Section 8.1: Arc Length
Welcome to Section 8.1, where we embark on a journey to understand and calculate the arc length of a curve. Arc length is the distance along a curve, and we can find it using calculus!
Imagine zooming in on a curve. As you zoom in closer and closer, the curve starts to look like a bunch of tiny straight lines. The Pythagorean theorem gives us a way to calculate the length of each of these tiny straight lines: $\sqrt{(\Delta x)^2 + (\Delta y)^2}$.
Now, let's formalize this idea with the Arc Length Formula. If $f'(x)$ is continuous on the interval $[a, b]$, then the length $L$ of the curve $y = f(x)$ from $x = a$ to $x = b$ is given by:
$$L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} \, dx$$This formula is derived by summing up the lengths of infinitely small line segments along the curve. Using Leibniz notation, we can also write the arc length formula as:
$$L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$Example 1: Arc Length of a Semicubical Parabola
Let's find the length of the arc of the semicubical parabola $y^2 = x^3$ between the points $(1, 1)$ and $(4, 8)$. First, rewrite the equation as $y = x^{3/2}$.
Next, find the derivative of $y$ with respect to $x$:
$$\frac{dy}{dx} = \frac{3}{2}x^{1/2}$$Now, square the derivative:
$$\left(\frac{dy}{dx}\right)^2 = \frac{9}{4}x$$Plug this into the arc length formula and integrate from $x = 1$ to $x = 4$:
$$L = \int_{1}^{4} \sqrt{1 + \frac{9}{4}x} \, dx$$Using u-substitution ($u = 1 + \frac{9}{4}x$, $du = \frac{9}{4} dx$), we find the arc length to be:
$$L = \frac{4}{9} \int_{13/4}^{10} u^{1/2} du = \frac{8}{27}\left[u^{3/2}\right]_{13/4}^{10} = \frac{8}{27} \left(10^{3/2} - \left(\frac{13}{4}\right)^{3/2}\right)$$Example 2: Arc Length of a Parabola
Find the length of the arc of the parabola $y^2 = x$ from $(0, 0)$ to $(1, 1)$. We can rewrite this as $x = y^2$. Then $\frac{dx}{dy} = 2y$.
The arc length is then given by:
$$L = \int_{0}^{1} \sqrt{1 + (2y)^2} \, dy = \int_{0}^{1} \sqrt{1 + 4y^2} \, dy$$Using a trigonometric substitution ($y = \frac{1}{2} \tan{\theta}$), and evaluating the integral, we find the arc length to be:
$$L = \frac{1}{4} \left[ \sec{\theta} \tan{\theta} + \ln{\left|\sec{\theta} + \tan{\theta}\right|} \right]_0^{\tan^{-1}2} = \frac{1}{4}(2\sqrt{5} + \ln(2+\sqrt{5}))$$Example 3: Arc Length of a Hyperbola
Let's set up the integral for the arc length of the hyperbola $xy = 1$ from the point $(1, 1)$ to the point $(2, \frac{1}{2})$. We have $y = \frac{1}{x}$, so $\frac{dy}{dx} = -\frac{1}{x^2}$.
The integral representing the arc length is:
$$L = \int_{1}^{2} \sqrt{1 + \left(-\frac{1}{x^2}\right)^2} \, dx = \int_{1}^{2} \sqrt{1 + \frac{1}{x^4}} \, dx$$This integral can be approximated using numerical methods like Simpson's Rule.
Keep practicing, and you'll master the art of finding arc lengths! Remember, calculus is a journey, not a destination. Embrace the challenge, and enjoy the process!