In the March 2003 issue of the College Mathematics Journal, Leon M Hall posed 12 questions related to the following figure:
```{julia}
#| echo: false
f(x) = x^2
fp(x) = 2x
a₀ = 7/8
q₀ = -a₀ - 1/(2a₀)
tangent(x) = f(a₀) + fp(a₀) * (x - a₀)
normal(x) = f(a₀) - (1 / fp(a₀)) * (x - a₀)
function make_plot(a₀=7/8, q₀=-a₀ - 1/2a₀)
plt = plot(; xlim=(-2,2), ylim=(-1, (1.5)^2),
xticks=nothing, yticks=nothing,
aspect_ratio=:equal, border=:none, legend=false)
f(x) = x^2
fp(x) = 2x
plot!(f, -1.5, 1.5)
plot!(zero)
tl = x -> f(a₀) + fp(a₀) * (x-a₀)
nl = x -> f(a₀) - 1/(fp(a₀)) * (x-a₀)
plot!(tl, -0.02, 1.6; linecolor=:black)
plot!(nl, -1.6, 1; linecolor=:black)
# add in right triangle
scatter!([a₀, q₀], f.([a₀, q₀]), markersize=5)
Δ = 0.01
annotate!([(a₀ + Δ, nl(a₀+Δ), "P", :bottom),
(q₀ - Δ, nl(q₀-Δ), "Q", :top)])
plt
end
make_plot()
```
The figure shows $f(x) = x^2$, the tangent line at $(a, f(a))$ (for $a > 0$), and the *normal* line at $(a, f(a))$. The questions all involve finding the value $a$ which minimizes a related quantity.
We set up some variables to work symbolically:
```{julia}
@syms a::positive x::real
f(x) = x^2
fp(x) = 2x
m = fp(a)
mᴵ = - 1/m
tl = f(a) + m * (x - a)
nl = f(a) + mᴵ * (x - a)
zs = solve(f(x) ~ nl, x)
q = only(filter(!=(a), zs))
```
----
The first question is simply:
> 1a. The $y$ coordinate of $Q$
```{julia}
#| echo: false
let
p = make_plot()
plot!(p, [q₀,q₀], [0,normal(q₀)], linewidth=5)
end
```
The value is $f(q)$
```{julia}
yvalue = f(q)
```
To minimize we solve for critical points:
```{julia}
cps = solve(diff(yvalue, a), a)
```
The lone critical point must be at a minimum. (Given the geometry of the problem, as $a$ goes to $\infty$ the height does too, and as $a$ goes to $0$ the height will also go to $\infty$. This can also be seen analytically, as $q = -a - 1/(2a)$ which goes to $-\infty$ when $a$ heads to $0$ or $\infty$.)
::: {.callout-note}
## We hide the code
In the remaining examples we don't show the code by default.
> 2c. The volume of the rotated solid formed by revolving the parabolic segment around the vertical line $k$ units to the right of $P$ or to the left of $Q$ where $k > 0$.
```{julia}
#| code-fold: true
#| code-summary: "Show the code"
@syms k::nonnegative
V = simplify(integrate(PI * (nl - f(x) - k)^2, (x, q, a)));
```
----
> 3. The $y$ coordinate of the centroid of the parabolic segment
# use parametric and 2π ∫ u(t) √(u'(t)^2 + v'(t)^2) dt
uu(x) = a - x
vv(x) = f(uu(x))
SA = 2PI * integrate(uu(x) * sqrt(diff(uu(x),x)^2 + diff(vv(x),x)^2), (x, q, a));
```
----
> 9. The height of the parabolic segment (i.e. the distance between the normal line and the tangent line to the parabola that is parallel to the normal line)
```{julia}
#| echo: false
# distance point to a line
# > solve(diff(x^2,x) ~ -1/fp(a), x) # mean value theorem
