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

use quarto, not Pluto to render pages

Browse Source
This commit is contained in:
jverzani
2022-07-24 16:38:24 -04:00
parent 93c993206a
commit 7b37ca828c
879 changed files with 793311 additions and 2678 deletions
Download Patch File Download Diff File Expand all files Collapse all files

140
CwJ/derivatives/first_second_derivatives.jmd
View File

@@ -416,17 +416,19 @@ g' + 0 - 0 - 0 + g'-sign
Consider the function $f(x) = x^2$. Over this function we draw some
secant lines for a few pairs of $x$ values:
```julia; hold=true; echo=false
f(x) = x^2
seca(f,a,b) = x -> f(a) + (f(b) - f(a)) / (b-a) * (x-a)
p = plot(f, -2, 3, legend=false, linewidth=5, xlim=(-2,3), ylim=(-2, 9))
plot!(p,seca(f, -1, 2))
a,b = -1, 2; xs = range(a, stop=b, length=50)
plot!(xs, seca(f, a, b).(xs), linewidth=5)
plot!(p,seca(f, 0, 3/2))
a,b = 0, 3/2; xs = range(a, stop=b, length=50)
plot!(xs, seca(f, a, b).(xs), linewidth=5)
p
```julia; echo=false
let
f(x) = x^2
seca(f,a,b) = x -> f(a) + (f(b) - f(a)) / (b-a) * (x-a)
p = plot(f, -2, 3, legend=false, linewidth=5, xlim=(-2,3), ylim=(-2, 9))
plot!(p,seca(f, -1, 2))
a,b = -1, 2; xs = range(a, stop=b, length=50)
plot!(xs, seca(f, a, b).(xs), linewidth=5)
plot!(p,seca(f, 0, 3/2))
a,b = 0, 3/2; xs = range(a, stop=b, length=50)
plot!(xs, seca(f, a, b).(xs), linewidth=5)
p
end
```
The graph attempts to illustrate that for this function the secant
@@ -573,11 +575,11 @@ One way to visualize the second derivative test is to *locally* overlay on a cri
```julia; hold=true;
f(x) = sin(x) + sin(2x) + sin(3x)
p = plot(f, 0, 2pi, legend=false, color=:blue, linewidth=3)
cps = fzeros(f', 0, 2pi)
h = 0.5
cps = find_zeros(f', (0, 2pi))
Δ = 0.5
for c in cps
parabola(x) = f(c) + (f''(c)/2) * (x-c)^2
plot!(parabola, c-h, c+h, color=:red, linewidth=5, alpha=0.6)
plot!(parabola, c - Δ, c + Δ, color=:red, linewidth=5, alpha=0.6)
end
p
```
@@ -637,16 +639,18 @@ find_zeros(k'', -3, 3)
A car travels from a stop for 1 mile in 2 minutes. A graph of its
position as a function of time might look like any of these graphs:
```julia; hold=true; echo=false
v(t) = 30/60*t
w(t) = t < 1/2 ? 0.0 : (t > 3/2 ? 1.0 : (t-1/2))
y(t) = 1 / (1 + exp(-t))
y1(t) = y(2(t-1))
y2(t) = y1(t) - y1(0)
y3(t) = 1/y2(2) * y2(t)
plot(v, 0, 2, label="f1")
plot!(w, label="f2")
plot!(y3, label="f3")
```julia; echo=false
let
v(t) = 30/60*t
w(t) = t < 1/2 ? 0.0 : (t > 3/2 ? 1.0 : (t-1/2))
y(t) = 1 / (1 + exp(-t))
y1(t) = y(2(t-1))
y2(t) = y1(t) - y1(0)
y3(t) = 1/y2(2) * y2(t)
plot(v, 0, 2, label="f1")
plot!(w, label="f2")
plot!(y3, label="f3")
end
```
All three graphs have the same *average* velocity which is just the
@@ -696,8 +700,8 @@ choices=[
"``(-5, -4.2)``",
"``(-5, -4.2)`` and ``(-2.5, 0)``",
"``(-4.2, -2.5)``"]
ans = 3
radioq(choices, ans)
answ = 3
radioq(choices, answ)
```
@@ -718,8 +722,8 @@ choices=[
"``(-25.0, 0.0)``",
"``(-5.0, -4.0)`` and ``(-4, -3)``",
"``(-4.0, -3.0)``"]
ans = 4
radioq(choices, ans)
answ = 4
radioq(choices, answ)
```
###### Question
@@ -740,8 +744,8 @@ choices=[
"``(-4.7, -3.0)``",
"``(-0.17, 0.17)``"
]
ans = 3
radioq(choices, ans)
answ = 3
radioq(choices, answ)
```
###### Question
@@ -763,8 +767,8 @@ choices=[
"``(-0.6, 0.6)``",
" ``(-3.0, -0.6)`` and ``(0.6, 3.0)``"
