Water is being filled at the rate of 1 cm³/sec in a right circular conical vessel (vertex downwards)

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

You absolutely could, but you would be left with two unknowns, $\frac{dr}{dt}$ and $\frac{dh}{dt}$. By using the similarity ratio $h=5r$ to substitute $h$ beforehand, you reduce the problem to a single variable $r$, making the differentiation much cleaner and less prone to algebraic errors.

Q: Wait, why is the slant height $l = \sqrt{r^2 + h^2}$? I thought it was just the height.

Remember, the slant height is the hypotenuse of the right-angled triangle formed by the radius and the vertical height of the water cone. The vertical height $h$ is just the depth; the slant height $l$ is the distance along the wall of the cone from the vertex to the water surface.

Q: Why do we reject the negative root when calculating the slant height?

In geometry, lengths like radius, height, and slant height represent physical distances, which must be positive. When we take the square root of $r^2 + h^2$, we are looking for a magnitude, so we only consider the positive root.

Question

Water is being filled at the rate of $1 	ext{ cm}^3/	ext{sec}$ in a right circular conical vessel (vertex downwards) of height $35 	ext{ cm}$ and diameter $14 	ext{ cm}$. When the height of the water level is $10 	ext{ cm}$, the rate (in $	ext{cm}^2/	ext{sec}$) at which the wet conical surface area of the vessel increases is

Options:

(A) 5

(B) $\frac{\sqrt{21}}{5}$

(C) $\frac{\sqrt{26}}{5}$

(D) $\frac{\sqrt{26}}{10}$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of Similarity

The first trap students fall into is treating the water cone and the vessel as separate entities. They are not. Because the vessel is a right circular cone, any water level creates a smaller cone that is perfectly similar to the vessel itself.

The vessel has a height $H = 35 	ext{ cm}$ and a diameter $D = 14 	ext{ cm}$, which gives us a radius $R = 7 	ext{ cm}$. The ratio of radius to height is constant:

$$ \frac{r}{h} = \frac{R}{H} = \frac{7}{35} = \frac{1}{5} $$

This relationship, $h = 5r$, is our golden key. It allows us to collapse the complexity of two variables into one.

### The Volume Dynamics

We know the volume of the water cone is $V = \frac{1}{3}\pi r^2 h$. If we differentiate this directly, we get a mess of product rules. Instead, let us substitute $h = 5r$ immediately.

Now, the volume becomes:

$$ V = \frac{1}{3}\pi r^2 (5r) = \frac{5}{3}\pi r^3 $$

Differentiating with respect to time $t$, we get:

$$ \frac{dV}{dt} = 5\pi r^2 \frac{dr}{dt} $$

Since we are given $\frac{dV}{dt} = 1 	ext{ cm}^3/	ext{sec}$, we can instantly find the rate at which the radius grows:

$$ \frac{dr}{dt} = \frac{1}{5\pi r^2} $$

Keep this in your pocket; we will need it soon.

### The Wet Surface Area

Now, the question asks for the rate of change of the wet conical surface area, $S$. The formula for the curved surface area is $S = \pi r l$, where $l$ is the slant height.

Using the Pythagorean theorem, $l = \sqrt{r^2 + h^2}$. Again, we use our golden key $h = 5r$:

$$ l = \sqrt{r^2 + (5r)^2} = \sqrt{26r^2} = r\sqrt{26} $$

Substituting this back into our area formula, we get:

$$ S = \pi r (r\sqrt{26}) = \sqrt{26}\pi r^2 $$

### The Final Synthesis

We are at the finish line. We need $\frac{dS}{dt}$. Differentiating $S = \sqrt{26}\pi r^2$ with respect to $t$, we apply the chain rule:

$$ \frac{dS}{dt} = 2\sqrt{26}\pi r \frac{dr}{dt} $$

Now, substitute the expression for $\frac{dr}{dt}$ we found earlier:

$$ \frac{dS}{dt} = 2\sqrt{26}\pi r \left( \frac{1}{5\pi r^2} ight) $$

Watch the magic happen. The $\pi$ terms cancel out, and one $r$ cancels out. We are left with:

$$ \frac{dS}{dt} = \frac{2\sqrt{26}}{5r} $$

At the moment when $h = 10 	ext{ cm}$, our radius $r$ is $\frac{10}{5} = 2 	ext{ cm}$. Plugging this in:

$$ \frac{dS}{dt} = \frac{2\sqrt{26}}{5(2)} = \frac{\sqrt{26}}{5} 	ext{ cm}^2/	ext{sec} $$

This is not just a number; it is the result of understanding how the geometry of the cone dictates the rate of change. You have mastered the flow.

Select Answer:

Visualized Solution

Visualizing the Conical Vessel

Defining Water Level Variables

Relating $r$ and $h$ via Similar Triangles

Expressing Volume in Terms of $r$

Differentiating Volume w.r.t Time

Finding $\frac{d r}{d t}$

Defining the Wet Surface Area

Expressing Surface Area in Terms of $r$

Rate of Change of Surface Area

Substituting $\frac{d r}{d t}$ into $\frac{d S}{d t}$

Evaluating at $h = 10 cm$

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Geometry of Similarity

The Volume Dynamics

The Wet Surface Area

The Final Synthesis

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