48  A dozen minima for a parabola

This section uses these packages:

using SymPy
using Plots
plotly()

Plots.PlotlyBackend()

---

In the March 2003 issue of the College Mathematics Journal, Leon M Hall posed 12 questions related to the following figure:

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:

@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))

$-a-\frac{1}{2a}$

---

The first question is simply:

1a. The $y$ coordinate of $Q$

The value is $f(q)$

yvalue = f(q)

${(-a-\frac{1}{2a})}^2$

To minimize we solve for critical points:

cps = solve(diff(yvalue, a), a)

$\left[\begin{array}{c}\frac{\sqrt{2}}{2}\end{array}\right]$

The lone critical point must be at a minimum. (Given the geometry of the problem, as $a$ goes to $\mathrm{\infty }$ the height does too, and as $a$ goes to $0$ the height will also go to $\mathrm{\infty }$. This can also be seen analytically, as $q=-a-1/(2a)$ which goes to $-\mathrm{\infty }$ when $a$ heads to $0$ or $\mathrm{\infty }$.)

We hide the code

In the remaining examples we don’t show the code by default.

---

1b. The length of the line segment $PQ$

lseg = sqrt((f(a) - f(q))^2 + (a - q)^2);

---

2a. The horizontal distance between $P$ and $Q$

hd = a - q;

---

2b. The area of the parabolic segment

A = simplify(integrate(nl - f(x), (x, q, a)));

---

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 $k>0$.

@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

We warm up with the $x$ coordinate, given by:

xₘ = integrate(x * (nl - f(x)), (x, q, a)) / A
simplify(xₘ)

$-\frac{1}{4a}$

a fact noted by the author.

yₘ = integrate( (1//2) * (nl^2 - f(x)^2), (x, q, a)) / A
yₘ = simplify(yₘ);

---

4. The length of the arc of the parabola between $P$ and $Q$

L = integrate(sqrt(1 + fp(x)^2), (x, q, a));

---

5. The $y$ coordinate of the midpoint ofthe line segment $PQ$

mp = nl(x => (a + q)/2);

---

6. The area of the trapezoid bound by the normal line, the $x$-axis, and the vertical lines through $P$ and $Q$.

trap = 1//2 * (f(q) + f(a)) * (a - q);

---

7. The area bounded by the parabola and the $x$ axis and the vertical lines through $P$ and $Q$

pa = integrate(x^2, (x, q, a));

---

8. The area of the surface formed by revolving the arc of the parabola between $P$ and $Q$ around the vertical line through $P$

# 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)

# find b through mean value theorem,
# then solve for point of intersection
b = only(solve(diff(f(x),x) ~ -1/fp(a), x))
b′ = only(solve(f(b) + fp(a)*(x-b) ~ nl, x))
segment_height = sqrt((b-b′)^2 + (f(b) - nl(x=>b′))^2);

---

10. The volume of the solid formed by revolving the parabolic segment around the $x$-axis

Vₓ = integrate(pi * (nl^2 - f(x)^2), (x, q, a));

---

11. The area of the triangle bound by the normal line, the vertical line through $Q$ and the $x$-axis

triangle = 1/2 * f(q) * (a - f(a)/(-1/fp(a)) - q);

---

12. The area of the quadrilateral bound by the normal line, the tangent line, the vertical line through $Q$ and the $x$-axis

# @syms x[1:4], y[1:4]
# v1, v2, v3 = [[x[i]-x[1],y[i]-y[1], 0] for i in 2:4]
# area = 1//2 * last(cross(v3,v2) + cross(v2, v1)) # 1/2 area of parallelogram
# print(simplify(area))
# -(x₁ - x₂)*(y₁ - y₃)/2 + (x₁ - x₃)*(y₁ - y₂)/2 - (x₁ - x₃)*(y₁ - y₄)/2 + (x₁ - x₄)*(y₁ - y₃)/2
tl₀ = a - f(a) / fp(a)
x₁,x₂,x₃,x₄ = (a,q,q,tl₀)
y₁, y₂, y₃, y₄ = (f(a), f(q), 0, 0)
quadrilateral = -(x₁ - x₂)*(y₁ - y₃)/2 + (x₁ - x₃)*(y₁ - y₂)/2 - (x₁ - x₃)*(y₁ - y₄)/2 + (x₁ - x₄)*(y₁ - y₃)/2;

---

The answers appear here in sorted order, some given as approximate floating point values:

article_answers = (1/(2sqrt(2)), 1/2, sqrt(3/10), 0.558480, 0.564641,
                   0.569723, 0.574646,
                   1/sqrt(3), 1/8^(1/4), 1/6^(1/4), .644004, 1/sqrt(2))

(0.35355339059327373, 0.5, 0.5477225575051661, 0.55848, 0.564641, 0.569723, 0.574646, 0.5773502691896258, 0.5946035575013605, 0.6389431042462724, 0.644004, 0.7071067811865475)