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

1. Geometric setup

Vessel: right circular cone

• Total height* $H = 35\,\text{cm}$
• Top radius* $R = \frac{14}{2} = 7\,\text{cm}$

When water has current height $h$, its current radius $r$ comes from similarity of triangles:

2. Volume–height relation

The water itself forms a smaller cone:

Plug $r = \frac{h}{5}$:

Differentiate wrt time $t$ to relate $dV/dt$ and $dh/dt$:

3. Finding $dh/dt$ at $h = 10\,\text{cm}$

Given filling rate $\displaystyle\frac{dV}{dt}=1\,\text{cm}^{3}\!\big/\!\text{s}$,

At $h = 10$:

4. Lateral surface area of the water cone

Slant height $l$ of the water cone:

Lateral (wet) area $A$:

5. Differentiate to relate $dA/dt$ and $dh/dt$

6. Substitute $h=10$ and $\displaystyle\frac{dh}{dt}=\frac{1}{4\pi}$

Hence the wet surface area is increasing at $\boxed{\dfrac{\sqrt{26}}{5}\,\text{cm}^{2}\!\big/\!\text{s}}$ when the water is 10 cm deep.

What’s going on here?

Imagine you are pouring water into an ice-cream cone that is standing on its tip. As the water level rises, more and more of the cone’s inner wall gets wet. The question asks how fast that wet part is growing right now when the water is 10 cm deep.

Big ideas in kid words

1. Cone shape shrinks as you go down – at the pointy end the circle is tiny, at the top it’s big.
2. Small cone looks like big cone – using Lego-like similarity, the little water cone has the same shape as the full vessel, just scaled down.
3. Volume tells us height speed – we know how fast the water amount (volume) is increasing, so we can figure out how fast the height is changing.
4. Surface = side wall area – take that new height speed and convert it to how fast the wet wall is spreading.

Game plan

• Turn volume change (easy, given) ➔ height change (by calculus).
• Turn height change ➔ surface-area change (more calculus). That’s all!

Step-by-Step Solution

1. Similarity for radius: $\displaystyle r = \dfrac{h}{5}$.
2. Volume as a function of $h$: [V = \dfrac{\pi}{75}h^{3}]
3. Differentiate: [\dfrac{dV}{dt} = \dfrac{\pi}{25}h^{2}\dfrac{dh}{dt}] Given $\dfrac{dV}{dt} = 1$, so [\dfrac{dh}{dt} = \dfrac{25}{\pi h^{2}}] At $h = 10$: [\dfrac{dh}{dt} = \dfrac{1}{4\pi},\text{cm/s}]
4. Lateral area function: [A = \dfrac{\pi\sqrt{26}}{25}h^{2}]
5. Differentiate $A$: [\dfrac{dA}{dt} = \dfrac{2\pi\sqrt{26}}{25}h\dfrac{dh}{dt}]
6. Substitute $h=10$, $dh/dt=1/(4\pi)$: [\dfrac{dA}{dt} = \dfrac{2\pi\sqrt{26}}{25}\times10\times\dfrac{1}{4\pi} = \dfrac{\sqrt{26}}{5},\text{cm}^{2}/\text{s}]

Final Answer: $\boxed{\dfrac{\sqrt{26}}{5}\,\text{cm}^{2}\!/\!\text{s}}$