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 $1 \text{ mm}^3/\text{min}$ . Find the rate at which level of perfume is dropping at an instant when level of perfume in the bottle is $10 \text{ mm}$ , if the semi-vertical angle of conical bottle is $\pi/6$ .

Q: 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 $1 \text{ mm}^3/\text{min}$. Find the rate at which level of perfume is dropping at an instant when level of perfume in the bottle is $10 \text{ mm}$, if the semi-vertical angle of conical bottle is $\pi/6$.

The level of perfume is dropping at a rate of **3/(100π) mm/min** ($\approx 0.00955 \text{ mm/min}$). **Setup:** For a cone with semi-vertical angle α, radius $r = h \cdot \tan\alpha$, so volume $V = (\pi/3)h^3\tan^2\alpha$. With $\alpha = \pi/6$, $\tan^2\alpha = 1/3$, giving $V = \pi h^3/9$. **Applying rates:** $dV/dt = (\pi h^2/3) \cdot dh/dt$. Substituting $dV/dt = -1 \text{ mm}^3/\text{min}$ and $h = 10 \text{ mm}$ gives $dh/dt = -3/(100\pi) \text{ mm/min}$, i.e., the level is dropping at 3/(100π) mm/min.

Q: How do you eliminate the radius r from the cone's volume formula?

Using the semi-vertical angle α, $r = h \cdot \tan\alpha$, so $r^2 = h^2\tan^2\alpha$. Substituting this into $V = (1/3)\pi r^2 h$ gives V purely as a function of h: $V = (\pi/3)\tan^2\alpha \cdot h^3$.

Inverted conical room-freshener bottle with semi-vertical angle 30 degrees and perfume level 10 mm

Application of Derivatives

Related Rates - Cone Volume

Official Answer

The level of perfume is dropping at a rate of 3/(100π) mm/min ( $\approx 0.00955 \text{ mm/min}$ ).

Setup: For a cone with semi-vertical angle α, radius $r = h \cdot \tan\alpha$ , so volume $V = (\pi/3)h^3\tan^2\alpha$ . With $\alpha = \pi/6$ , $\tan^2\alpha = 1/3$ , giving $V = \pi h^3/9$ .

Applying rates: $dV/dt = (\pi h^2/3) \cdot dh/dt$ . Substituting $dV/dt = -1 \text{ mm}^3/\text{min}$ and $h = 10 \text{ mm}$ gives $dh/dt = -3/(100\pi) \text{ mm/min}$ , i.e., the level is dropping at 3/(100π) mm/min.

related ratesrate of changecone volumesemi-vertical angledV/dtdh/dtimplicit differentiation with time

Marking Scheme

11 mark: correctly expressing $V = (\pi/3)h^3\tan^2(\pi/6) = \pi h^3/9$ by eliminating r.
21 mark: differentiating to get $dV/dt = (\pi h^2/3) dh/dt$ , substituting $dV/dt=-1$ , $h=10$ , and solving $dh/dt = -3/(100\pi) \text{ mm/min}$ .

Hint

Use $r = h \cdot \tan(\text{semi-vertical angle})$ to write V purely as a function of h, then differentiate with respect to time.

Quick Oral Answer

Since $r = h \cdot \tan(\pi/6)$ , I write $V$ purely in terms of $h$ as $\pi h^3/9$ , differentiate with respect to time, then substitute $dV/dt = -1$ and $h = 10$ to get $dh/dt = -3/(100\pi) \text{ mm/min}$ .

Analysis & Explanation

This is a classic related-rates (rate of change) problem combining cone geometry with time-differentiation.

Concept: Since the cone's radius and height are linked by the fixed semi-vertical angle ( $r = h \cdot \tan\alpha$ ), volume can be expressed purely in terms of h, allowing dV/dt and dh/dt to be related directly via the chain rule.

Key steps: Express $V(h) = (\pi/3)\tan^2\alpha \cdot h^3$ , differentiate with respect to time to get dV/dt in terms of $h^2 \cdot dh/dt$ , then substitute the known instantaneous values ( $dV/dt = -1 \text{ mm}^3/\text{min}$ since volume decreases, $h = 10 \text{ mm}$ ) to isolate dh/dt.

Exam trap: Students often forget to convert r into a pure function of h before differentiating, instead keeping two variables (r and h) and getting stuck without a second equation. Always eliminate r first using $r = h \cdot \tan(\text{semi-vertical angle})$ .

Real-world link: This models any conical container (perfume bottle, ice-cream cone, funnel) where fluid drains or evaporates at a known volumetric rate — engineers use identical related-rates reasoning to size drain valves or estimate depletion times.

Common Mistakes

1Keeping both r and h as separate variables in $V = (1/3)\pi r^2 h$ without eliminating r using the semi-vertical angle relation.
2Sign errors — forgetting that volume is decreasing, so $dV/dt$ should be substituted as $-1$ , not $+1$ .
3Using $\tan\alpha$ instead of $\tan^2\alpha$ when substituting $r^2 = h^2\tan^2\alpha$ into the volume formula.

Interesting Facts

This exact 'inverted cone' related-rates setup (funnel, ice-cream cone, or conical tank) is one of the most frequently recurring problem types across CBSE and NCERT calculus exercises.

Semi-vertical angle π/6 (30°) is a favourite exam choice because $\tan(\pi/6) = 1/\sqrt{3}$ gives a clean $\tan^2 = 1/3$ , simplifying the volume formula to $V = \pi h^3/9$ .

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Frequently Asked Questions

Why is $dV/dt$ taken as negative in this problem?

Because the perfume volume is decreasing over time (it is being sprayed out), so its rate of change with respect to time is negative: $dV/dt = -1 \text{ mm}^3/\text{min}$ .

How do you eliminate the radius r from the cone's volume formula?

Using the semi-vertical angle α, $r = h \cdot \tan\alpha$ , so $r^2 = h^2\tan^2\alpha$ . Substituting this into $V = (1/3)\pi r^2 h$ gives V purely as a function of h: $V = (\pi/3)\tan^2\alpha \cdot h^3$ .