A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its sem

Q: Why can't I just differentiate $V = \frac{1}{3}\pi r^2 h$ directly?

If you do that, you get $\frac{dV}{dt} = \frac{1}{3}\pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt})$. You have too many unknowns! You don't know $\frac{dr}{dt}$. By using the geometric constraint $r = \frac{3}{4}h$ first, you reduce the system to a single variable, making the differentiation clean and solvable.

Q: Wait, why do we need the slant height $l$ for the surface area?

The curved surface area of a cone is defined by the lateral surface, not the base. The formula is $S = \pi r l$. Since $l$ is the hypotenuse of the triangle formed by $r$ and $h$, it is also changing as the water level rises. You must express $l$ in terms of $h$ to keep your variables consistent.

Q: Is the rate of change of height $\frac{dh}{dt}$ constant?

Great question! Look at your derivative: $\frac{dV}{dt} = \frac{9\pi}{16}h^2 \frac{dh}{dt}$. Since $\frac{dV}{dt}$ is constant (6), $\frac{dh}{dt}$ depends on $h^2$. As the water level $h$ rises, the cross-sectional area of the water surface increases, so the height rises more slowly. $\frac{dh}{dt}$ is definitely not constant!

Question

A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is $	an^{-1} \frac{3}{4}$. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is ________

Sigma Percentile · Accepted Answer

### Analyzing the Geometric DNA

Every cone possesses a constant that defines its shape regardless of the volume of water contained within: the semi-vertical angle, $	heta$. We are given $	an 	heta = \frac{3}{4}$.

By examining the cross-section of the water, we observe a right-angled triangle where the vertical leg is the height $h$ and the horizontal leg is the radius $r$. The relationship is locked by the geometry:

$$\frac{r}{h} = 	an 	heta = \frac{3}{4} \Rightarrow r = \frac{3}{4}h$$

This relationship is our **golden key**. It allows us to collapse a two-variable problem into a single-variable reality, which is essential for solving dynamic systems efficiently.

### The Volume Dynamics

The volume of a cone is given by $V = \frac{1}{3}\pi r^2 h$. To avoid the complexity of differentiating with two variables, we substitute $r = \frac{3}{4}h$ into the volume formula:

$$V = \frac{1}{3}\pi \left( \frac{3}{4}h ight)^2 h = \frac{3\pi}{16}h^3$$

Now, the volume is purely a function of height. Differentiating with respect to time $t$ using the **Chain Rule**, we obtain:

$$\frac{dV}{dt} = \frac{9\pi}{16}h^2 \frac{dh}{dt}$$

Given $\frac{dV}{dt} = 6$ and evaluating at the moment $h = 4$, we solve for the rate of change of height:

$$6 = \frac{9\pi}{16}(4)^2 \frac{dh}{dt} \Rightarrow 6 = 9\pi \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{2}{3\pi}$$

### The Surface Area Evolution

The wet curved surface area $S$ is defined by $S = \pi r l$, where $l$ is the slant height. Using the Pythagorean theorem, $l = \sqrt{r^2 + h^2}$.

Substituting $r = \frac{3}{4}h$, we find the slant height in terms of $h$:

$$l = \sqrt{\left( \frac{3}{4}h ight)^2 + h^2} = \sqrt{\frac{25h^2}{16}} = \frac{5h}{4}$$

Substituting $r$ and $l$ back into the surface area formula yields:

$$S = \pi \left( \frac{3}{4}h ight) \left( \frac{5h}{4} ight) = \frac{15\pi}{16}h^2$$

### The Grand Finale

Differentiating $S$ with respect to $t$ gives us:

$$\frac{dS}{dt} = \frac{15\pi}{16} \cdot 2h \cdot \frac{dh}{dt} = \frac{15\pi}{8}h \frac{dh}{dt}$$

Finally, we substitute the known values $h = 4$ and $\frac{dh}{dt} = \frac{2}{3\pi}$ into the equation:

$$\frac{dS}{dt} = \frac{15\pi}{8} \cdot 4 \cdot \frac{2}{3\pi} = 5$$

The rate of change of the wet curved surface area is **$5 	ext{ m}^2/	ext{hr}$**. This result demonstrates the elegance of calculus: by respecting the geometry and applying the chain rule, we transform a complex dynamic system into a precise, satisfying solution.

Enter Numerical Value:

Visualized Solution

Visualizing the Cone

The Geometric Relation $r$ and $h$

Volume $V$ in Terms of Height $h$

Differentiating Volume $\frac{d V}{d t}$

Finding the Rate $\frac{d h}{d t}$

Wet Curved Surface Area $S$

Surface Area $S$ in Terms of $h$

Differentiating Surface Area $\frac{d S}{d t}$

Final Calculation for $\frac{d S}{d t}$

The Way Forward

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Geometric DNA

The Volume Dynamics

The Surface Area Evolution

The Grand Finale

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