daviddoji/CalculusWithJuliaNotes.jl
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity

some typos

Browse Source
This commit is contained in:
Fang Liu
2025-06-06 16:56:56 +08:00
parent 9dcafd7d7d
commit 8ede2b33fb
11 changed files with 34 additions and 33 deletions
Download Patch File Download Diff File Expand all files Collapse all files

10
quarto/integrals/arc_length.qmd
View File

@@ -493,14 +493,14 @@ What looks at first glance to be just a slightly more complicated equation is th
$$
\begin{align*}
s(u) &= \int_0^u \sqrt{(-\sin(t))^2 + b\cos(t)^2} dt\\
&= \int_0^u \sqrt{\sin(t)^2 + \cos(t)^2 + c\cos(t)^2} dt \\
&=\int_0^u \sqrt{1 + c\cos(t)^2} dt.
s(u) &= \int_0^u \sqrt{(-\sin(t))^2 + (b\cos(t))^2} dt\\
&= \int_0^u \sqrt{\sin(t)^2 + \cos(t)^2 + C\cos(t)^2} dt \\
&=\int_0^u \sqrt{1 + C\cos(t)^2} dt.
\end{align*}
$$
But, despite it not looking too daunting, this integral is not tractable through our techniques and has an answer involving elliptic integrals. We can work numerically though. Letting $a=1$ and $b=2$, we have the arc length is given by:
Where $C = 2c + c^2$ is a constant. But, despite it not looking too daunting, this integral is not tractable through our techniques and has an answer involving elliptic integrals. We can work numerically though. Letting $a=1$ and $b=2$, we have the arc length is given by:
```{julia}
@@ -710,7 +710,7 @@ For the latter claim, integrating in the $y$ variable gives
$$
\begin{align*}
\int_u^c (f(x)-h) dx &= \int_h^m (c - f_1^{-1}(y)) dy\\
&> \int_h^m (c - f_2^{-1}(y)) dy\\
&> \int_h^m (f_2^{-1}(y) - c) dy\\
&= \int_c^v (f(x)-h) dx
\end{align*}
$$