Question

Find the length of arc of the curve \(y^{4}=x^{5}\) from \(x=0\) to \(x=b\).

Short answer

Answer: \(L = \int_{0}^{b} \sqrt{1 + \frac{1}{16}x^{\frac{1}{2}}} dx\)

Step-by-step solution

Find the derivative of the curve with respect to x

Given the curve \(y^4 = x^5\), we will find its derivative with respect to x, which is denoted as \(\frac{dy}{dx}\). First, we will rewrite the equation as an explicit function of y in the form \(y = \sqrt[4]{x^5}\) to make it easier to compute the derivative. To find the derivative, apply the power rule: \(\frac{dy}{dx} = \frac{1}{4}x^{\frac{5}{4} - 1}\) Now simplify the exponent: \(\frac{dy}{dx} = \frac{1}{4}x^{\frac{1}{4}}\)

Apply the arc length formula

The arc length formula for a curve defined by \(y = f(x)\) from \(x=a\) to \(x=b\) is: \(L = \int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^2}dx\) We're given that \(a=0\) and \(b=b\). Plug in the derivative we found in Step 1: \(L = \int_{0}^{b} \sqrt{1 + (\frac{1}{4}x^{\frac{1}{4}})^2} dx\) Now square the term inside the square root: \(L = \int_{0}^{b} \sqrt{1 + \frac{1}{16}x^{\frac{1}{2}}} dx\)

Integrate the function

To find the length of the arc, we need to evaluate the integral: \(L = \int_{0}^{b} \sqrt{1 + \frac{1}{16}x^{\frac{1}{2}}} dx\) Unfortunately, this integral doesn't have an elementary solution, meaning it can't be expressed in terms of basic functions. However, we can leave it in the current form as it represents the exact length of the arc from \(x=0\) to \(x=b\) for the given curve. Therefore, the length of the arc of the curve \(y^4 = x^5\) from \(x=0\) to \(x=b\) is: \(L = \int_{0}^{b} \sqrt{1 + \frac{1}{16}x^{\frac{1}{2}}} dx\)

Key concepts

Derivative of a Curve

The derivative of a curve represents the rate at which the curve's y-value changes with respect to its x-value. In simpler terms, it tells us how steep the curve is at any given point. With the given curve, \(y^4 = x^5\), we're interested in how y changes as x changes, this is denoted as \(\frac{dy}{dx}\). This can be particularly challenging when the equation of the curve is not already solved for y, as in this case.

To find the derivative, we must first express y in terms of x, which yields \(y = \sqrt[4]{x^5}\). Applying the power rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\), we deduce that \(\frac{dy}{dx} = \frac{1}{4}x^{\frac{1}{4}}\). Understanding how to find the derivative of a curve is crucial because it sets the foundation for computing the arc length of the curve, among other applications in calculus.

Arc Length Formula

To find the length of the arching path a curve creates, we use the arc length formula. This formula is a key concept in calculus and is given by \(L = \int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^2}dx\), where \(\frac{dy}{dx}\) is the derivative of the curve. The limits of integration, 'a' and 'b', represent the interval over which we wish to calculate the length.

When we apply this formula to our curve \(y^4 = x^5\), we integrate from \(x=0\) to \(x=b\), using the derivative we obtained earlier. The integral, \(L = \int_{0}^{b} \sqrt{1 + (\frac{1}{4}x^{\frac{1}{4}})^2} dx\), represents the arc length. It's crucial to note that the square inside the square root in the arc length formula ensures that the value under the radical is always positive, which is essential as we cannot have a negative length.

Integration in Calculus

Integration is one of the two fundamental operations in calculus, alongside differentiation. It's used to find areas, volumes, central points, and many useful things, but in this context, we're using it to find the arc length of a curve.

The integral in the arc length formula does not always result in an expression that can be simplified to elementary functions. When we say 'elementary functions', we mean functions composed of algebraic, exponential, logarithmic, trigonometric, and constant functions. This is the case with the integral \(L = \int_{0}^{b} \sqrt{1 + \frac{1}{16}x^{\frac{1}{2}}} dx\), which, despite not simplifying nicely, still perfectly represents the length of the arc as a function of 'b'.

For some students, integration can be intimidating, but breaking it down into its steps - setting up the integral, identifying a strategy (such as substitution or integration by parts), and evaluating the result - can demystify the process. Understanding how to integrate functions involving square roots is particularly valuable in the study of arc lengths and other advanced topics in calculus.