To find the exact length of the curve given by the equation x^1 + 3y – y^3 = 1 and the line y = 9, we will first express the curve in terms of y and then calculate the length of the curve over the specified interval.
1. Start with the equation:
x + 3y – y^3 = 1
2. Rearranging gives us:
x = 1 – 3y + y^3
3. The formula for the length of a curve defined by y = f(x) from x = a to x = b is:
L = ∫ (sqrt(1 + (dy/dx)^2)) dx
4. Alternatively, if we have x = g(y), this formula changes to:
L = ∫ (sqrt(1 + (dx/dy)^2)) dy
5. Compute dx/dy:
From x = 1 – 3y + y^3, we differentiate:
dx/dy = -3 + 3y^2
6. Plug this into the length formula:
L = ∫ (sqrt(1 + (-3 + 3y^2)^2)) dy
7. We will evaluate this integral from the lower bound of y = 0 to the upper bound y = 9:
8. Now we simplify:
Let A = -3 + 3y^2
Then:
A^2 = (-3 + 3y^2)^2 = 9 – 18y^2 + 9y^4
Thus:
L = ∫ (sqrt(1 + 9 – 18y^2 + 9y^4)) dy = ∫ (sqrt(10 – 18y^2 + 9y^4)) dy
9. The exact evaluation of this integral can be complex and may require numerical or advanced calculus techniques.
In summary, the exact length of the curve from y = 0 to y = 9 requires evaluating the integral:
L = ∫ (sqrt(10 – 18y^2 + 9y^4)) dy from 0 to 9 which may be solved using numerical integration methods or software for precision.