# maths

posted by on .

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 - ,

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 - ,

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 - ,

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

let e^9b=x
ie 0 = 9x^2 + 9
9x^2 = -9
x^2 = -1
x= (-1)^1/2
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?

• maths - ,

Using these two points (0,2) & (8,7) can y find a & b using ghe formula y=aĆ·2(e^bx+e^-bx)