## University Calculus: Early Transcendentals (3rd Edition)

$\frac{ds}{dt}=1\frac{cm}{min}$
Given $s=20$ $\frac{dV}{dt}=1200$ $V=s^3$ on differentiating above, we get: $\frac{dV}{dt}=3s^2\frac{ds}{dt}$ $\frac{dV}{dt}=3\times20^2\frac{ds}{dt}$ $\frac{ds}{dt}=\frac{1200}{1200}=1$ thus, $\frac{ds}{dt}=1\frac{cm}{min}$