https://spherical-harmonics.hateblo.jp/entry/Todai/1912/Kougaku

[5] The altitude of a circular cone is 100 inches and decreasing at the rate of 10 inches per second ; and radius of the base is 50 inches and increasing at the rate of 5 inches per second. At what rate is the volume changing ?

2019.03.06記

[5] 円錐について、高さが100吋から秒速10吋で減少し、底面の半径が50吋から秒速5吋で増加するとき，体積の変化率は如何程か？

[解答]
円錐の体積は $V(t)=\dfrac{1}{3}\pi(100-10t)(50+5t)^2$ $=\dfrac{250}{3}\pi(1000-10t^2+100t-t^3)$ だから， $\displaystyle\lim_{t\to 0+0} V'(t)=\dfrac{25000}{3}\pi [吋^3/秒]$ となる．

■ もっと大雑把に
$V+dV=\dfrac{1}{3}\pi(100-10dt)(50+5dt)^2$ $=\dfrac{1}{3}\pi(100-10dt)(50+5dt)(50+5dt)$ $=\dfrac{1}{3}\pi\{250000+(-25000+25000+25000)dt\}$ $=V+\dfrac{25000}{3}\pi dt$
として求めても良い．