Question

Find the points on the curve where the tangent is horizontal or vertical. if you have a graphing device, graph the curve to check your work. \(x = {t^3} - 3t,y = {t^3} - 3{t^2}\)

Short answer

The horizontal and vertical tangent line are\((0,0),(2, - 4)\) and \(( - 2, - 2),(2, - 4)\)respectively.

Step-by-step solution

Find \(\frac{{dy}}{{dt}},\frac{{dx}}{{dt}}\)

For this question, we can compute from the function of x:

\(\begin{array}{l}x = {t^3} - 3t\\\frac{{dx}}{{dt}} = 3{t^2} - 3\end{array}\)

For this question, we can compute from the function of y:

\(\begin{array}{l}y = {t^3} - 3{t^2}\\\frac{{dy}}{{dt}} = 3{t^2} - 6t\end{array}\)

Find the point for horizontal tangent line

Substitute \(\frac{{dy}}{{dt}} = 0\) to find the horizontal line

\(\begin{array}{l}\frac{{dy}}{{dt}} = 0\\3{t^2} - 6t = 0\\3t(t - 2) = 0\\t = 0\,\,\,\,\,\,or\,\,\,\,\,\,t = 2\end{array}\)

Substitute \(t = 0\) and obtained point

\(\begin{array}{l}x = {t^3} - 3t\\x = 0 - 0\\x = 0\end{array}\)

\(\begin{array}{l}y = {t^3} - 3{t^2}\\y = 0 - 0\\y = 0\end{array}\)

Substitute \(t = 2\) and obtained point

\(\begin{array}{l}x = {t^3} - 3t\\x = 8 - 6\\x = 2\end{array}\)

\(\begin{array}{l}y = {t^3} - 3{t^2}\\y = 8 - 12\\y = - 4\end{array}\)

The curve has horizontal tangent at point \((0,0),(2, - 4)\)

Find the point for vertical tangent line

Substitute \(\frac{{dx}}{{dt}} = 0\) to find the vertical line

\(\begin{array}{l}\frac{{dx}}{{dt}} = 0\\3{t^2} - 3 = 0\\3(t + 1)(t - 1) = 0\\t = 1\,\,\,\,\,\,\,or\,\,\,\,\,\,t = - 1\end{array}\)

Substitute \(t = 1\) and obtained point

\(\begin{array}{l}x = {t^3} - 3t\\x = 1 - 3\\x = - 2\end{array}\)

\(\begin{array}{l}y = {t^3} - 3{t^2}\\y = 1 - 3\\y = - 2\end{array}\)

Substitute \(t = - 1\) and obtained point

\(\begin{array}{l}x = {t^3} - 3t\\x = - 1 + 3\\x = 2\end{array}\)

\(\begin{array}{l}y = {t^3} - 3{t^2}\\y = - 1 - 3\\y = - 4\end{array}\)

The curve has vertical tangent at point \(( - 2, - 2),(2, - 4)\)

Find the graph

The graph of the given curve can be given below:

Hence, the curve has vertical tangent at point \((0,0),(2, - 4)\) and horizontal tangent at point \(( - 2, - 2),(2, - 4)\) as shown in the following graph: