Problem 29 - Thomas Calculus

Question

Is there a smooth (continuously differentiable) curve \(y=f(x)\) whose length over the interval \(0 \leq x \leq a\) is always \(\sqrt{2} a\) ? Give reasons for your answer.

Step-by-Step Solution

Answer

Yes, the curve can be linear: \(y = x + C\) or \(y = -x + C\).

Step 1: Understand the Problem

We need to determine if there exists a function \(y=f(x)\) that is continuously differentiable such that the length of the curve over the interval \([0, a]\) is always \(\sqrt{2}a\). We are essentially looking for conditions on \(f(x)\) that satisfy this requirement.

Step 2: Formula for Arc Length

The arc length of a curve \(y=f(x)\) from \(x=0\) to \(x=a\) is given by the integral formula: \[ L = \int_0^a \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]We need \(L = \sqrt{2}a\).

Step 3: Set Up the Equation

We equate the arc length formula to the given constant length:\[ \int_0^a \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx = \sqrt{2}a \]This equation needs to be solved for \(f'(x) = \frac{dy}{dx}\).

Step 4: Analyze the Integral

The equation implies that: \[ \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \] must be a constant function over \([0, a]\) to yield a linear function with respect to \(a\). Specifically, \[ \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{2} \]

Step 5: Solve for the Derivative

From \( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{2} \), we have:\[ 1 + \left(\frac{dy}{dx}\right)^2 = 2 \] \[ \left(\frac{dy}{dx}\right)^2 = 1 \]Thus, \(\frac{dy}{dx} = \pm 1\).

Step 6: Derive the Function

The derivative \(\frac{dy}{dx} = \pm 1\) implies:If \(\frac{dy}{dx} = 1\), then \(y = x + C\).If \(\frac{dy}{dx} = -1\), then \(y = -x + C\),where \(C\) is a constant of integration.

Step 7: Check the Integral

Verify whether one of the solutions satisfies the problem condition:Using \(y = x + C\) over the interval\([0, a]\), we compute:\[ L = \int_0^a \sqrt{2} \, dx = \sqrt{2}a \]This satisfies the problem condition.

Key Concepts

Continuously Differentiable Function

A continuously differentiable function is a function that is both continuous and has a continuous derivative. In simpler terms, it means that the graph of the function has no abrupt changes in direction, and the rate of change (the derivative) is smooth without any jumps. Why is this important? Because when dealing with problems about curves and arc lengths, having a smooth graph allows us to use calculus to understand the curve's properties. This concept ensures that the tools of calculus, like derivatives and integrals, can be applied reliably to analyze the function's behavior.

Integral Calculus

Integral calculus is one branch of calculus that deals with the concept of integration. It is a way of adding things together in a smooth and continuous manner, typically used to find total values like area, volume, and, in our scenario, arc length.The formula for the arc length of a curve \(y=f(x)\) from \(x=0\) to \(x=a\) is given by:\[ L = \int_0^a \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]This integral tells us how to accumulate small distances along the curve between these two points. In this exercise, we aim for this resultant arc length to equal \(\sqrt{2}a\), providing us with constraints for our function \(f(x)\).

Derivative

The derivative of a function at a given point provides us with the rate at which the function's value changes at that point. In simpler terms, it tells us the slope or steepness of the curve at any particular point.For our problem, the derivative \(\frac{dy}{dx}\) of the function \(y = f(x)\) is crucial because the arc length formula involves \(\left(\frac{dy}{dx}\right)^2\). By understanding how the derivative behaves, specifically its value over the interval, we can determine the function form. Solving \(\left(\frac{dy}{dx}\right)^2 = 1\) leads us to discover that \(\frac{dy}{dx} = \pm 1\), which means our function is linear, either sloping upwards or downwards.

Constant Function

A constant function is a function that produces the same value regardless of the input. In terms of graphing, it appears as a straight, horizontal line.In the context of this calculation, the relationship \(\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{2}\) implies that this value is constant over the interval. Instead of having a changing slope, the constant \(\sqrt{2}\) relationship guides the form of the function, which in turn impacts the final computation of the arc length for \(y = x + C\) or \(y = -x + C\). These forms satisfy the criteria that result in a constant arc length of \(\sqrt{2}a\) over \([0, a]\).