Kenneth Kuttler

370 CHAPTER 17. SOME PHYSICAL APPLICATIONS

13. Suppose you havex′′+a2x = by′, y′′+a2y =−bx′ (17.23)

and x(0) = x′ (0) = y(0) = y′ (0) = 0. Show that x(t) = y(t) = 0. Show this impliesthere is only one solution to the initial value problem 17.21 and 17.22. Hint: If youhad two solutions to 17.21 and 17.22, x̃, ỹ and x̂, ŷ, consider x = x̂− x̃ and y = ŷ− ỹand show x,y satisfies 17.23. To show the first part, multiply the first equation by x′

the second by y′ add and obtain the following using the product rule.

ddt

((x′)2

+(y′)2

+a2 (x2 + y2))= 0

Thus the inside is a constant. From the initial condition, this constant can only be 0.

14. This problem is about finding the equation of a hanging chain. Consider the follow-ing picture of a portion of this chain.

T (x)

T (x)cosθ

T (x)sinθ

ρl(x)g

In this picture, ρ denotes the density of the chain which is assumed to be constantand g is the acceleration due to gravity. T (x) and T0 represent the magnitude of thetension in the chain at x and at 0 respectively, as shown and l(x) is the length of thechain up to x. Let the bottom of the chain be at the origin as shown. If this chaindoes not move, then all these forces acting on it must balance. In particular,

T (x)sinθ = l (x)ρg, T (x)cosθ = T0.

Therefore, dividing these yields tan(θ) = sinθ

cosθ= l (x)

≡c︷︸︸︷ρg/T0. Now letting y(x) de-

note the y coordinate of the hanging chain corresponding to x, tanθ = y′ (x) . There-fore, this yields y′ (x) = cl (x) . From formula for the length of a graph, explain why

l′ (x) =√

1+ y′ (x)2. Explain why y′′ (x) = cl′ (x) = c√

1+ y′ (x)2. Now let z(x) =

y′ (x) and explain why z′(x)√1+z2

= c. Therefore,∫ z′(x)√

1+z2dx = cx+d. Change variables

and verify that sinh−1 (y′ (x)) = cx+ d. Now verify that y(x) = 1c cosh(cx+d)+ k

which is the equation of a catenary.

37013.14.CHAPTER 17. SOME PHYSICAL APPLICATIONSSuppose you havex"+ax= by’, y" +a’y = —by! (17.23)and x(0) =x’ (0) = y(0) = y’ (0) = 0. Show that x(t) = y(t) = 0. Show this impliesthere is only one solution to the initial value problem 17.21 and 17.22. Hint: If youhad two solutions to 17.21 and 17.22, ¥,¥ and £,%, consider x = -—*X and y= y-—JFand show x,y satisfies 17.23. To show the first part, multiply the first equation by x’the second by y’ add and obtain the following using the product rule.(()?+ (y')? +? (x? +y°)) =0Thus the inside is a constant. From the initial condition, this constant can only be 0.adtThis problem is about finding the equation of a hanging chain. Consider the follow-ing picture of a portion of this chain.T (x) sin @“ | pulseIn this picture, 9 denotes the density of the chain which is assumed to be constantand g is the acceleration due to gravity. T (x) and 7p represent the magnitude of thetension in the chain at x and at 0 respectively, as shown and /(x) is the length of thechain up to x. Let the bottom of the chain be at the origin as shown. If this chaindoes not move, then all these forces acting on it must balance. In particular,T (x) sin@ =1 (x) pg, T (x)cos@ = Th.=cTherefore, dividing these yields tan(@) = we = 1(x) pg/To. Now letting y(x) de-note the y coordinate of the hanging chain corresponding to x, tan@ = y’ (x). There-fore, this yields y’ (x) = cl (x). From formula for the length of a graph, explain whyI’ (x) = 1/1+y' (x)?. Explain why y” (x) = cl! (x) =cy/1+y! (x)?. Now let z(x) =y’ (x) and explain why aes =c. Therefore, [ are dx = cx+d. Change variablesand verify that sinh! (y’ (x)) = cx-+d. Now verify that y(x) = 4cosh(cx+d) +kwhich is the equation of a catenary.