Posted by giules on Tuesday, May 10, 2011 at 9:23pm.
Catenary; y= a/2(e^bx +e^-bx)
goes through the origin (0,0) and has point (-9,18) & (9,18)
solve for a and b.
- maths - Reiny, Tuesday, May 10, 2011 at 11:08pm
sub in (0,0)
0 = a/2(e^0 + e^)
0 = a/2(2)
a = 0
something is not right here, a catenary of the above form does not pass through (0,0)
Only if a = 0, but then the equation would collapse to
y = 0
- maths - giules, Tuesday, May 10, 2011 at 11:33pm
What about if we move the turning point (0,0) to (9,0) and the other points are (0,18) and (18,18, what then?
- maths - giules, Tuesday, May 10, 2011 at 11:50pm
If we use the above points,I think
y= a/2(e^bx +e^-bx)
18 = a/2 (e^b0 + e^-b0)
18 = a/2 (e^0 + e^0)
18 = a/2 (1 + 1)
18 = a/2 (2)
18 = a
To find b;
then substitute in y= a/2(e^bx +e^-bx)
at point (9,0)
0= 18/2(e^b9 +e^-9b)
0= 9(e^9b +e^-9b)
0 = 9e^9b + 9e^-9b
0 = 9e^9b + 9/e^9b (Xe^9b)
0 = 9(e^9b)^2 + 9
ie 0 = 9x^2 + 9
9x^2 = -9
x^2 = -1
Still it doesn't work out
or do I differentiate d(y)/d for a gradient of 0/ Still doesn't work.
What am I missing here?
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