# Chapter 2 - Differentiation - 2.6 Exercises - Page 153: 17

$\frac{dh}{dt}=\frac{8}{405\pi}feet/min$

#### Work Step by Step

Step-1: The following data are given to us, $\frac{dV}{dt} = 10 cm^3/min$ $diameter\approx 3h\implies radius=1.5h$ The relation between volume of the cone and its height is, $$V = \frac{1}{3}\pi r^2 h$$ Since $r=1.5h$, $$V = \frac{1}{3}\pi \times (2.25)h^3=0.75\pi h^3$$ Step-2: Differentiate the above equation with respect to $h$, $$\frac{dV}{dt}=(2.25)\pi h^2 \frac{dh}{dt}$$ $$\implies \frac{dh}{dt}=\frac{dV}{dt} \times \frac{1}{(2.25)\pi h^2}=\frac{10}{(2.25)\pi h^2}$$ Step-3: When $h=15$feet, $$\frac{dh}{dt}=\frac{10}{(2.25)\pi (15)^2}=\frac{8}{405\pi}feet/min$$

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