A dam of capacity V (less than πb²h/2) is to be constructed on level ground next to a long straight wall which runs from (−b, 0) to (b, 0). This is to be achieved by joining the ends of a new wall, of height h, to those of the existing wall. Show that, in order to minimise the length L of new wall

Question

A dam of capacity V (less than [latex]\pi[/latex]b²h/2) is to be constructed on level ground next to a long straight wall which runs from (−b, 0) to (b, 0). This is to be achieved by joining the ends of a new wall, of height h, to those of the existing wall. Show that, in order to minimise the length L of new wall to be built, it should form part of a circle, and that L is then given by

[latex]\int_{-b}^b \frac{d x}{\left(1-\lambda^2 x^2\right)^{1 / 2}},[/latex]

where [latex]\lambda[/latex] is found from

[latex]\frac{V}{h b^2}=\frac{\sin ^{-1} \mu}{\mu^2}-\frac{\left(1-\mu^2\right)^{1 / 2}}{\mu}[/latex]

and µ = [latex]\lambda[/latex]b.

Accepted Answer

The objective is to chose the wall profile, y = y(x), so as to minimise

[latex]L=\int_{-b}^b \sqrt{(d x)^2+(d y)^2}=\int_{-b}^b\left(1+y^{\prime 2} ight)^{1 / 2} d x[/latex]

subject to the constraint that the capacity of the dam formed is

[latex]V=h \int_{-b}^b y \ d x.[/latex]

For this constrained variation problem we consider the minimisation of

[latex]K=\int_{-b}^b\left[\left(1+y^{\prime 2} ight)^{1 / 2}-\lambda y ight] d x,[/latex]

where [latex]\lambda[/latex] is a Lagrange multiplier.

Since x does not appear in the integrand, a first integral of the E–L equation is

[latex]\begin{aligned}\left(1+y^{\prime 2} ight)^{1 / 2}-\lambda y-y^{\prime} \frac{y^{\prime}}{\left(1+y^{\prime 2} ight)^{1 / 2}} & =k, \\frac{1}{\left(1+y^{\prime 2} ight)^{1 / 2}} & =k+\lambda y .\end{aligned}[/latex]

Rearranging this and integrating gives

[latex]\begin{aligned}\frac{1}{(k+\lambda y)^2}-1 & =y^{\prime 2} ,\\frac{(k+\lambda y) d y}{\sqrt{1-(k+\lambda y)^2}} & =d x \\Rightarrow \quad-\frac{\sqrt{1-(k+\lambda y)^2}}{\lambda} & =x+c .\end{aligned}[/latex]

This result can be arranged in a more familiar form as

[latex]\lambda^2(x+c)^2+(k+\lambda y)^2=1.[/latex]

This is the equation of a circle that is centred on (−c, −k/[latex]\lambda[/latex]); from symmetry c = 0. Further, since (±b, 0) lies on the curve, we must have

[latex]\lambda^2 b^2+k^2=1,[/latex] (*)

giving a connection between the Lagrange multiplier and one of the constants of integration. The length of the wall is given by

[latex]L=\int_{-b}^b\left(1+y^{\prime 2} ight)^{1 / 2} d x=\int_{-b}^b \frac{1}{k+\lambda y} d x=\int_{-b}^b \frac{1}{\left(1-\lambda^2 x^2 ight)^{1 / 2}} d x.[/latex]

The remaining constraint determines the value of [latex]\lambda[/latex] and is that

[latex]\begin{aligned}\frac{V}{h} & =\int_{-b}^b y d x=\frac{1}{\lambda} \int_{-b}^b\left(\sqrt{1-\lambda^2 x^2}-k ight) d x \& =\frac{1}{\lambda} \int_{-b}^b\left(\sqrt{1-\lambda^2 x^2}-\sqrt{1-\lambda^2 b^2} ight) d x, ext { using(*), } \\frac{\lambda V}{h} & =\left[x \sqrt{1-\lambda^2 x^2} ight]_{-b}^b-\int_{-b}^b \frac{-\lambda^2 x x}{\sqrt{1-\lambda^2 x^2}} d x-\left[x \sqrt{1-\lambda^2b^2} ight]_{-b}^b \& =\lambda^2 \int_{-b}^b \frac{x^2}{\sqrt{1-\lambda^2 x^2}} d x.\end{aligned}[/latex]

To evaluate this integral we set [latex]\lambda x[/latex] = sin θ and µ = [latex]\lambda b[/latex] = [latex]\sin \phi[/latex], to give

[latex]\begin{aligned}\frac{\lambda V}{h} & =\int_{-\phi}^\phi \frac{\sin ^2 heta \cos heta}{\lambda \cos heta} d heta, \\frac{\lambda^2 V}{h} &=\int_{-\phi}^\phi \frac{1}{2}(1-\cos 2 heta) d heta \& =\phi-\frac{2}{4} \sin 2 \phi, \\frac{\mu^2 V}{h b^2} & =\sin ^{-1} \mu-\frac{1}{2} 2\mu\left(1-\mu^2 ight)^{1 / 2}, \\frac{V}{h b^2} & =\frac{\sin ^{-1} \mu}{\mu^2}-\frac{\left(1-\mu^2 ight)^{1 / 2}}{\mu} .\end{aligned}[/latex]

This equation determines µ and hence [latex]\lambda[/latex].

Question 22.13: A dam of capacity V (less than πb²h/2) is to be constructed ...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions