Single-Variable-Calculus · Unit 11 · Video 1 · Interactive Practice

The Hanging Ring: Where It Settles, and Why the Two Pieces of String Make Equal Angles

IKey Formulas

Formula	Name	What it tells you
$\sqrt{x^2 + y^2} + \sqrt{(x-a)^2 + (y-b)^2} = L$	String-length constraint	Ring at $(x, y)$ , pins at $(0, 0)$ and $(a, b)$ ; it defines the ring's height $y = y(x)$ implicitly
$\dfrac{x + y\,y'}{\sqrt{x^2 + y^2}} + \dfrac{x - a + (y - b)\,y'}{\sqrt{(x-a)^2 + (y-b)^2}} = 0$	Implicit derivative	Chain rule on each square root, remembering $y$ depends on $x$ ; the constant $L$ differentiates to $0$
$y' = 0:\ \ \dfrac{x}{\sqrt{x^2 + y^2}} = \dfrac{a - x}{\sqrt{(x-a)^2 + (y-b)^2}}$	Lowest point	Each side is a sine: $\sin\alpha = \sin\beta$ , and with $0\degree < \alpha, \beta < 90\degree$ this gives $\alpha = \beta$
$T_1 \sin\alpha = T_2 \sin\beta \ \Rightarrow\ T_1 = T_2$	Horizontal force balance	Equal angles with the vertical mean equal tension in the two pieces

Key Insight: The ring's possible positions form an ellipse with the pins as foci. At the lowest point the two pieces make equal angles with the horizontal tangent; turning the ellipse moves that fact to every point, so a ray from one focus always reflects to the other.

IIVisualization 1 — Where Does the Ring Settle?

Along the taut string the ring's height is a function $y(x)$ ; it settles where $y' = 0$ .

IIIVisualization 2 — From $y' = 0$ to Equal Angles and Equal Tension

With pins at $(0, 0)$ and $(6, 2)$ and $L = 10$ , the ring rests at $(2.25, -3)$ .

Step 1

Step 1 — The constraint, from two right triangles

\sqrt{2.25^2 + 3^2} = 3.75, \qquad \sqrt{3.75^2 + 5^2} = 6.25

\sqrt{x^2 + y^2} + \sqrt{(x-6)^2 + (y-2)^2} = 3.75 + 6.25 = 10 = L

Step 2 — Differentiate implicitly

\frac{d}{dx}\left(x^2 + y^2\right)^{1/2} = \tfrac12\left(x^2 + y^2\right)^{-1/2}\left(2x + 2y\,y'\right) = \frac{x + y\,y'}{\sqrt{x^2 + y^2}}

\frac{x + y\,y'}{\sqrt{x^2 + y^2}} + \frac{x - 6 + (y - 2)\,y'}{\sqrt{(x-6)^2 + (y-2)^2}} = 0

Step 3 — Lowest point: $y' = 0$

\frac{x}{\sqrt{x^2 + y^2}} = \frac{6 - x}{\sqrt{(x-6)^2 + (y-2)^2}}

\frac{2.25}{3.75} = 0.6 = \frac{3.75}{6.25}\ \checkmark

Step 4 — Read the sides as sines

\sin\alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2.25}{3.75}, \qquad \sin\beta = \frac{3.75}{6.25}

\sin\alpha = \sin\beta = 0.6,\ \ 0\degree < \alpha, \beta < 90\degree \ \Rightarrow\ \alpha = \beta \approx 36.87\degree

Step 5 — Horizontal forces cancel

T_1 \sin\alpha = T_2 \sin\beta

\alpha = \beta,\ \sin\alpha \ne 0 \ \Rightarrow\ T_1 = T_2

IVVisualization 3 — Hole in One

Tee at one focus, hole at the other: every shot meets the wall at equal angles and drops in.

Aim θ 200° Ball 0%

VQuiz Questions

Problem 1 · Equal Sines at the Lowest Point

Given: The ring rests at its lowest point. The left piece of string has length $5$ , and the ring sits $x = 3$ to the right of the left pin. The right piece has horizontal leg $a - x = 6$ . Find the length of the right piece.

✅ Correct!

\sin\beta = \sin\alpha = \tfrac35

, so the right piece is

6 \div \tfrac35 = 10

❌ Not quite.

6 \div \tfrac45

uses

\cos\alpha

. The horizontal leg is opposite the angle with the vertical, so it pairs with the sine,

\tfrac35

❌ Not quite.

6 \cdot \tfrac35

would be shorter than its own leg. A hypotenuse is the leg divided by the sine.

❌ Not quite.

6

is the horizontal leg; the piece of string is the hypotenuse of its right triangle, so it is longer.

❌ Not quite. Use

\sin\alpha = \sin\beta

at the lowest point.

Show solution

In the left triangle the horizontal leg $x = 3$ is opposite $\alpha$ and the hypotenuse is the left piece:

\sin\alpha = \frac{x}{\sqrt{x^2 + y^2}} = \frac{3}{5}.

At the lowest point $y' = 0$ gives $\sin\alpha = \sin\beta$ , so in the right triangle

\sin\beta = \frac{a - x}{\text{right piece}} = \frac{6}{\text{right piece}} = \frac35 \quad\Rightarrow\quad \text{right piece} = 10.

The string has total length $L = 5 + 10 = 15$ .

Problem 2 · Differentiating the First Square Root

Given: $y$ is defined implicitly as a function of $x$ by the string constraint. Find $\dfrac{d}{dx}\sqrt{x^2 + y^2}$ .

✅ Correct! The chain rule gives

\tfrac12(x^2+y^2)^{-1/2}(2x + 2y\,y')

, and the factors of

2

cancel.

