Mathematical curves and Hooke’s chain theory
1 √1+𝑝 2
∫
𝑑𝑝=∫𝑘𝑑𝑥 (9)
Solve (9):
ln(|√1+𝑝 2 +𝑝|) = 𝑎𝑥 +𝑐,𝑟𝑒𝑚𝑜𝑣𝑒 𝑡ℎ𝑒 𝑚𝑜𝑑 𝑠𝑖𝑔𝑛 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 √1+𝑝 2 > |𝑝|
ln(√1+𝑝 2 + 𝑝) = 𝑎𝑥 + 𝑐
When 𝑥 =0 , 𝑝 =0 (because the curve is flat at x = 0), 𝑙𝑛1= 𝑐 , 𝑐 =0
ln(√1+𝑝 2 +𝑝) = 𝑎𝑥 (10)
Now eliminate logarithm:
√1+𝑝 2 +𝑝 = 𝑒 𝑎𝑥 (11)
Attempt to eliminate roots:
(√1+𝑝 2 + 𝑝) (√1 + 𝑝 2 −𝑝)= 𝑒 𝑎𝑥 (√1+𝑝 2 −𝑝),𝑡ℎ𝑒𝑛 𝑒𝑥𝑝𝑎𝑛𝑑 𝑏𝑟𝑎𝑐𝑘𝑒𝑡𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒:
1= 𝑒 𝑎𝑥 √1+𝑝 2 −𝑒 𝑎𝑥 𝑝,𝑟𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑒 1+𝑒 𝑎𝑥 𝑝 = 𝑒 𝑎𝑥 √1+𝑝 2 ,𝑠𝑞𝑢𝑎𝑟𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠: 1+𝑒 2𝑎𝑥 𝑝 2 +2𝑒 𝑎𝑥 𝑝 = 𝑒 2𝑎𝑥 +𝑝 2 𝑒 2𝑎𝑥 ,𝑟𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑒 𝑎𝑔𝑎𝑖𝑛,1+2𝑒 𝑎𝑥 𝑝 = 𝑒 2𝑎𝑥 (12)
Rearrange for p:
𝑒 2𝑎𝑥 −1 2𝑒 𝑎𝑥
𝑒 𝑎𝑥 −𝑒 −𝑎𝑥 2
𝑝 =
=
𝑑𝑦 𝑑𝑥
Substitute in p as
and integrate:
𝑒 𝑎𝑥 −𝑒 −𝑎𝑥 2
𝑒 𝑎𝑥 −𝑒 −𝑎𝑥 2𝑎 𝑒 𝑎𝑥 −𝑒 −𝑎𝑥 2𝑎
𝑑𝑦 𝑑𝑥
=
, 𝑦 =
+𝑑,𝑎𝑡 𝑥 = 0,𝑦 = 0,𝑠𝑜 𝑑 = 0
cosh𝑎𝑥 𝑎
𝑦 =
=
3
This can be used to model all strings or chains that hang under their own weight with their two ends fixed horizontally.
3 See Differential Equations with Applications and Historical Notes 3 rd edition 88-94.
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