Chapter 7.6, Problem 64E

(a)

To determine

To Find: To find the equation of parabola from given information.

Answer to Problem 64E

The equation is $y=-0.294t^2+3.765t-0.659$

Explanation of Solution

Given information:Equation of parabola $y=a{t}^2+bt+c$

Formula used: Matrix Algebra.

Since t represent the year with $t=4$ corresponding to 2014 and y represent the travellers.

Therefore,

$t$	$y$
4	9.697
5	10.816
6	11.347

Also, the given equation of parabola is $y=a{t}^2+bt+c$

Put $t=4\text{and}y=9.697$ then

  $\begin{array}{l}$ 9.697=a{(4)}^2+b(4)+c$16a+4b+c=9.697\mathrm{...}(1)\end{array}$

Put $t=5\text{and}y=10.816$

  $\begin{array}{l}$ 10.816=a{(5)}^2+b(5)+c$25a+5b+c=10.816\mathrm{...}(2)\end{array}$

Put $t=6\text{and}y=11.347$

  $\begin{array}{l}$ 11.347=a{(6)}^2+b(6)+c$36a+6b+c=11.347\mathrm{...}(3)\end{array}$

Now, the given systems of equations are

  $\begin{array}{l}16a+4b+c=9.697\\ 25a+5b+c=10.816\\ 36a+6b+c=11.347\end{array}$

Let

  $A=\left[\begin{array}{ccc}16& 4& 1\\ 25& 5& 1\\ 36& 6& 1\end{array}\right],X=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]\text{and}B=\left[\begin{array}{c}9.697\\ 10.816\\ 11.347\end{array}\right]$

Such that,

  $\begin{array}{l}AX=B\\ \text{multiplying}\text{by}{A}^{-1}\text{on}\text{both}\text{sides}\\ A^{-1}(AX)=A^{-1}B\\ X=A^{-1}B\end{array}$

Where

  $A^{-1}=\frac{1}{\det (A)}{\left[A_{ij}\right]}^T$

Calculation of determinant and cofactors of matrix A

  $\begin{array}{l}\det (A)=\left|\begin{array}{ccc}16& 4& 1\\ 25& 5& 1\\ 36& 6& 1\end{array}\right|\\ \text{Expanding}\text{along}{\text{1}}^{\text{st}}\text{row}\\ \det (A)=16(5-6)-4(25-36)+1(150-180)\\ \det (A)=-16+44-30=-2\ne 0\end{array}$

Thus, $A^{-1}$ exists, now the cofactors are

  $\begin{array}{l}{A}_{11}=5-6=-1\\ A_{12}=-(25-36)=11\\ A_{13}=150-180=-30\\ A_{21}=-(4-6)=2\\ A_{22}=16-36=-20\\ A_{23}=-(96-144)=48\\ A_{31}=4-5=-1\\ A_{32}=-(16-25)=9\\ A_{33}=80-100=-20\end{array}$

Now,

  $\left[\begin{array}{l}a\\ b\\ c\end{array}\right]={\left[\begin{array}{ccc}16& 4& 1\\ 25& 5& 1\\ 36& 6& 1\end{array}\right]}^{-1}\left[\begin{array}{l}9.697\\ 10.816\\ 11.347\end{array}\right]$

  $\left[\begin{array}{l}a\\ b\\ c\end{array}\right]=\frac{1}{-2}{\left[\begin{array}{ccc}-1& 11& -30\\ 2& -20& 48\\ -1& 9& -20\end{array}\right]}^T\left[\begin{array}{l}9.697\\ 10.816\\ 11.347\end{array}\right]$

  $\left[\begin{array}{l}a\\ b\\ c\end{array}\right]=\frac{1}{-2}\left[\begin{array}{ccc}-1& 2& -1\\ 11& -20& 9\\ -30& 48& -20\end{array}\right]\left[\begin{array}{c}9.697\\ \begin{array}{l}10.816\\ 11.347\end{array}\end{array}\right]$

  $\left[\begin{array}{l}a\\ b\\ c\end{array}\right]=\frac{1}{-2}\left[\begin{array}{c}(-1)\times 9.697+2\times 10.816+(-1)\times 11.347\\ \begin{array}{l}11\times 9.697+(-20)\times 10.816+9\times 11.347\\ (-30)\times 9.697+48\times 10.816+(-20)\times 11.347\end{array}\end{array}\right]$

  $\left[\begin{array}{l}a\\ b\\ c\end{array}\right]=\frac{1}{-2}\left[\begin{array}{l}-9.697+21.632-11.347\\ 106.667-216.32+102.123\\ -290.91+519.16+8-226.94\end{array}\right]$

  $\left[\begin{array}{l}a\\ b\\ c\end{array}\right]=\left[\begin{array}{l}\raisebox{1ex}{$0.588$}\!\left/ \!\raisebox{-1ex}{$-2$}\right.\\ \raisebox{1ex}{$-7.53$}\!\left/ \!\raisebox{-1ex}{$-2$}\right.\\ \raisebox{1ex}{$1.318$}\!\left/ \!\raisebox{-1ex}{$-2$}\right.\end{array}\right]$

  $\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}-0.294\\ 3.765\\ -0.659\end{array}\right]$

Thus, the values are $a=-0.294,b=3.765\text{and}c=-0.659$

Hence the required parabolic equation is $y=-0.294t^2+3.765t-0.659$

(b)

To determine

To Graph: To graph the given equation.

Explanation of Solution

Given information:Given

Year	Travellers (in thousands) $y$
2014	9.697
2015	10.816
2016	11.347

Graph:

  

Interpretation:

From the graph, the x-axis shows the year x and y-axis shows yearly travellers in thousands. Also, it can be seen that given quadratic equation is a good fit on the graph.

(c)

To determine

To Find: To find the number of travellers from 2017 to 2020

Answer to Problem 64E

The answer is

Year	International travellers in thousands
2017	11.29
2018	10.645
2019	9.412
2020	7.621

Explanation of Solution

Given information:Given $y=-0.294t^2+3.765t-0.659$

Calculation:

Since given equation is

  $y=-0.294t^2+3.765t-0.659$

Put $t=7$

  $\begin{array}{l}$ y=-0.294{(7)}^2+3.765(7)-0.659$y=-0.294\times 49+26.355-0.659\\ y=-14.406+25.696\\ y=11.29\end{array}$

In 2017, the number of travellers will be 11.29 thousand

Similarly, put $t=8$

  $\begin{array}{l}$ y=-0.294{(8)}^2+3.765(7)-0.659$y=-0.294\times 64+30.12-0.659\\ y=-18.816+29.461\\ y=10.645\end{array}$

In 2018, the number of travellers will be 10.645 thousand

Put $t=9$

  $\begin{array}{l}$ y=-0.294{(9)}^2+3.765(7)-0.659$y=-0.294\times 81+33.885-0.659\\ y=-23.814+33.226\\ y=9.412\end{array}$

In 2019, the number of travellers will be 9.412 thousand

Put $t=10$

  $\begin{array}{l}$ y=-0.294{(10)}^2+3.765(7)-0.659$y=-0.294\times 100+37.65-0.629\\ y=-29.4+37.021\\ y=7.621\end{array}$

In 2020, the number of travellers will be 7.621 thousand

(d)

To determine

To Explain:The above results are reasonable or not.

Explanation of Solution

Given information:Given equation $y=-0.294t^2+3.765t-0.659$ with graph and table

By comparing the graph and the table, it can be clearly seen that the number of international travellers declined from 2017.

Thus, the given answers are reasonable.