Question

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

Short answer

Yes, a smooth curve, \(y=f(x)\), can have a length of \(a\sqrt{2}\) over the interval \(0 \leq x \leq a\). A function with a derivative \(f'(x)\) of either 1 or -1 in the interval will satisfy this condition.

Step-by-step solution

Identify Given Information

The problem gives us the length of the curve over an interval and asks us to infer what the function could be. The length of any curve from \(x=0\) to \(x=a\) is \(a\sqrt{2}\).

Arc Length Formula Setup

To find out if there's any curve which satisfies this condition, we will start by substituting the given length into the formula for the arc length, this gives us \[a\sqrt{2} = \int_{0}^{a} \sqrt{1+(f'(x))^2} \, dx\]. Let's assume \(f'(x)=k\) is a constant, because the derivative of the function doesn't depend on x in the interval mentioned.

Evaluate The Integral

Now, if we evaluate the integral using our assumption \(f'(x)=k\), it turns into \[a\sqrt{2} = \int_{0}^{a} \sqrt{1+k^2} \, dx = a\sqrt{1+k^2}.\] We can form an equation by equating the two expressions of arc length, i.e., \(a\sqrt{1+k^2} = a\sqrt{2}.\]

Solve for \(k\)

Dividing by \(a\) (as long as \(a\) is not zero) and taking square on both sides, we have \[1+k^2=2 \implies k^2 = 1 \implies k=1, -1.\] Thus any smooth function with a derivative of either 1 or -1 over the given interval will always give the curve length as \(a\sqrt{2}.\] So this verifies that such a curve, \(y=f(x)\), can exist.

Key concepts

Arc Length Formula

Calculating the length of a curve on a graph, often known as the arc length, is a fundamental concept in calculus that measures the distance along a curve between two points. The arc length formula is a powerful tool that comes to use when trying to determine the length of a segment along a smooth curve. The formula is given by the integral \[ L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx \] where \( a \) and \( b \) are the endpoints of the interval on the x-axis, \( f(x) \) is the function representing the curve, and \( f'(x) \) is its derivative. This formula finds the infinitesimally small lengths along the curve (differentials) and sums them up over the interval to get the total length.

When applying the arc length formula to a specific example, such as the one provided, it's essential to correctly setup the integral by identifying the function and its interval. Should the derivative of the function turn out to be constant, which is simplistic but convenient for calculations, then the integral simplifies considerably and the calculation becomes more straightforward. In the example given, the problem stipulates that the curve length is always \( a\sqrt{2} \), offering a clear path to verifying whether such a curve exists by simply using the arc length formula and equating it with the given length.

Integral Calculus

Integral calculus is a branch of mathematics that deals with the accumulation of quantities and the area under and between curves. When students learn about integrals, they are often taught about antiderivatives and area under the curve first. Integrals, symbolized by the elongated S (\( \int \)), represent the area under a curve from one point to another on a graph. The concept of integrals is not only foundational for computing areas but also for finding the arc length, volume, and many other quantities.

In the context of the exercise provided, the integral calculus is used to solve for the length of a smooth curve by setting up and evaluating the arc length integral. It is important to remember that when evaluating the integral over an interval, the function's derivative is squared and added to one inside the square root, then integrated with respect to \( x \). For the function in the exercise, we assumed the derivative, \( f'(x) \), to be constant over the interval, which made it easier to evaluate the integral. Thus, integral calculus is the framework for solving these types of problems, demonstrating how integrals are used for practical applications beyond simply finding areas.

Derivatives

In calculus, derivatives represent the rate at which a function is changing at any given point on its curve, often thought of as the slope of the tangent line at any point along a function's graph. Derivatives are a form of linear approximation that give an idea of how a function behaves locally. The notation \( f'(x) \) or \( \frac{df}{dx} \) is typically used to represent the derivative of a function \( f(x) \). Understanding derivatives is vital since they are used to find extrema, solve rates of change problems, and in the case of arc lengths, describe the gradient of a curve.

The exercise in question cleverly uses the concept of derivatives by assuming that \( f'(x) = k \), a constant. This assumption points to the fact that the slope of the curve is consistent over the interval \( [0, a] \). This simplifies the integration process as the derivative does not change with \( x \), hence the integral simplifies to a solvable equation. With the derivative squared in the arc length formula, we find \( k \) such that when inserted into the equation, it satisfies the condition for the curve's length. Calculating the derivative is the first step in using the arc length formula and highlights the interconnectedness of different calculus concepts.