frac{dy}{dt} = 0$: $3t^2 - 2 = 0$ $3t^2 = 2$ $t^2 = \frac{2}{3}$ $t = \pm \sqrt{\frac{2}{3}}$

Thus, the values of $t$ where the tangent line is horizontal are: $t = \pm \sqrt{\frac{2}{3}}$

Step 3: Determine when the tangent line is vertical

A tangent line is vertical when $\frac{dx}{dt} = 0$ (provided $\frac{dy}{dt} \neq 0$).

Set $\frac{dx}{dt} = 0$: $2t = 0$ $t = 0$

Thus, the value of $t$ where the tangent line is vertical is: $t = 0$

Summary

• The tangent line to the curve is horizontal at $t = \pm \sqrt{\frac{2}{3}}$.
• The tangent line to the curve is vertical at $t = 0$.

Would you like a more detailed explanation or have any questions?

Here are five related questions for further exploration:

1. How can we find the equation of the tangent line at a specific value of $t$?
2. What are the coordinates of the points where the tangent is horizontal or vertical?
3. How can we determine the concavity of the curve at these points?
4. What would change if the parametric equations were different (e.g., $x(t) = t^3 - 5$, $y(t) = t^2 - 2$)?
5. How can we find the arc length of the curve over a given interval?

Tip: To determine whether a parametric curve intersects itself, solve the system of equations $x(t_1) = x(t_2)$ and $y(t_1) = y(t_2)$ for different parameter values $t_1$ and $t_2$.