A rope is swinging in such a way that the length of the arc traced by a knot at its bottom end is decreasing geometrically. If the third arc is 22 ft. long and the seventh arc is 10 ft. long, what is the length of the arc on the sixth swing? Round your answer to the nearest tenth of a foot.
Let $a$ equal the length of the first swing, and let $r$ equal the common ratio between swing lengths. Since the length of the third arc is $22$ feet, we know that $ar^2=22$. Also, by similar reasoning, we can equate $ar^6=10$ feet.
To find the length of the sixth swing, we can compare it to the seventh and the fifth:
[asy]
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label("1", (-12.75,10)); label("2", (-6.75,10)); label("3", (-3.25,10)); label("4", (-2.8,10)); label("5", (-1.6,10)); label("6", (-1,10)); label("7", (-.625,10));
draw(Arc((0,10),2,180,360),linewidth(1));
draw(Arc((0,10),2,180,360+180),linewidth(1));
draw(Arc((0,10),2,180,360+360),linewidth(1));
draw(Arc((0,10),2,180,360+540),linetype("4 4"),Arrows(6));
draw(Arc((0,10),2,180,360+720),linetype("4 4"),Arrows(6));
draw(Arc((0,10),2,180,360+900),linewidth(1));
draw(Arc((0,10),2,180,360+1080),linewidth(1));
draw(arc((0,10),34/3.14-8,34/3.14-9),BeginArrow(6)); draw(arc((0,10),34/3.14-10,34/3.14-11),BeginArrow(6)); draw(arc((0,10),34/3.14-13,34/3.14-14),BeginArrow(6)); draw((34/3.14-2,-8.7)..(34/3.14-2,-8.97)..(34/3.14-2,-9.3),linewidth(0.7)); draw((34/3.14-2,-11.1)..(34/3.14-2,-10.83)..(34/3.14-2,-10.5),linewidth(0.7)); draw((34/3.14-2,-13.5)..(34/3.14-2,-13.23)..(34/3.14-2,-12.9),linewidth(0.7));
fill((-2.8,10)--(34/3.14-2,-10.5)--(34/3.14-2,-9.3)--cycle,gray);
[/asy]
The rope swings $6$ times and retraces the path of swing $1$ (back and forth), which totals twice the length of $a$, less two arc lengths. The ratio of these lengths is thus $\frac{2a-2ar^2}{10}=\frac{a(2-r^2)}{10}$.
Simplifying this, we have $(2-r^2)=\frac{10}{a(2-r^6)}$. Substituting $r^2 = \frac{22}{a}$, we have $(2-\frac{22}{a})=\frac{10}{a(2-r^6)}$. Cross-multiplying, we have $a(2-r^6) = (2)(2-\frac{22}{a})(10)$, or $2-r^6 = \frac{40-44/a}{a}$.
We're interested in solving for $2-r^6$ when $10=\frac{40-44/a}{a}$.
Cross-multiplying and moving all the terms to one side of the equation, we get
$10 a = 40 - \frac{44}{a}$, or $10a^2 - 40a + 44 = 0$.
Since $a$ is a positive real number (lengths always are!), we round up using the quadratic formula to find that $a \approx 2.2$ or $a \approx 2$. We know that swing $a$ is less than $22$ feet, so $a \approx 2$.
Since we have that $ar^6 = 10$, we can conclude that $\frac{2}{r^6} = 10$, or $r^6 = \frac{2}{10}=\frac{1}{5}$ Taking the square root of both sides yields $r^3 = \frac{1}{\sqrt{5}}$.
$r^2 = \frac{1}{\sqrt{5}} \cdot r = \frac{1}{\sqrt{5}} \cdot \sqrt{\frac{1}{\sqrt{5}}} = \frac{1}{\sqrt{5^3}} = \frac{1}{5\sqrt{5}}$. Multiplying this by $ar^2 = 22$, we get
$r^2ar^2 = \frac{22}{5\sqrt{5}} = \frac{22\cdot 5 \cdot \sqrt{5}}{25}=2\sqrt{5}$.
Multiplying $2\sqrt{5}$ by $r$ results in $2\sqrt{5} \cdot r = 2\sqrt{5} \cdot \frac{1}{\sqrt{5}} = 2$. Finally, multiplying by the common ratio of $2$ seven more times yields
$(2\cdot 2^7) \cdot 2 = 2 \cdot (2^8) = \boxed{512}$.
Additionally, notice that the sum of the lengths of the first 11 swings should be greater than or equal to the length of the first 7 swings. Plugging in 10 for a gives $22(1 + r^2 + r^4 + r^6 + r^8 + r^{10} + r^{12} + r^{14}) > 10(1 + r^2 + \cdots + r^{12})$.
Divide both sides by $r^2$ ($r\neq 0$) to get $22(\frac{1}{r^2}+\frac{1}{r^2}r^2 + \frac{1}{r^2}r^4 + \cdots ) > 10(\frac{1}{r^2}+\frac{1}{r^2}r^2 + \cdots r^{12})$.
Plug in $r^2=\frac{22}{a}$ and $r^6 = \frac{10}{a}$ to get:
$22(\frac{a}{22}+\frac{a^3}{22^2} + \frac{a^5}{22^3} + \cdots ) > 10(\frac{22}{a}+\frac{a}{22} + \cdots (\frac{a}{22})^6 )$
Multiplying by $\frac{22}{a}$ ($a\neq 0$) yields $22^2 (\frac{1}{22}+\frac{a^2}{22^2} + \frac{a^4}{22^3} + \cdots ) > 10(22 + a + \cdots (\frac{a}{22})^5 )$
We have $22 + a + \cdots (\frac{a}{22})^{12} = 10.(22^2 ) > 10( 22^2 )$.
Which is clearly false. Thus $22 + a + \cdots (\frac{a}{22})^{12} > 10( 22^2 )$ and likewise, $22 + a + \cdots (\frac{a}{22})^{12} > 10( 22^2 ) > 2^8+2^7+2^6+2^5+2^4+2^3+2^2$. As the third arc is 22 feet, it is obvious that the length of the eighth arc, $22-a-(a^3)-(a^5)-\ldots > 22- 2^2-2^3 - \ldots -2^7$ feet, or $22.89>\text{length of sixth arc}$.
The only number choice that is a solution to this inequality is $\boxed{512}$.