Please help with this question, and explain?!?

a railroad track runs perfectly straight for exactly one mile assume that the two ends are fixed but the track expands by one foot and buckles up in the shape of an arc of a circle how far is the middle of that arc off the ground

Please help with this question, and explain?!?
Draw the situation and work on one side of the middle line.


Let the angle at the centre of the circle be A.





You have 2640.5 = R*A (providing A is in radians)


Also 2640 = R*sinA (A still in radians)





These together give


2640.5/A = 2640/sinA


2640.5*sinA = 2640*A





This equation cannot be solved exactly by algebra but since we assume the angle is small it is reasonable to replace sinA by its approximation of A - (A^3)/6 (only in radians).





This gives


2640.5*(A - (A^3)/6) = 2640*A


1 - (A^2)/6 = 2640/2640.5


(A^2)/6 = 1 - 2640/2640.5 = 0.00018936


A^2 = 6*0.00018936 = 0.0013615


A = sqrt(0.0013615) = 0.033707





This is a small angle and justifies the approximation used.





Using this figure we can now work out that the radius of the circle is 2640.5/0.033707 = 78337.3 ft





We can now use pythogoras to find the distance from the centre of the circle to the ground level = 78293.8 ft


This gives the height of the centre of the arc as


78337.3 - 78293.8 = 44.5 ft.





Three comments on your additional notes.





1 Two top contributors using different methods come to exactly the same answer.





2 Yes, the result is surprising, but that doesn't make it wrong.


Old fashioned railway tracks were laid in short stretches with gaps between them for precisely this reason. Modern track uses longer lengths but it is fixed down very securely so that although it wants to expand it can't. I have seen photographs of old track which experienced freak hot weather. It had buckled into a very curly S shape and that's with a lot less than a mile length.





3 I have a lifetime's experience of teaching advanced mathematics. What do you have - intution? If you think that we are both wrong then show us why!
Reply:Thank you for the time it took you to help me with this. I still cant wrap my brain around the 45 foot height. I am in integral calculus right now, but this is for another class. any insight as to how i can make my brain believe this haha?:) Report It

Reply:I think that since it is extended by 1 foot, the arc would have a 1 foot perimeter. The arc would also be half the circumference of a whole circle. What you are trying to find in this problem is the radius of the circle. So use:


C=2πr


(1 foot)*2=2πr


You multiply C by 2 because the arc is only half the circumference, not the whole.


Solve for r and that is your answer.
Reply:See the link below for all symbols used.





Using the approximation:





s² = c² + (16/3)h²





solve for h





h ≈ 44.5 ft





Alternately, we can solve two equations in two unknowns to find R and θ.





5280 = 2 * R *sin(θ/2)


5281 = R * θ





R ≈ 78335.08050


θ ≈ 0.06741551762





Then use:





R = h + R * cos(θ/2)





and solve for h.





h ≈ 44.5 ft



home theater