A telephone wire hangs from two points P, Q 60m apart P, Q are on the same level . the mid point of the telephone wire is 3m below the level of P, Q. Assuming that it hangs in form of a curve , find it equation. please help. thanks for your time.


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Question Number 6169 by sanusihammed last updated on 17/Jun/16

$A t e l e p h o n e w i r e h a n g s f r o m t w o p o i n t s P, Q 60 m a p a r t$ $P, Q a r e o n t h e s a m e l e v e l . t h e m i d p o i n t o f t h e t e l e p h o n e$ $w i r e i s 3 m b e l o w t h e l e v e l o f P, Q . A s s u m i n g t h a t i t h a n g s i n$ $f o r m o f a c u r v e, f i n d i t e q u a t i o n .$ $p l e a s e h e l p . t h a n k s f o r y o u r t i m e .$
	Answered by Yozzii last updated on 17/Jun/16

	$L e t c u r v e b e q u a d r a t i c w h e r e 0 ⩽ x ⩽ 60$ $f (x) = {a x}^{2} + b x, a > 0, P (0, 0), Q (60, 0) (P & Q l i e o n x - a x i s)$ $\Rightarrow f (60) = 0 o r 3600 a + 60 b = 0$ $b = - 60 a . . . . . . . . (i)$ $A t x = 0.5 ∣ P Q ∣= 30, f^{'} (30) = 0 i f$ $w e a s s u m e a u n i f o r m w i r e l i e s b e t w e e n$ $P & Q .$ $∴ 2 a (30) + b = 0 \Rightarrow b = - 60 a (n o n e w i n f o r m a t i o n b a s e d o n q u a d r a t i c a s s u m p t i o n .)$ $∴ A t x = 30, f (30) = - 3 .$ $∴ 900 a + 30 b = - 3$ $300 a + 10 b = - 1 . . . . . . . . . . (i i)$ $F r o m (i), i n (i i) 300 a + 10 (- 60 a) = - 1$ $- 300 a = - 1 \Rightarrow a = \frac{1}{300} \Rightarrow b = \frac{- 60}{300} = \frac{- 1}{5} .$ $∴ f (x) = \frac{1}{300} x^{2} - \frac{1}{5} x (0 ⩽ x ⩽ 60)$ $- - - - - - - - - - - - - - - - - - - - - - - -$ $I n g e n e r a l, i n x - y C a r t e s i a n c o - o r d i n a t e s,$ $w e c a n l e t P (a, b) a n d Q (a + 60, b) w h e r e$ $a, b \in R .$ $T h i s m e a n s t h a t o n e c a n c h o o s e a n y a r b i t r a r y$ $p l a c e i n t h e x - y p l a n e t o p u t t h i s c u r v e$ $f o r x \in [a, a + 60] .$ $A s s u m i n g t h e w i r e i s u n i f o r m, t h e$ $e x a m p l e a b o v e i n d i c a t e s t h a t t h e$ $q u a d r a t i c c u r v e i s a s u i t a b l e m o d e l .$ $L e t f (x) = {c x}^{2} + d x (c > 0) .$ $A t x = a, f (x) = b ∴ b = {c a}^{2} + d a$ $o r d = \frac{b - {c a}^{2}}{a} (a \neq 0) . . . . . . . (i)$ $A t x = a + 60, f (a + 60) = b$ $∴ c {(a + 60)}^{2} + d (a + 60) = b$ ${c a}^{2} + c (120 a + 3600) + d a + 60 d = b$ $120 c (a + 30) + 60 d = 0$ $2 c (a + 30) + d = 0$ $2 c a + 60 c + \frac{b}{a} - c a = 0 (f r o m (i))$ $c a + 60 c = \frac{- b}{a}$ $c = \frac{- b}{a (a + 60)}$ $\Rightarrow d = \frac{b}{a} - a \times \frac{- b}{a (a + 60)}$ $d = \frac{b}{a} (1 + \frac{a}{a + 60}) = \frac{b (a + 60 + a)}{a (a + 60)}$ $d = \frac{2 b (a + 30)}{a (a + 60)}$ $S i n c e c > 0 \Rightarrow - 60 < a < 0 & b > 0 o r b < 0 & a > 0 o r b < 0 & a < - 60 .$ $∴ f (x) = \frac{b x}{a (a + 60)} (- x + 2 a + 60)$ $A t x = a + 30,$ $f (a + 30) = \frac{b (a + 30) (- a - 30 + 2 a + 60)}{a (a + 60)}$ $f (a + 30) = \frac{b {(a + 30)}^{2}}{a (a + 60)}$ $f (a + 30) - b = \frac{900 b}{a (a + 60)}$ $B u t f (a + 30) - f (a) = - 3 .$ $\Rightarrow 300 b = - a (a + 60)$ $b = \frac{- a (a + 60)}{300} .$ $∴ f (x) = \frac{- a (a + 60)}{300} \times \frac{x}{a (a + 60)} (- x + 2 a + 60)$ $f (x) = \frac{x (x - 2 a - 60)}{300} = \frac{1}{300} x^{2} - \frac{x (a + 30)}{150}$ $f (a) = \frac{a (a - 2 a - 60)}{300} = \frac{- a (a + 60)}{300}$ $f (a + 60) = \frac{(a + 60) (- a)}{300} = b$ $f^{'} (x) = \frac{x - a - 30}{150} \Rightarrow a t m i d p o i n t, f^{'} (x) = 0$ $o r x = a + 30 .$ $W e m a y c h o o s e t o s h i f t t h i s c u r v e$ $b y a n a m o u n t h v e r t i c a l l y s o t h a t i n$ $g e n e r a l$ $f (x) = \frac{1}{300} x^{2} - \frac{x (a + 30)}{150} + h (a ⩽ x ⩽ a + 60, a, h \in R) .$ $a = 0 a n d h = 0 \Rightarrow f (x) = \frac{1}{300} x^{2} - \frac{1}{5} x a s w a s f o u n d .$
	Commented by sanusihammed last updated on 17/Jun/16

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