Question

Find the length of the curve \(y=\sqrt{16-x^{2}}\) \mathrm{\\{} b e t w e e n ~ \(x=0\) and \(x=2\).

Short answer

The length of the curve is \(\frac{2\pi}{3}\).

Step-by-step solution

Identify the Formula for Arc Length

To find the length of a curve defined by a function \(y=f(x)\) between \(x=a\) and \(x=b\), we use the formula: \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] where \(\frac{dy}{dx}\) is the derivative of the function.

Find the Derivative of the Function

The given function is \(y = \sqrt{16 - x^2}\). First, find the derivative \(\frac{dy}{dx}\) using the chain rule. Let \(u = 16 - x^2\), then \(y = u^{1/2}\).The derivative \(\frac{dy}{dx}\) is given by \[ \frac{dy}{dx} = \frac{1}{2}u^{-1/2} \times \frac{du}{dx} \] \(\frac{du}{dx} = -2x\), so \[ \frac{dy}{dx} = \frac{1}{2} (16-x^2)^{-1/2} \times (-2x) = \frac{-x}{\sqrt{16-x^2}} \].

Plug Derivative into the Arc Length Formula

Substitute \(\frac{dy}{dx}\) into the arc length formula:\[ L = \int_{0}^{2} \sqrt{1 + \left( \frac{-x}{\sqrt{16-x^2}} \right)^2} \, dx \]Simplify the expression inside the square root:\[ L = \int_{0}^{2} \sqrt{1 + \frac{x^2}{16-x^2}} \, dx \]Factor and simplify:\[ L = \int_{0}^{2} \frac{\sqrt{16}}{\sqrt{16-x^2}} \, dx = \int_{0}^{2} \frac{4}{\sqrt{16-x^2}} \, dx \].

Evaluate the Integral

Recognize that the integral \(\int \frac{1}{\sqrt{a^2-x^2}} \ dx\) is an inverse trigonometric function, specifically \(\sin^{-1}(\frac{x}{a})\). Here, \(a = 4\), so we have:\[ L = 4 \int_{0}^{2} \frac{1}{\sqrt{16-x^2}} \, dx = 4 \left[ \sin^{-1}\left(\frac{x}{4}\right) \right]_0^2 \].

Substitute and Simplify the Result

Substitute the limits of integration:\[ L = 4 \left( \sin^{-1}\left(\frac{2}{4}\right) - \sin^{-1}(0) \right) \]Simplify:\[ L = 4 \left( \sin^{-1}\left(\frac{1}{2}\right) \right) \]The angle whose sine is \(\frac{1}{2}\) is \(\frac{\pi}{6}\), thus:\[ L = 4 \times \frac{\pi}{6} = \frac{2\pi}{3} \].

Key concepts

Arc Length

When you want to find the arc length of a curve that is defined by a function, you use a particular formula that involves calculus. Arc length helps us measure the distance along a curve.
This is different from just measuring a straight line between the endpoints of the curve. Instead, it accounts for all the twists and turns along its path.
For a function given by a curve from point \(x = a\) to \(x = b\), use the formula:

• \( L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \)

Here, \( \frac{dy}{dx} \) represents the derivative of the function with respect to x. This means first you'll need to find the slope of the curve at any given point before you plug it into the formula.
Learning about arc length is crucial in fields such as physics or engineering, where understanding the path of motion or the shape of objects is important.

Integral Calculation

Integrals are a key concept in calculus and are used to find quantities like areas under curves, volumes, and in this case, arc lengths.
The integral used in our arc length problem is computed by summing up tiny sections of the curve from \(x = 0\) to \(x = 2\).

• After finding the derivative, \(\frac{dy}{dx}\), substitute into the arc length formula.
• Simplify the expression to make it easier to integrate.

In this exercise, simplifying \(\sqrt{1 + \frac{x^2}{16-x^2}}\) leads us to a cleaner expression: \(\frac{4}{\sqrt{16-x^2}}\).
This simplifies the integration step and can help in many practical situations, like computing the arc lengths for design and construction projects. Understanding how to evaluate integrals is key for solving these types of problems effectively.

Inverse Trigonometric Functions

Inverse trigonometric functions flip the regular trig functions (like sine, cosine, and tangent) so that inputs and outputs are reversed. They are used extensively in calculus for integration, especially when integrating functions that involve square roots.
In our exercise, we deal with an integral involving the format \(\frac{1}{\sqrt{a^2-x^2}}\), which corresponds to the inverse sine function, also noted as \(\sin^{-1}(x/a)\).

• They help us simplify the integration process, especially in geometrical contexts like our arc length calculation.
• Recognizing these standard forms helps make evaluating complex integrals easier.

To solve the exercise, recognize
that of our integral simplifies to \(4 \sin^{-1}(\frac{x}{4})\), and we evaluate this from \(x = 0\) to \(x = 2\).
With \(\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}\), we find that the length of the curve can be calculated to be \(\frac{2\pi}{3}\). This connection of trigonometric functions to geometry is a potent tool in calculus.