daviddoji/CalculusWithJuliaNotes.jl
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some typos.
This commit is contained in:
Fang Liu
2023-06-04 16:24:38 +08:00
parent f10e045d29
commit 01e9e42661
2 changed files with 11 additions and 11 deletions
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quarto/differentiable_vector_calculus/polar_coordinates.qmd
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@@ -315,7 +315,7 @@ p
As well, see this part of a [Wikipedia](http://en.wikipedia.org/wiki/Polar_coordinate_system#Integral_calculus_.28area.29) page for a figure.
Imagine we have $a < b$ and a partition $a=t_0 < t_1 < \cdots < t_n = b$. Let $\phi_i = (1/2)(t_{i-1} + t_{i})$ be the midpoint. Then the wedge of radius $r(\phi_i)$ with angle between $t_{i-1}$ and $t_i$ will have area $\pi r(\phi_i)^2 (t_i-t_{i-1}) / (2\pi) = (1/2) r(\phi_i)(t_i-t_{i-1})$, the ratio $(t_i-t_{i-1}) / (2\pi)$ being the angle to the total angle of a circle. Summing the area of these wedges over the partition gives a Riemann sum approximation for the integral $(1/2)\int_a^b r(\theta)^2 d\theta$. This limit of this sum defines the area in polar coordinates.
Imagine we have $a < b$ and a partition $a=t_0 < t_1 < \cdots < t_n = b$. Let $\phi_i = (1/2)(t_{i-1} + t_{i})$ be the midpoint. Then the wedge of radius $r(\phi_i)$ with angle between $t_{i-1}$ and $t_i$ will have area $\pi r(\phi_i)^2 (t_i-t_{i-1}) / (2\pi) = (1/2) r(\phi_i)^2(t_i-t_{i-1})$, the ratio $(t_i-t_{i-1}) / (2\pi)$ being the angle to the total angle of a circle. Summing the area of these wedges over the partition gives a Riemann sum approximation for the integral $(1/2)\int_a^b r(\theta)^2 d\theta$. This limit of this sum defines the area in polar coordinates.
> *Area of polar regions*. Let $R$ denote the region bounded by the curve $r(\theta)$ and bounded by the rays $\theta=a$ and $\theta=b$ with $b-a \leq 2\pi$, then the area of $R$ is given by:
@@ -554,7 +554,7 @@ A point in Cartesian coordinates is given by $(-12, -5)$. In has a polar coordin
#| hold: true
#| echo: false
x,y = -12, -5
r1, theta1 = sqrt(x^2 + y^2), atan(y,x)
r1, theta1 = sqrt(x^2 + y^2), atan(y,x) + 2pi
numericq(r1)
```
@@ -565,7 +565,7 @@ What is $\theta$?
#| hold: true
#| echo: false
x,y = -12, -5
r1, theta1 = sqrt(x^2 + y^2), atan(y,x)
r1, theta1 = sqrt(x^2 + y^2), atan(y,x) + 2pi
numericq(theta1)
```
@@ -626,8 +626,8 @@ allows the slope to be computed when $y$ and $x$ are the Cartesian form of the p
$$
\frac{dy}{d\theta} = \frac{d}{d\theta}(2\cos(\theta) \cdot \cos(\theta)),~ \text{ and }
\frac{dx}{d\theta} = \frac{d}{d\theta}(2\sin(\theta) \cdot \cos(\theta)).
\frac{dy}{d\theta} = \frac{d}{d\theta}(2\sin(\theta) \cdot \cos(\theta)),~ \text{ and }
\frac{dx}{d\theta} = \frac{d}{d\theta}(2\cos(\theta) \cdot \cos(\theta)).
$$
Numerically, what is the slope of the tangent line when $\theta = \pi/4$?
@@ -637,8 +637,8 @@ Numerically, what is the slope of the tangent line when $\theta = \pi/4$?
#| hold: true
#| echo: false
r(theta) = 2cos(theta)
g(theta) = r(theta)*cos(theta)
f(theta) = r(theta)*sin(theta)
g(theta) = r(theta)*sin(theta)
f(theta) = r(theta)*cos(theta)
c = pi/4
val = D(g)(c) / D(f)(c)
numericq(val)
@@ -729,7 +729,7 @@ radioq(choices, answ)
###### Question
Find the area of a lobe of the [eight](http://www-history.mcs.st-and.ac.uk/Curves/Eight.html) curve traced out by $r(\theta) = \cos(2\theta)\sec(\theta)^4$ from $-\pi/4$ to $\pi/4$. Do this numerically.
Find the area of a lobe of the [eight](http://www-history.mcs.st-and.ac.uk/Curves/Eight.html) curve traced out by $r(\theta) = \sqrt{\cos(2\theta)\sec(\theta)^4}$ from $-\pi/4$ to $\pi/4$. Do this numerically.
```{julia}
@@ -757,7 +757,7 @@ numericq(val)
###### Question
Find the arc length of a lobe of the [eight](http://www-history.mcs.st-and.ac.uk/Curves/Eight.html) curve traced out by $r(\theta) = \cos(2\theta)\sec(\theta)^4$ from $-\pi/4$ to $\pi/4$. Do this numerically.
Find the arc length of a lobe of the [eight](http://www-history.mcs.st-and.ac.uk/Curves/Eight.html) curve traced out by $r(\theta) = \sqrt{\cos(2\theta)\sec(\theta)^4}$ from $-\pi/4$ to $\pi/4$. Do this numerically.
```{julia}