in trouble...again (1 Viewer)

[pip88]

i am unsure what i have to do for this question its a long one. i need help from all you maths champs!

a flexible chain when suspended from its ends, hangs in a curve that looks a bit like a parabola. this curve is called a catenary. you are to investigate how good an approximation this is

A) assume the equation of the catenary is y =1/2 (e^x + e^-x) and the end points are at x =+-3 fine the coordinates of the lowest point of the catenary?

B) find the equation of a parabola through the same three points.

c)graph the parabola and the catenary on the same set of axes. describe their differencs

d) use a graphical or other method to estimate the maxium difference between the y coordinates from x=-3 to x=3

i have already done A,B & C. but i cannot do d
i find the equation from the parabola to be 1.007518(x^2) +1
i thought could do D by doind a table of values, any other ideas

i need help! thanks anyone who can help :wave:

 

[hyparzero]

for d) it doesn't matter if you use the parabolic function y = 1.007518(x^2) +1 or your hyperbolic cosine function caternary f(x) = 1/2 (e^x + e^-x) as the maximum difference between the y-coordinates from x= +- 3 should be exactly the same.

Hence the maximum difference would be

cosh(3) -1 = 9.067661996

 

[hyparzero]

And by no way is this in the course of 2Unit maths, so it shouldn't be posted here.

 

[iLoveJessika]

hey pip just wondering if u could help me out with part a b and c... lol... thanks heaps

 

[hyparzero]

catenarys aren't in the syllabus, so you wouldn't need to know

 

[iLoveJessika]

yea but we have been set this assessment task that is partially on catenarys... so kinda do need to know..
could u go into some more depth with ur explanation of part d??

 

[hyparzero]

All catenarys are given by the hyperbolic cosine function

f(x) = acosh(x/a) = 1/2 (e^x/a + e^-x/a)

for part a) the lowest point of a catenary is always the vertex, which is at (0,a) in an orthogonal coordinate system. Hence, in this case its at (0,1) since a = 1 (given)

b) If the endpoints are at, x =+-3 all you have to do is to substitute the x-values into the catenary function to find the corresponding y-values.

Hence, the parabola must cut the following points:
(3, cosh(3)) ; (0,1) ; (-3,cosh(-3))

simply use (x-k)² = 4a(y-h) <- careful, as "a" does not correspond to "a" value of the general catenary function.

simpify, you get y= 1.007518(x^2) +1

c) If you draw both the parabola and the catenary, they would look exactly the same, except sections between the vertex to the endpoints are very very slightly wider.
A catenary basically can be described as a complex parabola

d) The maxium difference between the y coordinates from x=-3 to x=3 would equal to the y-value of the endpoint minus the y-value of the vertex.

 

[abefisher]

for part d i just got the tables for both equasions(y=cosh(x), y=1.007518(x^2) +1) and made the x values of both corrispond to compare the y values. you minus one form the other and find the absolute value of the result. (eg y=coshx, when x=2.7, y=[FONT=&quot]7.47347 and y=1.007518(x^2)+1 when x=2.7, y=[/FONT][FONT=&quot]8.34481. therefore the difference between y values is |[/FONT][FONT=&quot]7.47347-[/FONT][FONT=&quot]8.34481|=[/FONT]0.87134) i think this is right haha. btw this assignment sucks