A one lane highway runs through a tunnel in the shape of one half a sine curve cycle

The sine curve equation, y = 10·sin(x·π/24), that models the entrance of the

tunnel with a cross section that is the shape of half of a sine curve and the

height of the tunnel at the edge of the road, (approximately 7.07 ft.) are

found by applying the following steps

(a) The equation for the sine curve is y = 10·sin(x·π/24)

(b) The height of the tunnel at the edge of the road is approximately 7.07 feet

The reason for the above answers are presented as follows;

(a) From a similar question posted online, the missing part of the question

is, what is the height of the tunnel at the edge of the road

The known parameters;

The shape of the tunnel = One-half sine curve cycle

The height of the road at its highest point = 10 ft.

The opening of the tunnel at road level = 24 ft.

The unknown parameter;

The equation of the sine curve that fits the opening

Method;

Model the sine curve equation of the tunnel using the general equation of a sine curve;

The general equation of a sine curve is y = A·sin(B·(x - C) + D

Where;

y = The height at point x

A = The amplitude = The distance from the centerline of the sine wave to the top of a crest

Therefore;

The amplitude, A = The height of half the sine wave = The height of the tunnel = 10 ft.

D = 0, C = 0 (The origin, (0, 0) is on the left end, which is the central line)

The period is the distance between successive points where the curve passes through the center line while rising to a crest

Therefore

The period, T = 2·π/B = 2 × Opening at the road level = 2 × 24 ft. = 48 ft.

T = 48 ft.

We get;

48 = 2·π/B

B = 2·π/48 = π/24

By plugging in the values for A, B, C, and D, we get;

y = 10·sin((π/24)·(x - 0) + 0 = 10·sin(x·π/24)

The equation of the sine curve that fits the opening is y = 10·sin(x·π/24)

(b) The height of the tunnel at the edge of the road is given by substituting

the value of x at the edge of the road into the equation for the sine curve

as  follows;

The width of the shoulders = 6 feet

∴ At the edge of the road, x = 0 + 6ft = 6 ft., and 6 ft. + 12 ft. = 18 ft.

Therefore, we get;

y = 10 × sin(6·π/24) = 10 × sin(π/4) = 5×√2

y = 10 × sin(18·π/24) = 10 × sin(3·π/4) = 5×√2

The height of the, y, tunnel at the edge of the road where, x = 6, and 18 is y = 5·√2 feet ≈ 7.07 ft.

Learn more about the sine curve here;

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.