A room freshner bottle in the shape of an inverted cone sprays the perfume at regular intervals such that volume of the perfume in the bottle decreases at the steady rate of 1 mm3/min. Find the rate at which level of perfume is dropping at an instant when level of perfume in the bottle is 10 mm, if the semi-vertical angle of conical bottle is $\frac{\pi}{6}$
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Detailed Solution
Step 1: Define Variables and Geometry
Let $h$ be the height of the perfume level, $r$ be the radius of the surface of the perfume, and $V$ be the volume of the perfume. The bottle is an inverted cone with semi-vertical angle $\alpha = \frac{\pi}{6}$. From the geometry of the cone, we have the relation: $$r = h \tan(\alpha) = h \tan\left(\frac{\pi}{6}\right) = \frac{h}{\sqrt{3}}$$
Step 2: Express Volume in terms of $h$
The volume of a cone is given by $V = \frac{1}{3}\pi r^2 h$. Substituting $r = \frac{h}{\sqrt{3}}$: $$V = \frac{1}{3}\pi \left(\frac{h}{\sqrt{3}}\right)^2 h = \frac{1}{3}\pi \left(\frac{h^2}{3}\right) h = \frac{\pi h^3}{9}$$
Step 3: Differentiate with respect to time $t$
Given $\frac{dV}{dt} = -1$ mm$^3$/min. Differentiating $V = \frac{\pi h^3}{9}$ with respect to $t$: $$\frac{dV}{dt} = \frac{\pi}{9} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{3} \cdot \frac{dh}{dt}$$
Step 4: Calculate the rate of change
Substitute $h = 10$ and $\frac{dV}{dt} = -1$: $$-1 = \frac{\pi (10)^2}{3} \cdot \frac{dh}{dt}$$ $$-1 = \frac{100\pi}{3} \cdot \frac{dh}{dt}$$ $$\frac{dh}{dt} = -\frac{3}{100\pi}$$ The level is dropping at a rate of $\frac{3}{100\pi}$ mm/min.
Final Answer: \frac{3}{100\pi} \text{ mm/min}
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