Question

Find the area enclosed by the loop of the curve in Exercise \(27.\)

Short answer

The area enclosed by the loop of the curve is \(\frac{{81}}{{20}}\) square units.

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.

Draw graph\({\rm{y = }}{{\rm{t}}^{\rm{2}}}{\rm{ + t + 1}}\):

First plot the graph of\(y = {t^2} + t + 1\).

Then, from the graph,

\(\begin{aligned}{c}\frac{{s + t}}{2} = - \frac{1}{2}\\ \Rightarrow \;\;\;s = - t - 1\\ \Rightarrow \;\;\;y(t) = y( - t - 1)\,\end{aligned}\)

Where,\(\;t > - \frac{1}{2}\).

Find the values of parameters

Here we need to find 2 values of parameter such that:

\(\begin{aligned}{l}y(t) = y(s)\\x(t) = x(s)\end{aligned}\)

So, we have \(s = - t - 1\) where\(t > - \frac{1}{2}\).

\(\begin{aligned}{c}x(t) = x(s)\\x(t) = x( - t - 1)\\{t^3} - 3t = {( - t - 1)^3} - 3( - t - - 1)\\{t^3} - 3t = - {t^3} - 3{t^2} - 3t - 1 + 3t + 3\\2{t^3} + 3{t^2} - 3t - 2 = 0\\2{t^2}(t + 2) - t(t + 2) - (t + 2) = 0\\(t + 2)\left( {2{t^2} - t - 1} \right) = 0\\(t + 2)(2t(t - 1) + t - 1) = 0\\(t + 2)(2t + 1)(t - 1) = 0\end{aligned}\)

\(\begin{aligned}{c} \Rightarrow t\left. { = - 2\;{\rm{(rejected, because }} - 2 \not > - \frac{1}{2}} \right)\\t = - \frac{1}{2}\;\;\;\left( {{\rm{ rejected, because }} - \frac{1}{2} \not > - \frac{1}{2}} \right)\\ \Rightarrow t = 1\end{aligned}\)

By using the obtained value of t, find the value of s.

\(\begin{aligned}{c}s = - t - 1\\ = - 1 - 1\\ = - 2\end{aligned}\)

Therefore, calculated value is \( - 2\).

Find the required area

Below is graph of the given curve \((x,y) = \left( {{t^3} - 3t,{t^2} + t + 1} \right)\):

Let us simplify the area \(A = \int_{t = - 2}^{t = 1} y \;{\rm{d}}x\), where \({\rm{d}}x = \frac{d}{{dt}}\left( {{t^3} - 3t} \right) = 3{t^2} - 3\).

\(\begin{aligned}{c}A = \int_{ - 2}^1 {\left( {{t^2} + t + 1} \right)} \left( {3{t^2} - 3} \right){\rm{d}}x\\ = 3\int_{ - 2}^1 {\left( {{t^2} + t + 1} \right)} \left( {{t^2} - 1} \right){\rm{d}}x\\ = 3\int_{ - 2}^1 {\left( {{t^4} + {t^3} - t - 1} \right)} {\rm{d}}x\\ = 3\left( {\frac{1}{5}{t^5} + \frac{1}{4}{t^3} - \frac{1}{2}{t^2} - t} \right)_{ - 2}^1\\ = 3\left( {\frac{1}{5} + \frac{1}{4} - \frac{1}{2} - 1 + \frac{{32}}{5} - 4 + 2 - 2} \right)\\ = \frac{{81}}{{20}}\end{aligned}\)

Therefore, the calculated value is \(\frac{{81}}{{20}}\) square units.