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.
• