Question

Use a graph to estimate the coordinates of the lowest point on the curve \(x = {t^3} - 3t,y = {t^2} + t + 1\). Then use calculus to find the exact coordinates.

Short answer

The exact coordinate for the minimum point is \(\left( {\frac{{11}}{8},\frac{3}{4}} \right)\).

Step-by-step solution

Definition of the parametric equation

A parametric equation in mathematics specifies a set of numbers as functions of one or more independent variables known as parameters.

Sketch the graph

The given curve equations are \(x = {t^3} - 3t,y = {t^2} + t + 1\).

A graph of the given parametric representation is shown below:

Note that the minimum point is approximately near to \(\left( {1.5,0.5} \right)\).

Check the graph

Let us solve the given problem.

At minimum point on the graph the value of \(y\) will be minimum.

That is the expression \(x\left( t \right) = {t^3} - 3t,y\left( t \right) = {t^2} + t + 1\) will shows the minimum point.

Find \(\frac{{dy}}{{dx}}\) by using \(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\). So differentiate \(x\left( t \right) = {t^3} - 3t,y\left( t \right) = {t^2} + t + 1\) with respect to t.

\(\begin{aligned}{l}\frac{{dx}}{{dt}} = 3{t^2} - 3{\rm{ }}......\left( 1 \right)\\\frac{{dy}}{{dt}} = 2t + 1{\rm{ }}......\left( 2 \right)\end{aligned}\)

Then, divide equation (1) by (2).

\(\begin{aligned}{c}\frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{2t + 1}}{{3{t^2} - 3}}\\\frac{{dy}}{{dx}} = \frac{{2t + 1}}{{3{t^2} - 3}}\end{aligned}\)

Put \(\frac{{dy}}{{dx}} = 0\).

\(\begin{aligned}{c}\frac{{2t + 1}}{{3{t^2} - 3}} = 0\\2t + 1 = 0\\t = - \frac{1}{2}\end{aligned}\)

Hence, minimum value will be occurred when \(t = - \frac{1}{2}\). Substitute \(t = - \frac{1}{2}\) in \(\left( {x\left( t \right) = {t^3} - 3t,{\rm{ }}y\left( t \right) = {t^2} + t + 1} \right)\).

\(\left( {{{\left( { - \frac{1}{2}} \right)}^3} - 3\left( { - \frac{1}{2}} \right),{{\left( { - \frac{1}{2}} \right)}^2} + \left( { - \frac{1}{2}} \right) + 1} \right) = \left( {\frac{{11}}{8},\frac{3}{4}} \right)\)

Therefore, the required graph is:

Hence, the coordinate of the minimum point is \(\left( {\frac{{11}}{8},\frac{3}{4}} \right)\).