Problem 64 - Essential Calculus

Question

Find the arc length of the graph of \(y=\sqrt{16-x^{2}}\) over the interval [0,4]

Step-by-Step Solution

Answer

The arc length of the graph of the function \(y=\sqrt{16-x^{2}}\) over the interval [0,4] is \(2\pi\).

Step 1: Differentiate the function

The first step is to obtain the derivative of the function \(f(x)=\sqrt{16-x^{2}}\). Applying the chain rule, we obtain \(f'(x)=-\frac{x}{\sqrt{16-x^{2}}}\).

Step 2: Insert the derivative into the arc length formula

To apply the arc length formula, square the derivative and add 1: \(1 + [f'(x)]^{2}=1+\left(\frac{x^{2}}{16-x^{2}}\right)= \frac{16}{16-x^{2}}\).

Step 3: Integrate

Now, we should integrate this from 0 to 4 to find the arc length: \(L=\int_{0}^{4} \sqrt{\frac{16}{16-x^{2}}} dx\). Make a substitution \(x=4sin(\theta)\), \(dx=4cos(\theta)d\theta\). The integral becomes \(L= \int_{0}^{\frac{\pi}{2}} 4 d\theta = 2\pi\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. In simpler terms, when you have a function nested inside another function, the chain rule helps you take the derivative of that complex setup.
For example, consider the function given in the exercise:

• The function is expressed as a square root function: \(y = \sqrt{16 - x^2}\).

To find the derivative, you first identify the outer and inner functions. Here, the outer function is the square root operation, and the inner function is \(16 - x^2\).
Applying the chain rule involves these steps:

• Take the derivative of the outer function while keeping the inner function unchanged. This gives \(\frac{1}{2\sqrt{16-x^2}}\).
• Then, multiply by the derivative of the inner function \(-2x\).

The result is the derivative: \(f'(x) = -\frac{x}{\sqrt{16-x^2}}\). This derivative is crucial for finding the arc length as it leads to the integrand used in the formula.

Trigonometric Substitution

Trigonometric substitution is a technique used in calculus to simplify the integration of expressions, especially those involving square roots. When encountering an expression like \(\sqrt{16 - x^2}\), trigonometric substitution is an excellent tool to make integration more straightforward.
Here's how it works:

• You substitute \(x = 4\sin(\theta)\) since the structure resembles the Pythagorean identity \(sin^2(\theta) + cos^2(\theta) = 1\).

Why choose \(4\sin(\theta)\)? Because you aim to leverage the identity \(16 - x^2 = 16(1 - \sin^2(\theta)) = 16\cos^2(\theta)\).
Upon substituting, you also need to change \(dx\):

• Differentiate \(x = 4\sin(\theta)\) to get \(dx = 4\cos(\theta)d\theta\).

This transforms the integral into a simpler form, \(L = \int_{0}^{\frac{\pi}{2}} 4 d\theta\), making it directly solvable and leading to the arc length of \(2\pi\). This substitution simplifies the problem significantly, showing just how useful this technique can be.

Integration

Integration plays a central role in finding arc lengths, as it allows you to calculate the total length along a curve by summing infinitesimally small parts. The arc length formula is given by:\[L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx\]With the function \(y=\sqrt{16-x^2}\) in our exercise, once you have the derivative, you incorporate it into this formula. After replacing \(f'(x)\) and simplifying, you arrive at an expression ready to be integrated.
In our step-by-step solution:

• We first manipulate \(1 + [f'(x)]^2\) to get \(\frac{16}{16-x^2}\).
• The problem became more manageable with trigonometric substitution, simplifying the integral \(\int \sqrt{\frac{16}{16-x^2}} dx\).

Transforming the variable via trigonometric substitution, the integral reduces to a basic trigonometric integral. The integration becomes straightforward, ultimately yielding an arc length of \(2\pi\). Regardless of the problem's complexity, integration techniques like these help solve it efficiently.