User:Sloverlord/Sandbox

$\begin{align} y & = kx^2 \\ 143 & = k(640)^2 \\ & = 409600k \\ k & = \frac{143}{409600} \\ k \approx .00035 \\ \end{align}$

$y = .00035x^2 \,$

$\int_{-640}^{640} \sqrt{1 + \left(f'(x)\right)^2}\, dx \quad f'(x) = .0007x$

$\int_{-640}^{640} \sqrt{1 + (.00000049x^2)} \, dx$

$\approx 1321.612$

To determine the next problem, we know that the new parabola is also of the form

$f(x) = kx^2 \,$

for some k. Therefore,

$f'(x) = 2kx \,$

So the arc length of the new parabola is

$1387.69 = \int_{-640}^{640} \sqrt{1 + (2kx)^2} \, dx$

$k = .0005752 \,$

$y = (.0005752)(640)^2 \,$

$y = 235.6 \,$

$235.6 - 143 = 92.6 \mathrm{ft.} \,$