A right circular cone with radius R and height H contains a liquid which evaporates at a rate propor

Q: Why can't we just use energy conservation to solve this?

Energy conservation is a powerful tool for closed systems where energy is conserved. However, this is an open system where mass (liquid) is leaving the cone due to evaporation. We must rely on the definition of the rate of change of volume and the geometric constraints of the cone.

Q: Does the radius $R$ really not matter? That feels counter-intuitive.

It does feel strange! But mathematically, the surface area $S$ grows as $r^2$, and the volume $V$ grows as $r^2 h$. When you differentiate the volume, the $r^2$ terms from the geometry and the evaporation law cancel out perfectly. The time taken depends only on the height $H$ and the evaporation constant $k$.

Q: Why is the rate of change of volume negative?

The negative sign is essential because the liquid is evaporating, meaning the volume is decreasing over time. If you omit the negative sign, you would be calculating the time for the cone to fill up, not empty, and you would end up with a negative time value, which is physically impossible.

Question

A right circular cone with radius R and height H contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = k > 0). Find the time after which the cone is empty.

Sigma Percentile · Accepted Answer

### The Geometry of Change

Imagine you are standing before a perfect right circular cone, filled to the brim with a mysterious liquid. As time ticks away, this liquid begins to vanish, evaporating into the air.

This is not just a simple problem of subtraction; it is a dynamic process where the geometry of the container dictates the rate of the change. To solve this, we must first bridge the gap between the static shape of the cone and the dynamic process of evaporation.

### The Bridge: Similar Triangles

At any arbitrary time $t$, the liquid forms a smaller cone inside the larger one. Let the height of this liquid be $h$ and its radius be $r$.

Because the liquid conforms to the shape of the cone, the ratio of the radius to the height must remain constant. By the property of similar triangles, we have the elegant relationship:

$$\frac{r}{h} = \frac{R}{H} \implies r = \frac{R}{H} h$$

This is our first key insight. It allows us to express the radius $r$ solely in terms of the height $h$. Without this, we would be trapped with two variables, $r$ and $h$, making our differential equation unsolvable.

### The Physics: The Law of Evaporation

The problem states that the liquid evaporates at a rate proportional to its surface area in contact with air. The surface of the liquid is a circle with radius $r$, so its area $S$ is $\pi r^2$.

The rate of change of volume is therefore:

$$\frac{dV}{dt} = -k S = -k \pi r^2$$

The negative sign is our physical reminder that the volume is decreasing. Now, we must express the volume $V$ of the liquid cone in terms of $h$.

The volume of a cone is $V = \frac{1}{3} \pi r^2 h$. Substituting our expression for $r$, we get:

$$V = \frac{1}{3} \pi \left( \frac{R}{H} h \right)^2 h = \frac{1}{3} \pi \frac{R^2}{H^2} h^3$$

### The Calculus: The Beauty of Cancellation

Now, we differentiate this volume with respect to time $t$ using the chain rule:

$$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{1}{3} \pi \frac{R^2}{H^2} h^3 \right) = \pi \frac{R^2}{H^2} h^2 \frac{dh}{dt}$$

We now have two expressions for $\frac{dV}{dt}$. Equating them gives us:

$$\pi \frac{R^2}{H^2} h^2 \frac{dh}{dt} = -k \pi \left( \frac{R}{H} h \right)^2$$

Look closely at this equation. The terms $\pi$, $\frac{R^2}{H^2}$, and $h^2$ appear on both sides. They cancel out completely! We are left with the remarkably simple differential equation:

$$\frac{dh}{dt} = -k$$

### The Conclusion: A Surprising Result

This tells us that the height of the liquid decreases at a constant rate, regardless of the cone's radius. To find the total time $T$ to empty the cone, we integrate from the initial height $H$ to $0$:

$$\int_{H}^{0} dh = \int_{0}^{T} -k dt \implies -H = -kT \implies T = \frac{H}{k}$$

The radius $R$ has vanished from our final answer! This means that whether the cone is wide or narrow, if the height is the same, it will take the exact same time to empty.

This is the elegance of physics—finding the simple truth hidden within the complexity. The final time required is **$T = \frac{H}{k}$**.

Animated Solution for Mathematics - Differentiation: A right circular cone with radius R and height H contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant = k > 0). Find the time after which the cone is empty.

Visualized Solution

Visualizing the Evaporating Cone

The Rate of Evaporation

Expressing Surface Area

Geometric Relationship

Volume of the Liquid

Simplifying the Volume Expression

Rate of Change of Volume

Simplified Rate of Change

Equating the Two Expressions for $\frac{d V}{d t}$

The Differential Equation for $h$

Setting up the Integral

Evaluating the Integral

Final Time to Empty the Cone

The Sigma Insight: Tangents, Normals and Rate Measure

The Geometry of Change

The Bridge

The Physics

The Calculus

The Conclusion

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