Answer:
15450.4 pounds
Step-by-step explanation:
Given the distance between the poles is D = 3737 feet, and
f(x) = 10 + 0.1x^3/2
= 10 + 1/10 x^3/2
We need to find the arc length of the cable and this is given by:
L = integral of √(1 + [f'(x)]²) dx, between D=0 and D/2
Where f'(x) = d(f(x))/DX
f'(x) = 3/2 × 1/10x^1/2 = 3/20 √x
Hence, L = integral of (√(1 + [3/20√x]²))dx, between D=0 and D/2
L = integral(√(1 + (9/400)x))dx; D=0 and D/2
L = integral(√(1 +9x/400))dx; D=0 and D/2
Let u= 1+ (9x/400) ; du/dx = 9/400
dx = (400/9)du
L= integral (√u) × +(400/9)du; btw D=0 and D/2
L = (400/9) × integral (√u)du
L= (400/9) × (u^3/2)÷(3/2)
L = (800/27)[u^3/2]
Now we replace 1 + (9x/400)
and evaluate between D=0 and D/2 = 1868.5 ft
L= (800/27)[(1 + (9× 1868.5/400)-(1 + 0)]
L=( 800/27) × (9× 1868.5/400)
L = (800 27) × 42.04
L=1246 feet (approximately)
The weight = 12.4 × 1246
Weight = 15450.4 pounds
