Q. 14 - Precalculus Enhanced with Graphing Utilities

Question

A parabolic arch has a span of $120ft$ and a maximum height of $25ft$. Choose suitable rectangular coordinate axes and find an equation of the parabola. Then calculate the height of the arch at points $10ft, 20ft, 40ft$ from the center.

Step-by-Step Solution

Answer

The height of the arch at the point $10ft$ from the center is $24.31ft$.

The height of the arch at the point $20ft$ from the center is $22.22ft$.

The height of the arch at the point $40ft$ from the center is $18.75ft$.

Step 1. Given information.

Consider the given question,

Maximum height of the arch is $25ft$.

The vertex form of the equation of a parabola facing downwards is $f\left(x\right)=-a{\left(x-h\right)}^2+k ...... \left(i\right)$

Where $\left(h,k\right)$is the vertex of the parabola.

Draw the figure,

Step 2. Substitute 0 , 25 in equation (i), followed by simplification.

Substitute $\left(0,25\right)$ in equation (i),

$$\begin{gathered} f\left(x\right)=-a{\left(x-0\right)}^2+25 \\ f\left(x\right)=-a{x}^2+25 ...... \left(ii\right) \end{gathered}$$

Substitute $f\left(60\right)=0$ in equation (ii),

$$\begin{gathered} 0=-a{\left(60\right)}^2+25 \\ a=\frac{25}{3600} \end{gathered}$$

Substitute the value of a in equation (ii),

$f\left(x\right)=-\frac{25}{3600}{x}^2+25 ...... \left(iii\right)$

Thus, the equation of the parabolic arch is $f\left(x\right)=-\frac{25}{3600}{x}^2+25$.

Step 3. Find the height of the arch at point 10 f t ,   20 f t ,   30 f t .

Substitute $x=10$ in equation (iii),

$$\begin{gathered} f\left(10\right)=-\frac{25}{3600}{\left(10\right)}^2+25 \\ f\left(10\right)=24.31ft \end{gathered}$$

Substitute $x=20$ in equation (iii),

$$\begin{gathered} f\left(20\right)=-\frac{25}{3600}{\left(20\right)}^2+25 \\ f\left(20\right)=22.22ft \end{gathered}$$

Substitute $x=30$in equation (iii),

$$\begin{gathered} f\left(30\right)=-\frac{25}{3600}{\left(30\right)}^2+25 \\ f\left(30\right)=18.75ft \end{gathered}$$