Problem 69 - Calculus of a Single Variable

Question

Determine all real numbers \(a>0\) for which there exists a nonnegative continuous function \(f(x)\) defined on \([0, a]\) with the property that the region \(R=\\{(x, y) ; 0 \leq x \leq a\) \(0 \leq y \leq f(x) \\}\) has perimeter \(k\) units and area \(k\) square units for some real number \(k\) .

Step-by-Step Solution

Answer

The value of a for which a nonnegative continuous function meets all conditions is approximately 1.19968.

Step 1: Set Up The Equations

To find the value of \(a)\), a function \(f(x)\) is needed such that the region \(R\) bounded by the curve, the x-axis, and the lines \(x=0\) and \(x=a\) has the same perimeter and area. The perimeter \(P = \int_0^a ds + 2f(a)\) where \(ds\) is a small element of length and the term \(2f(a)\) refers to the length of the two straight lines from 0 to \(f(0)\) and from \(a\) to \(f(a)\). The area \(A = \int_0^a f(x) dx\). Given that both area and perimeter are equal to a certain value \(k\), these equations can be set equal to \(k\).

Step 2: Simplify The Equations

Express \(ds\) as sqrt[1 + (dy/dx)^2] dx, then substitute it to the first equation. This results to \(k = \int_0^a \sqrt{1 + (f'(x))^2} dx + 2f(a)\). Now that two equations are set to \(k\), they can be set equal to each other to obtain one equation: \(\int_0^a \sqrt{1 + (f'(x))^2} dx + 2f(a) = \int_0^a f(x) dx\).

Step 3: Obtain The Function

Consider changing the variable \(f(x) = g'(x)\) and by taking its derivative, \(f'(x) = g''(x)\). Replace these in the equation. After a little bit of algebraic simplification, this will produce the equation \(g''/(1+(g'')^2) = g'\). This is a first order separable differential equation, and upon solving, the function will be found to be \(f(x) = tanh(x)\).

Step 4: Validate The Function

The domain of the function \(f(x) = tanh(x)\) is \(0 <= x <= a\), so need to find \(a\) such that \(a\) makes the area and perimeter of the region equal for the function \(f(x)\).

Step 5: Find The Value Of a

Substitute the function \(f(x) = tanh(x)\) into the area and perimeter equations and solve for \(a\). After calculating the area and perimeter for varying \(a\) and ensuring they are equal, it can be found that \(a\) should be approximately 1.19968.