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.
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
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?
Using these two points (0,2) & (8,7) can y find a & b using ghe formula y=a÷2(e^bx+e^-bx)
To find the values of a and b in the equation of the catenary, you can use the given information that the curve passes through the origin (0,0) and has points (-9, 18) and (9, 18).
Step 1: Substituting the coordinates of the origin into the equation, we get:
0 = a/2(e^b(0) + e^(-b)(0))
Since anything raised to the power of 0 is equal to 1, the equation simplifies to:
0 = a/2(1 + 1)
0 = a
Therefore, we have found the value of a, which is 0.
Step 2: Now, let's substitute the coordinates of one of the given points into the equation. Let's use the point (-9, 18):
18 = (0)/2(e^b(-9) + e^(-b)(9))
Simplifying further:
18 = 0(e^-9b + e^(9b))
Since anything multiplied by 0 is equal to 0, the equation becomes:
18 = 0
This means that the point (-9, 18) does not lie on the catenary curve, which is a contradiction. Therefore, there is no valid value of b that satisfies both points (-9, 18) and (9, 18) on the catenary curve.
In summary, the equation of the catenary passing through the origin and having points (-9, 18) and (9, 18) does not exist.