Welcome back to Professor Baker's Math Class! In this lesson, we are moving beyond simply calculating the area under a curve and starting to look at the physical properties of the curves themselves. Specifically, we are tackling Chapter 8-1: Arc Length.
Have you ever wondered how to find the exact length of a curved line? While we can use the distance formula for straight lines, curves require the power of Calculus. This set of notes introduces the integral formulas required to measure the length of a smooth curve over an interval.
1. The Arc Length Formulas
The concept of arc length relies on the Pythagorean theorem applied to infinitely small segments of the curve. Depending on how the function is defined, we have two primary formulas:
- With respect to $x$: If $y = f(x)$ and $f'$ is continuous on $[a, b]$:
$$L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx$$ - With respect to $y$: If $x = g(y)$ and $g'$ is continuous on $[c, d]$:
$$L = \int_c^d \sqrt{1 + [g'(y)]^2} \, dy$$
2. Applying the Formula: Basic Integration
In our first example from the notes, we look at a semicubical parabola, $y^2 = x^3$, or $y = x^{3/2}$.
The key to these problems is handling the algebra under the radical. Once we find the derivative $f'(x) = \frac{3}{2}x^{1/2}$, we square it to get $\frac{9}{4}x$. This sets up a standard u-substitution integral:
$$L = \int_1^4 \sqrt{1 + \frac{9}{4}x} \, dx$$This is a great warm-up problem because the term under the square root is linear, making the $u$-sub straightforward.
3. Advanced Techniques: Trig Substitution
Not every arc length problem is that simple. In Example 2 ($y^2 = x$), setting up the integral leads us to a more complex radical: $\sqrt{1 + 4y^2}$.
When you see a sum of squares under a radical like $\sqrt{a^2 + u^2}$, it is a signal to use Trigonometric Substitution. In this case, we use the substitution:
- $2y = \tan \theta$
- $dy = \frac{1}{2} \sec^2 \theta \, d\theta$
Warning: This specific problem leads to the integral of $\sec^3 \theta$. As shown in the handwritten notes, this requires Integration by Parts and is known as a "cyclic" integral. It is a rigorous algebraic workout, but it leads to an exact answer involving natural logs: $\frac{1}{4}(2\sqrt{5} - \ln(\sqrt{5}+2))$.
4. The "Perfect Square" Trick
Finally, we look at the Arc Length Function with the curve $y = x^2 - \frac{1}{8}\ln x$.
A common pattern in Calculus 2 textbook problems is the "perfect square trick." When calculating $1 + [f'(x)]^2$, the algebra often simplifies miraculously if you are careful. In this example, the expression under the radical becomes a perfect square:
$$1 + \left(2x - \frac{1}{8x}\right)^2 = \left(2x + \frac{1}{8x}\right)^2$$This eliminates the square root entirely, making the integration trivial! Always keep an eye out for this algebraic simplification—it usually means you are on the right track.
Review the attached PDF to see the step-by-step derivations for these examples. Keep practicing those integration techniques!