Question

A parabolic tunnel has a width of 20 feet and a height of 12 feet at the center. Find the cross-sectional area of the parabolic tunnel.

A. Illustrate the parabolic tunnel. Show the measurements.

B. What function represents the parabolic tunnel? Show your solution.

C. Compute for the cross-sectional area of the parabolic tunnel. Use the Definite Integral. Show your solution.

No troll.

Answer 1

A. See attached graph

B.
`\boxed{y = 0.12x^2+ 12}`

C.
`\boxed{160\;square\;feet}`

Explanation:

Consider the figure which shows the cross-sectional view of the parabolic tunnel as viewed from the entrance

A. See attached graph

B. The parabola is drawn with the y-axis as the axis of symmetry.

The x-intercepts are at (-10, 0) and (10, 0)

The vertex is at (0, 12)

The vertex form equation for a parabola is
y = a(x - h)² + k

where (h, k) is the vertex - the (x, y) coordinates of the vertex

Here h = 0, k = 12

So the parabola equation is:

y = a(x - 0)² + 12

y = ax² + 12

To find a, plug in one of the x-intercepts, say (10, 0)

We get

0 = a(10)² + 12

0 = 100a + 12

100a = -12

a = -12/100 = -0.12

Therefore the equation of the parabola is

`\boxed{y = 0.12x^2+ 12}`

C. The cross-sectional area is given by the formula:

D. To find the area of the parabola between x = -10 and x = 10 which forms the base of the parabola, you can use the formula

`A = (2)/(3)bh\\\\`

where b is the base length and h is the height

Using the values for b = 20 and h = 12 we get


`A = (2)/(3)\cdot 20 \cdot 12 = 160 \textrm { square feet}`

Using the definite integral, let's find the area between x = -10 and x = 10

I am not going through every step but leading you to the final answer

`\int _(-10)^(10)\left(\:-0.12x^(2\:)+\:12\right)dx\\\\\\\textrm{We have}\\\\0.12\cdot \int _(-10)^(10)x^2dx=0.12\left[(x^3)/(3)\right]_(-10)^(10)\\\\`

`=0.04\left[x^3\right]_(-10)^(10)\\\\= 80\\\\`

Therefore

`-0.12\cdot \int _(-10)^(10)x^2dx = -80`

`\int _(-10)^(10)12dx = \left[12x\right]_(-10)^(10) = 240`

So total area = -80 + 240 = 160 square feet

Ans: Cross sectional area =
`\boxed{160\;square\;feet}`