A cable hangs between two poles of equal height and 3737 feet apart.
At a point on the ground directly under the cable and xx feet from the point on the ground halfway between the poles the height of the cable in feet is h(x)=10+(0.1)(x1.5).The cable weighs 12.4 pounds per linear foot.

Find the weight of the cable.

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

x= 149. 24+7=31. 180-31=149

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