Question

A dam of capacity \(V\) (less than \(\left.\pi b^{2} h / 2\right)\) 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 $$ \int_{-b}^{b} \frac{d x}{\left(1-\lambda^{2} x^{2}\right)^{1 / 2}} $$ where \(\lambda\) is found from $$ \frac{V}{h b^{2}}=\frac{\sin ^{-1} \mu}{\mu^{2}}-\frac{\left(1-\mu^{2}\right)^{1 / 2}}{\mu} $$ and \(u=\lambda b\)

Short answer

\[L = \int_{-b}^{b} \frac{d x}{(1 - \lambda^2 x^2)^{1/2}} \] The minimization condition relates the volume V and height h with \(\lambda and\) \(\b\).

Step-by-step solution

Identify Variables and Given Information

We are given the capacity of the dam as V, the height h, and the bounds of the wall from \((-b, 0)\) to \((b, 0)\). The problem involves minimizing the length L of the new wall to be constructed.

Setting Up the Shape of the New Wall

The optimal shape for minimizing length is part of a circle. Assume the new wall forms a semicircle. The equation for the semicircle can be represented as: \[(x^2 + y^2 = R^2 \text{ with the center at the origin})\]

Form the Length of the New Wall as an Integral

We need to find the length of the semicircle from \-b\ to \b\ and derive the integral for L. The length (L) of a semicircle from x = -b to x = b is given by the integral: \[\begin{aligned} L = \int_{-b}^b \frac{dx}{(1 - \lambda^2 x^2)^{1/2}} \end{aligned} \] where \(\lambda = \frac{1}{R}\)

Volume Capacity Constraint

Given that the volume is less than half the volume of the semicircle with height h, relate the volume V to the height h and radius R using the provided formula involving \(\lambda = \frac{1}{R}\): \[\frac{V}{h b^{2}} = \frac{\text{sin}^{-1} \mu}{\mu^{2}} - \frac{(1 - \mu^2)^{1/2}}{\mu} \] where \(\mu = \lambda b\)

Relate \(\lambda\) and \(\b\)

Finally, use \(\b = \lambda b\) to solve for \(\b\). This parameter \(\b\) leads us to the expression for the length L in terms of b and \(\b\)

Key concepts

integral calculus

Integral calculus is a branch of mathematics that focuses on the concept of integration. In this exercise, we're interested in finding the length of a new wall, which involves calculating an integral. When we minimize the length of a new wall to be built, the wall forms part of a circle. The arc length of this circle segment is what we need to calculate.
To find this length, we use the integral formula: oindent \[ L = \int_{-b}^{b} \frac{dx}{(1 - \lambda^2 x^2)^{1/2}} \] where oindent \ \( \lambda = \frac{1}{R} \).
The integral helps us summarize the small lengths of the wall into a total length L between the points oindent \(-b\) and oindent \(b\). This kind of integral is common in geometry for finding lengths, areas, and volumes.
Integration allows us to add up tiny infinitesimal elements to find more complex shapes' exact size or extent in one dimension.

geometry of circles

The geometry of circles plays a vital role in solving this optimization problem. When we need to minimize the length of a new dam wall, we assume the wall forms part of a circle. This shape is used because circles have the shortest perimeter for a given area. By setting the new wall as a semicircle, the dam will be more efficient in terms of materials used.
The equation for a circle centered at the origin is oindent \( x^2 + y^2 = R^2 \).
For a circle segment or arc, we only look at a section between oindent \(-b\) and oindent \(b\). Here, oindent \(R\) is the radius, and substituting oindent \(y\) to find the coordinate values along the circle helps us visualize the dam's wall better.
The connection between the height oindent \(h\) of the wall and radius oindent \(R\) also plays into solving for the integral, giving us the overall length.

volume constraints

In this exercise, the volume constraint is critical. The dam's capacity, oindent \(V\), must be less than oindent \(\left. \pi b^2 h / 2 \right)\). This constraint helps define the wall's height oindent \(h\) and the limit oindent \(b\). We use this information to derive oindent \(\lambda \) from the equation: oindent \[ \frac{V}{h b^{2}} = \frac{\sin ^{-1} \mu}{\mu^{2}} - \frac{(1 - \mu^2)^{1/2}}{\mu} oindent \] where oindent \ \( \mu = \lambda b\).
By analyzing this constraint, we ensure our solution applies to practical engineering limitations. The expression relates the wall's capacity, height, and distance, which provides us a valuable check and balance. This ensures the structurally optimal design is functional and efficient.