❌ Close, but… this treats

y

as a constant. Since

y = y(x)

, the derivative of

y^2

2y\,y'

, not

0

❌ Not quite. That is only the outer derivative; it still has to be multiplied by the derivative of the inside,

2x + 2y\,y'

❌ Not quite. The

\tfrac12

from the power

\tfrac12

was dropped; it cancels the

2

❌ Not quite. Write the root as

(x^2 + y^2)^{1/2}

and apply the chain rule.

Show solution

Write the root as a power and use the chain rule, remembering $\frac{d}{dx}y^2 = 2y\,y'$ :

\frac{d}{dx}\left(x^2 + y^2\right)^{1/2} = \frac12\left(x^2 + y^2\right)^{-1/2}\left(2x + 2y\,y'\right) = \frac{x + y\,y'}{\sqrt{x^2 + y^2}}.

At the lowest point $y' = 0$ and this becomes $\dfrac{x}{\sqrt{x^2 + y^2}} = \sin\alpha$ , but only after differentiating correctly.

Problem 3 · Locate the Lowest Point

Given: Pins at $(0, 0)$ and $(a, b) = (8, 3)$ , string length $L = 10$ . At the lowest point $\sin\alpha = \sin\beta$ . Find $\sin\alpha$ and the ring's resting point $(x, y)$ .

What is $\sin\alpha$ at the lowest point?

Where does the ring rest?

✅ Correct!

r_1 = 2.5

r_2 = 7.5

, so

(x, y) = (2.5 \cdot \tfrac45,\ -2.5 \cdot \tfrac35) = (2, -\tfrac32)

❌ Check $\sin\alpha$ .

\dfrac{x}{r_1} = \dfrac{a - x}{r_2}

are equal ratios, so they also equal

\dfrac{x + (a - x)}{r_1 + r_2} = \dfrac{a}{L}

❌ Not quite.

\tfrac{b}{a}

is the slope of the line between the pins. The sine is a horizontal leg over a piece of string:

\tfrac{a}{L}

❌ Check the point. It must satisfy

r_1 + r_2 = 10

; with

\cos\alpha = \tfrac35

, the vertical legs give

b = (r_2 - r_1)\cos\alpha

❌ Close, but…

(4, -3)

is the answer for level pins,

b = 0

. With the right pin higher, the ring settles closer to the lower pin.

❌ Not quite. That is on the wrong side: the ring settles horizontally closer to the lower end, and this point is not on the string (

r_1 + r_2 \ne 10

Show solution

Step 1: the sine. Let $r_1, r_2$ be the two pieces. At the lowest point

\sin\alpha = \frac{x}{r_1} = \frac{a - x}{r_2} = \frac{x + (a - x)}{r_1 + r_2} = \frac{a}{L} = \frac{8}{10} = \frac45,

so $\cos\alpha = \tfrac35$ .

Step 2: the pieces. The vertical legs are $-y = r_1\cos\alpha$ and $b - y = r_2\cos\alpha$ . Subtracting, $b = (r_2 - r_1)\cos\alpha$ :

r_2 - r_1 = \frac{3}{3/5} = 5, \qquad r_1 + r_2 = 10 \ \Rightarrow\ r_1 = 2.5,\ r_2 = 7.5.

Step 3: the point. $x = r_1\sin\alpha = 2$ , $y = -r_1\cos\alpha = -\tfrac32$ .

Check: $\sqrt{2^2 + 1.5^2} = 2.5$ and $\sqrt{(2-8)^2 + (-1.5-3)^2} = \sqrt{56.25} = 7.5$ ; the sum is $10$ ✓. The ring sits left of the midpoint $x = 4$ , toward the lower pin.

Problem 4 · The Mini-Golf Ellipse

Given: A mini-golf wall is the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ , with the tee at the focus $(-4, 0)$ and the hole at the focus $(4, 0)$ . A ball leaves the tee in some direction, bounces once off the wall, and drops into the hole. Find the total distance the ball rolls.

✅ Correct! Tee → wall → hole is a "string" from focus to focus: its length is the major axis,

2 \cdot 5 = 10

, for every shot.

❌ Not quite.

8

is the straight distance between the foci; a ball that bounces off the wall travels farther.

❌ Not quite.

6

is the minor axis. The sum of distances to the foci equals the major axis.

❌ Not quite. The bounce point varies, but the sum of its distances to the two foci is the same at every point of an ellipse.

❌ Not quite. Think of the path as the string with its ends pinned at the foci.

Show solution

Here the semi-axes are $5$ and $3$ , and the foci sit at $\pm c$ with $c^2 = 25 - 9 = 16$ , so $c = 4$ : the tee and the hole are the foci.

By the reflection property, a shot from one focus bounces once at some point $P$ on the wall and heads straight to the other focus. Its path is $|PF_1| + |PF_2|$ , the length of the string that draws the ellipse:

|PF_1| + |PF_2| = 2 \cdot 5 = 10.

For example, $P = (5, 0)$ gives $9 + 1 = 10$ , and $P = (0, 3)$ gives $5 + 5 = 10$ . Every shot rolls exactly $10$ .

Solved: 0 / 4

The Hanging Ring: Where It Settles, and Why the Two Pieces of String Make Equal Angles

IKey Formulas

IIVisualization 1 — Where Does the Ring Settle?

IIIVisualization 2 — From y′=0y' = 0y′=0 to Equal Angles and Equal Tension

IVVisualization 3 — Hole in One

VQuiz Questions

IIIVisualization 2 — From $y' = 0$ to Equal Angles and Equal Tension