Chapter 10.6, Problem 61E

The Arch of a Bridge The opening of a railway bridge over a roadway is in the shape of a parabola. A surveyor measures the heights of three points on the bridge, as shown in the figure. He wishes to find an equation of the form

$y=a{x}^2+bx+c$

to model the shape of the arch.

1. (a) Use the surveyed points to set up a system of linear equations for the unknown coefficients a, b, and c.
2. (b) Solve the system using Cramer’s Rule.

To determine

The system for linear equations from given measures.

Answer to Problem 61E

The linear equations for the system are $\{\begin{array}{c}100a+10b+c=25\\ 225a+15b+c=33.75\\ 1600a+40b+c=40\end{array}$.

Explanation of Solution

Given:

The shape of the railway bridge over a roadway is in shape of parabola,

$y=a{x}^2+bx+c$ (1)

The coordinates of railway bridge from the Figure (1) are,

$\left(10,25\right),\left(15,33.75\right)\text{and}\left(40,40\right)$.

Compare the above coordinates with $\left(x_1,y_1\right),\left(x_2,y_2\right),\left(x_3,y_3\right)$

Substitute 10 as $x_1$ and 25 as $y_1$ in equation (1).

$\begin{array}{l}25=a{\left(10\right)}^2+b\left(10\right)+c\\ 25=100a+10b+c\end{array}$

The first equation is,

$100a+10b+c=25$ (2)

Substitute 15 for $x_2$ and $33.75$ as $y_2$ in equation (1).

$\begin{array}{l}33.75=a{\left(15\right)}^2+b\left(15\right)+c\\ 33.75=225a+15b+c\end{array}$

Thus, the second equation is

$225a+15b+c=33.75$ (3)

Substitute 40 for $x_3$ and $40$ as $y_3$ in equation (1).

$\begin{array}{l}40=a{\left(40\right)}^2+b\left(40\right)+c\\ 40=1600a+40b+c\end{array}$

Thus, the third equation is

$1600a+40b+c=40$ (4)

Thus the linear equations for the system are $\{\begin{array}{c}100a+10b+c=25\\ 225a+15b+c=33.75\\ 1600a+40b+c=40\end{array}$.

To determine

To find: The solution of the linear system using Cramer’s rule.

Answer to Problem 61E

The value of x is $0.45$, y is $3$ and z is 0.

Explanation of Solution

Given:

From part (a) the linear system is $\{\begin{array}{c}100a+10b+c=25\\ 225a+15b+c=33.75\\ 1600a+40b+c=40\end{array}$.

Evaluate the left hand side of equation for determinant.

$\left|D\right|=\left|\begin{array}{ccc}100& 10& 1\\ 225& 15& 1\\ 1600& 40& 1\end{array}\right|$

Apply $C_1\to C_1-100C_3$ and $C_2\to C_2-10C_3$ to above determinant.

$\begin{array}{c}\left|D\right|=\left|\begin{array}{ccc}100-100& 10-10& 1\\ 225-100& 15-10& 1\\ 1600-100& 40-10& 1\end{array}\right|\\ \left|D\right|=\left|\begin{array}{ccc}0& 0& 1\\ 125& 5& 1\\ 1500& 30& 1\end{array}\right|\end{array}$

Expand along $R_1$ the above determinant.

$\begin{array}{l}\left|D\right|=1\left|\begin{array}{cc}125& 5\\ 1500& 30\end{array}\right|\\ \left|D\right|=\left(125\times 30\right)-\left(5\times 1500\right)\\ \left|D\right|=-3750\end{array}$

Now replace the terms of first column of $\left|D\right|$ with terms of right side of equations for $\left|D_a\right|$,

$\left|D_a\right|=\left|\begin{array}{ccc}25& 10& 1\\ 33.75& 15& 1\\ 40& 40& 1\end{array}\right|$

Apply $C_1\to C_1-25C_3$ and $C_2\to C_2-10C_3$ in the above determinant.

$\begin{array}{c}\left|D_a\right|=\left|\begin{array}{ccc}25-25& 10-10& 1\\ 33.75-25& 15-10& 1\\ 40-25& 40-10& 1\end{array}\right|\\ \left|D_a\right|=\left|\begin{array}{ccc}0& 0& 1\\ 8.75& 5& 1\\ 15& 30& 1\end{array}\right|\end{array}$

Expand along $R_1$ the above determinant,

$\begin{array}{c}\left|D_a\right|=\left|\begin{array}{cc}8.75& 5\\ 15& 30\end{array}\right|\\ =\left(8.75\times 30\right)-\left(15\times 5\right)\\ =187.5\end{array}$

Similarly,

$\begin{array}{c}\left|D_b\right|=\left|\begin{array}{ccc}100& 25& 1\\ 225& 33.75& 1\\ 1600& 40& 1\end{array}\right|\\ =-11250\end{array}$

Similarly,

$\begin{array}{c}\left|D_c\right|=\left|\begin{array}{ccc}100& 10& 25\\ 225& 15& 33.75\\ 1600& 40& 40\end{array}\right|\\ =0\end{array}$

Now, from Cramer’s rule for solution,

$\begin{array}{c}a=\frac{\left|D_a\right|}{\left|D\right|}\\ =\frac{\left|187\right|}{\left|3750\right|}\left[\because D_a=187,D=3750\right]\\ =0.45\end{array}$

Similarly,

$\begin{array}{c}b=\frac{\left|D_b\right|}{\left|D\right|}\\ =\frac{\left|-11250\right|}{\left|3750\right|}\left[D_b=-11250,D=3750\right]\\ =3\end{array}$

Similarly,

$\begin{array}{c}c=\frac{\left|D_c\right|}{\left|D\right|}\\ =\frac{0}{3750}\left[\because D_c=0,D=3750\right]\\ =0\end{array}$

Thus, the value of a is $0.45$, b is $3$ and c is 0.