Problem 25 - 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

No such smooth curve exists, since linear functions do not satisfy the arc length condition \( \sqrt{2}a \).

Step 1: Understanding the Problem

The problem asks if there exists a function \( y = f(x) \) defined over the interval \([0, a]\) that has a constant arc length \( \sqrt{2}a \) for any non-negative value of \( a \). The arc length of a curve \( y = f(x) \) is calculated using the formula: \[ L = \int_{0}^{a} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \].

Step 2: Formula for Arc Length

Substitute the given arc length condition into the arc length formula: \[ \sqrt{2}a = \int_{0}^{a} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]. We need to determine if such a function \( f(x) \) exists that satisfies this equation for all \( a \geq 0 \).

Step 3: Consider Constant Functions

First, consider the simplest case where \( f(x) \) is a constant function, say \( f(x) = c \). In this case, \( \frac{dy}{dx} = 0 \) and the length becomes \[ L = \int_{0}^{a} \sqrt{1 + 0} \, dx = \int_{0}^{a} 1 \, dx = a \]. This does not equal \( \sqrt{2}a \), suggesting \( f(x) \) cannot be constant.

Step 4: Differentiating the Arc Length Condition

Differentiate both sides of \( \sqrt{2}a = \int_{0}^{a} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \) with respect to \( a \). The left side becomes \( \sqrt{2} \) and the right becomes \( \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \) at \( x = a \). Thus, \( \sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \sqrt{2} \), which simplifies to \( \left( \frac{dy}{dx} \right)^2 = 1 \).

Step 5: Solving for Derivative

Given \( \left( \frac{dy}{dx} \right)^2 = 1 \), we have \( \frac{dy}{dx} = \pm 1 \). This means the curve could only be \( y = x + c \) or \( y = -x + c \), where \( c \) is a constant. However, both cases lead to linear functions which do not satisfy the arc length constant \( \sqrt{2}a \).

Step 6: Conclusion

The previously derived forms \( y = x + c \) and \( y = -x + c \) result in lengths \( L = a \) for \( y = x+c \) and \( L = a \) for \( y = -x+c \). Thus, no such function \( f(x) \) of the form \( y = f(x) \) exists that can maintain an arc length of \( \sqrt{2}a \) over every interval \([0, a]\).

Key Concepts

Differentiation

Differentiation is a key concept in calculus that allows us to calculate the rate of change of one variable with respect to another. In simple terms, differentiation lets us find the derivative, which is a mathematical tool to find how a function changes as its input changes. This rate of change is often represented as \( \frac{dy}{dx} \), where \( y = f(x) \) is the function dependent on \( x \).
In the context of calculating the arc length of a curve, differentiation plays a crucial role. To find out how much the curve bends or slopes, we use \( \frac{dy}{dx} \), the derivative of \( y \, = \, f(x) \) with respect to \( x \). This helps determine the shape and length of the curve over a certain interval.
In our problem, the expression for arc length \( \int_{0}^{a} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \) involves using the derivative to account for any changes in the curve's direction.

Integration

Integration is the counterpart to differentiation in calculus. While differentiation deals with rates of change, integration focuses on accumulation – summing up small pieces to get a total.
In calculating the arc length of a curve, we use integration to accumulate infinitely many small line segments along the curve from \( x = 0 \) to \( x = a \).
The formula \( L = \int_{0}^{a} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \) is central to our problem. Here, we integrate the square root of \( 1 + (\frac{dy}{dx})^2 \), with respect to \( x \), over the interval from \( 0 \) to \( a \). This expression accounts for both the horizontal and vertical changes in the curve, thereby helping to compute its total length.
Understanding integration in this way allows us to solve problems involving curves, motion, and other phenomena where total measures over an interval are essential.

Calculus Problem Solving

Calculus problem solving often involves a strategic approach to tackle complex issues effectively. It combines the foundational concepts of differentiation and integration to understand the behavior of functions.
In our problem of determining whether a function exists that gives a constant arc length \( \sqrt{2}a \), we systematically explored various possibilities using these calculus tools.

• We first expressed the problem mathematically using the arc length formula.
• We investigated simple cases, such as constant functions, to see if they could be solutions.
• We differentiated the arc length condition concerning \( a \), which led us to an equation involving the derivative: \( \frac{dy}{dx} = \pm 1 \).

These steps showcased how calculus aids in approaching and dissecting problems, helping us conclude that no such function exists that uniformly satisfies the given arc length condition. By combining differentiation and integration, calculus enables us to navigate the intricacies involved in analyzing curves and other functions.