Answer:
a) s = 2a sinh(b/a)
b) 18 feet
Step-by-step explanation:
The arc length is:
s = ∫ √(1 + (dy/dx)²) dx
y = a cosh(x/a)
y = ½ a (e^(x/a) + e^(-x/a))
dy/dx = ½ a (1/a e^(x/a) − 1/a e^(-x/a))
dy/dx = ½ (e^(x/a) − e^(-x/a))
(dy/dx)² = ¼ (e^(2x/a) − 2 + e^(-2x/a))
Plugging in:
s = ∫ √(1 + ¼ (e^(2x/a) − 2 + e^(-2x/a))) dx
s = ∫ ½ √(4 + e^(2x/a) − 2 + e^(-2x/a)) dx
s = ∫ ½ √(e^(2x/a) + 2 + e^(-2x/a)) dx
s = ∫ ½ √(e^(x/a) + e^(-x/a))² dx
s = ∫ ½ (e^(x/a) + e^(-x/a)) dx
s = ½ (a e^(x/a) − a e^(-x/a))
s = ½ a (e^(x/a) − e^(-x/a))
Evaluate from x=-b to x=b:
s = ½ a (e^(b/a) − e^(-b/a)) − ½ a (e^(-b/a) − e^(b/a))
s = ½ a e^(b/a) − ½ a e^(-b/a) − ½ a e^(-b/a) + ½ a e^(b/a)
s = a e^(b/a) − a e^(-b/a)
s = a (e^(b/a) − e^(-b/a))
s = 2a sinh(b/a)
We know that s = 48, and b = 20. We also know y = 30 when x = b.
48 = 2a sinh(20/a)
24 = a sinh(20/a)
30 = a cosh(20/a)
Using the given identity:
cosh²(20/a) − sin²(20/a) = 1
a² cosh²(20/a) − a² sin²(20/a) = a²
30² − 24² = a²
a² = 324
a = 18
Therefore, the wire will be 18 feet above the ground at the lowest point.
