# The Mathematical Ninja and the Variable Volume

The student, at the third time of asking, navigated the perilous straits of negative powers and fractions of $\pi$ and came to rest, exhausted, on the answer: “$r^3 = \frac{500}{\pi}$,” he said.

The Mathematical Ninja stopped poking him with the foam sword (going soft? perhaps. Or perhaps this student needed nudging more than skewering.) “I suppose you want to calculate that?”

“Um… am I allowed to?”

“HA!” Another poke. “Well, OK. It turns out $r \approx 5.4$, but OK.”

The student sighed. “Go on, then.”

“First, take out the factor of 125, making $r = 5 \sqrt[3]{\frac{4}{\pi}}$.”

“I think I see that,” lied the student.

“Now, because $\pi \approx \frac{22}{7}$, what’s inside the cube root? It’s about $\frac{28}{22}$, or $\frac{14}{11}$.”

“And you happen to know the cube root of $\frac{14}{11}$ because of course you do.”

“I don’t, but I can approximate it - I know that the cube root of $1+3x$ is about $1+x$ for smallish $x$ - and what’s under the root is $1 + \frac{3}{11}$.”

“So the cube root is $1 + \frac{1}{11}$? That’s neat.”

”$\frac{12}{11}$ is neater. So we have something in the region of $5\times \frac{12}{11}$, which is 5.45. However, the cube root is a slight overestimate (($\br{1+3x}^\frac{1}{3} \approx 1 + x - x^2 + \frac{5}{3}x^3 + …$)), so I’d round down.”

“I think I’d use the calculator,” said the student.

“Just because this is a foam sword doesn’t mean I won’t make you walk the plank,” said the Mathematical Ninja.