Key Concepts and Formulas

• Related Rates: Using derivatives to find the rate of change of one quantity in terms of the rate of change of other related quantities.
• Volume of a Cone: $V = \frac{1}{3}\pi r^2 h$, where $r$ is the radius and $h$ is the height.
• Curved Surface Area of a Cone: $A = \pi r l$, where $r$ is the radius and $l$ is the slant height ($l = \sqrt{r^2 + h^2}$).

Step-by-Step Solution

Step 1: Understand the Geometry and Define Variables

We have a right circular conical vessel with its vertex downwards. Water is being filled into it. As water fills, it forms a smaller cone similar to the vessel's shape.

• Vessel Dimensions:
  • Height of the vessel, $H = 35 \text{ cm}$.
  • Diameter of the vessel, $D = 14 \text{ cm}$, so Radius of the vessel, $R = D/2 = 7 \text{ cm}$.
• Water Cone Dimensions (at any instant $t$):
  • Let $h$ be the height of the water level.
  • Let $r$ be the radius of the water surface.
  • Let $l$ be the slant height of the wet conical surface.

Establishing a relationship between $r$ and $h$: Since the water cone is similar to the vessel's cone (due to the vertex being downwards), we can use similar triangles: $\frac{r}{h} = \frac{R}{H}$ Substituting the given dimensions of the vessel: $\frac{r}{h} = \frac{7}{35} = \frac{1}{5}$ This gives us a crucial relationship: $r = \frac{h}{5}$. This allows us to express any quantity related to the water cone in terms of just one variable, $h$.

Step 2: Express Volume in Terms of a Single Variable ($h$) and Differentiate

The volume of water in the conical vessel at any instant $t$ is given by: $V = \frac{1}{3}\pi r^2 h$ Substituting $r = \frac{h}{5}$ into the volume equation, we get: $V = \frac{1}{3}\pi \left(\frac{h}{5}\right)^2 h = \frac{\pi}{75}h^3$ We are given that water is being filled at a rate of 1 cm$^3$/sec, which means $\frac{dV}{dt} = 1$. Now, we differentiate the volume equation with respect to time $t$: $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{\pi}{75}h^3\right) = \frac{\pi}{75} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi}{25}h^2 \frac{dh}{dt}$ Since $\frac{dV}{dt} = 1$, we have: $1 = \frac{\pi}{25}h^2 \frac{dh}{dt}$ Therefore, $\frac{dh}{dt} = \frac{25}{\pi h^2}$

Step 3: Express the Wet Conical Surface Area in Terms of $h$ and Differentiate

The wet conical surface area, $A$, is given by: $A = \pi r l$ We know $r = \frac{h}{5}$ and $l = \sqrt{r^2 + h^2}$. Substituting $r = \frac{h}{5}$ into the expression for $l$: $l = \sqrt{\left(\frac{h}{5}\right)^2 + h^2} = \sqrt{\frac{h^2}{25} + h^2} = \sqrt{\frac{26h^2}{25}} = \frac{h}{5}\sqrt{26}$ Now, substituting $r = \frac{h}{5}$ and $l = \frac{h}{5}\sqrt{26}$ into the surface area equation: $A = \pi \left(\frac{h}{5}\right) \left(\frac{h}{5}\sqrt{26}\right) = \frac{\pi\sqrt{26}}{25}h^2$ Differentiate $A$ with respect to time $t$: $\frac{dA}{dt} = \frac{d}{dt}\left(\frac{\pi\sqrt{26}}{25}h^2\right) = \frac{\pi\sqrt{26}}{25} \cdot 2h \cdot \frac{dh}{dt} = \frac{2\pi\sqrt{26}}{25}h \frac{dh}{dt}$

Step 4: Substitute and Solve for $\frac{dA}{dt}$

We found that $\frac{dh}{dt} = \frac{25}{\pi h^2}$. Substitute this into the equation for $\frac{dA}{dt}$: $\frac{dA}{dt} = \frac{2\pi\sqrt{26}}{25}h \left(\frac{25}{\pi h^2}\right) = \frac{2\sqrt{26}}{h}$ We are asked to find $\frac{dA}{dt}$ when $h = 10$ cm. Substituting $h = 10$: $\frac{dA}{dt} = \frac{2\sqrt{26}}{10} = \frac{\sqrt{26}}{5} \text{ cm}^2/\text{sec}$

Common Mistakes & Tips

• Units: Always keep track of units. This helps in dimensional analysis and spotting errors.
• Similar Triangles: Recognizing similar triangles is key in relating the variables.
• Chain Rule: Remember to apply the chain rule correctly when differentiating with respect to time.

Summary

The problem involves finding the rate of change of the wet conical surface area of a vessel as water is poured into it at a constant rate. By establishing a relationship between the radius and height of the water cone using similar triangles, expressing the volume and surface area in terms of a single variable, and differentiating with respect to time, we found the rate of change of the height and subsequently the rate of change of the surface area. When the height of the water level is 10 cm, the rate at which the wet conical surface area of the vessel increases is $\frac{\sqrt{26}}{5}$ cm$^2$/sec.

Final Answer

The final answer is $\boxed{\frac{\sqrt{26}}{5}}$, which corresponds to option (C).