]
ans = 4
radioq(choices, ans)
answ = 4
radioq(choices, answ)
```
@@ -786,8 +790,8 @@ choices = [
"That the critical point at ``0`` is a relative maximum",
"That the critical point at ``0`` is a relative minimum"
]
ans = 2
radioq(choices, ans, keep_order=true)
answ = 2
radioq(choices, answ, keep_order=true)
```
###### Question
@@ -801,8 +805,8 @@ choices = [
" ``f(x)`` is continuous and differentiable at ``2`` and has a critical point",
" ``f(x)`` is continuous and differentiable at ``2`` and has a critical point that is a relative minimum by the second derivative test"
]
ans = 3
radioq(choices, ans, keep_order=true)
answ = 3
radioq(choices, answ, keep_order=true)
```
@@ -810,22 +814,26 @@ radioq(choices, ans, keep_order=true)
Find the smallest critical point of $f(x) = x^3 e^{-x}$.
```julia; hold=true; echo=false
f(x)= x^3*exp(-x)
cps = find_zeros(D(f), -5, 10)
val = minimum(cps)
numericq(val)
```julia; echo=false
let
f(x)= x^3*exp(-x)
cps = find_zeros(D(f), -5, 10)
val = minimum(cps)
numericq(val)
end
```
###### Question
How many critical points does $f(x) = x^5 - x + 1$ have?
```julia; hold=true; echo=false
f(x) = x^5 - x + 1
cps = find_zeros(D(f), -3, 3)
val = length(cps)
numericq(val)
```julia; echo=false
let
f(x) = x^5 - x + 1
cps = find_zeros(D(f), -3, 3)
val = length(cps)
numericq(val)
end
```
###### Question
@@ -833,11 +841,13 @@ numericq(val)
How many inflection points does $f(x) = x^5 - x + 1$ have?
```julia; hold=true; echo=false
f(x) = x^5 - x + 1
cps = find_zeros(D(f,2), -3, 3)
val = length(cps)
numericq(val)
```julia; echo=false
let
f(x) = x^5 - x + 1
cps = find_zeros(D(f,2), -3, 3)
val = length(cps)
numericq(val)
end
```
###### Question
@@ -850,8 +860,8 @@ choices = [
"No, the second derivative test is possibly inconclusive",
"Yes"
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
@@ -865,15 +875,17 @@ choices = [
"No, the second derivative test is possibly inconclusive if ``c=0``, but otherwise yes",
"Yes"
]
ans = 2
radioq(choices, ans)
answ = 2
radioq(choices, answ)
```
###### Question
```julia; hold=true; echo=false
f(x) = exp(-x) * sin(pi*x)
plot(D(f), 0, 3)
```julia; echo=false
let
f(x) = exp(-x) * sin(pi*x)
plot(D(f), 0, 3)
end
```
The graph shows $f'(x)$. Is it possible that $f(x) = e^{-x} \sin(\pi x)$?
@@ -931,8 +943,8 @@ choices = [
"The critical points are at ``x=1`` (a relative minimum), ``x=2`` (not a relative extrema), and ``x=3`` (a relative minimum).",
"The critical points are at ``x=1`` (a relative minimum), ``x=2`` (a relative minimum), and ``x=3`` (a relative minimum).",
]
ans=1
radioq(choices, ans)
answ=1
radioq(choices, answ)
```
##### Question
@@ -945,8 +957,8 @@ choices = [
"The function is decreasing over ``(-\\infty, 1)`` and increasing over ``(1, \\infty)``",
"The function is negative over ``(-\\infty, 1)`` and positive over ``(1, \\infty)``",
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
##### Question
@@ -957,8 +969,8 @@ While driving we accelerate to get through a light before it turns red. However,
choices = ["A zero of the function",
"A critical point for the function",
"An inflection point for the function"]
ans = 3
radioq(choices, ans, keep_order=true)
answ = 3
radioq(choices, answ, keep_order=true)
```
###### Question