$5\mathrm{.}16$ Water is flowing in a trapezoidal channel at a rate of $Q=2030\frac{m^3}{s}.$ The critical depth $y$ for such a channel must

satisfy the equation:

$0=1-\frac{Q^2}{g{A}_c^3}B$

where $g=9\mathrm{.}81\frac{m}{s^2},A_c=$ the cross$-$sectional area $(m^2),$ and $B=$ the width of the channel at the surface $(m).$ For this

case, the width and the cross$-$sectional area can be related to depth $y$ by

$B=3+y,$ and $,A_c=3y+\frac{y^2}{2}$

Solve for the critical depth using $($a$)$ the graphical method, $($b$)$ bisection, and $($c$)$ false position. For $($b$)$ and $($c$),$ use

initial guesses of $x_l-0\mathrm{.}5$ and $x_u-3,$ and iterate until the approximate error falls below $1\%$ or the number of

iterations exceeds $10.$ Discuss your results. See modifications below.

Use a rate of $Q=30\frac{m^3}{s}$ and solve for the critical depth using:

a$.(2$ pts$)$ Bisection method, use initial bounds of $x_L=0\mathrm{.}5$ and $x_U=3\mathrm{.}0$

b$.(2$ pts$)$ Secant method, use initial guesses of $x_0=1\mathrm{.}0,x_1=1\mathrm{.}1.$

c$.(1$ pt$)$ using Excel Solver

Iterate until approximate relative error falls below ${}_s=1\%(\mathrm{.}01)$ and $f_{\text{t}\text{o}\text{l}}$ within $+-0\mathrm{.}01$ of zero