Key Concepts and Formulas

• 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.
• Related Rates: Using the chain rule to find the rate of change of one quantity in terms of the rates of change of other related quantities.

Step-by-Step Solution

Step 1: Understand the Geometry and Establish Relationships

We are given a right circular cone with semi-vertical angle $\theta$ such that $\tan \theta = \frac{3}{4}$. Let $r$ be the radius of the water surface and $h$ be the depth of the water. From the geometry of the cone, we have:

$\tan \theta = \frac{r}{h} = \frac{3}{4}$

This gives us the relationship:

$r = \frac{3}{4}h \quad \text{ or } \quad h = \frac{4}{3}r \quad \text{(Equation 1)}$

We are also given that $\frac{dV}{dt} = 6 \, \text{m}^3/\text{hr}$. We want to find $\frac{dA}{dt}$ when $h = 4 \, \text{m}$.

Step 2: Express Volume in Terms of r and Differentiate

The volume of the water in the cone is given by:

$V = \frac{1}{3}\pi r^2 h$

Substitute $h = \frac{4}{3}r$ from Equation 1 into the volume equation:

$V = \frac{1}{3}\pi r^2 \left(\frac{4}{3}r\right) = \frac{4}{9}\pi r^3 \quad \text{(Equation 2)}$

Now, differentiate both sides of Equation 2 with respect to time $t$:

$\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{9}\pi r^3\right) = \frac{4}{9}\pi \cdot 3r^2 \frac{dr}{dt} = \frac{4}{3}\pi r^2 \frac{dr}{dt} \quad \text{(Equation 3)}$

Step 3: Calculate $\frac{dr}{dt}$ when h = 4 m

We are given $\frac{dV}{dt} = 6 \, \text{m}^3/\text{hr}$. When $h = 4 \, \text{m}$, we can find $r$ using Equation 1:

$r = \frac{3}{4}h = \frac{3}{4}(4) = 3 \, \text{m}$

Substitute $\frac{dV}{dt} = 6$ and $r = 3$ into Equation 3:

$6 = \frac{4}{3}\pi (3^2) \frac{dr}{dt} = 12\pi \frac{dr}{dt}$

Solving for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{6}{12\pi} = \frac{1}{2\pi} \, \text{m/hr}$

Step 4: Express Curved Surface Area in Terms of r and Differentiate

The curved surface area of the cone is given by $A = \pi r l$, where $l$ is the slant height. The slant height is related to $r$ and $h$ by $l = \sqrt{r^2 + h^2}$. Since $h = \frac{4}{3}r$, we have:

$l = \sqrt{r^2 + \left(\frac{4}{3}r\right)^2} = \sqrt{r^2 + \frac{16}{9}r^2} = \sqrt{\frac{25}{9}r^2} = \frac{5}{3}r$

Substitute this into the surface area formula:

$A = \pi r \left(\frac{5}{3}r\right) = \frac{5}{3}\pi r^2 \quad \text{(Equation 4)}$

Differentiate both sides of Equation 4 with respect to time $t$:

$\frac{dA}{dt} = \frac{d}{dt}\left(\frac{5}{3}\pi r^2\right) = \frac{5}{3}\pi \cdot 2r \frac{dr}{dt} = \frac{10}{3}\pi r \frac{dr}{dt} \quad \text{(Equation 5)}$

Step 5: Calculate $\frac{dA}{dt}$ when h = 4 m

When $h = 4 \, \text{m}$, we have $r = 3 \, \text{m}$ and $\frac{dr}{dt} = \frac{1}{2\pi} \, \text{m/hr}$. Substitute these values into Equation 5:

$\frac{dA}{dt} = \frac{10}{3}\pi (3) \left(\frac{1}{2\pi}\right) = \frac{10}{3}\pi \cdot 3 \cdot \frac{1}{2\pi} = 5 \, \text{m}^2/\text{hr}$

Common Mistakes & Tips

• Substituting too early: Always differentiate with respect to time before substituting the given value of $h$ (or $r$). Substituting early will treat the variable as a constant.
• Incorrectly calculating slant height: Make sure to relate the slant height correctly to the radius and height using the Pythagorean theorem.
• Forgetting the chain rule: Remember to multiply by $\frac{dr}{dt}$ or $\frac{dh}{dt}$ when differentiating terms involving $r$ or $h$ with respect to $t$.

Summary

We used related rates to find the rate of change of the curved surface area of the water in a conical tank. We related the volume and surface area to the radius and height, used the given information about the semi-vertical angle to express everything in terms of a single variable, and then differentiated with respect to time to find the desired rate. The rate at which the wet curved surface area of the tank is increasing when the depth of water is 4 meters is 5 square meters per hour.

The final answer is $\boxed{5}$.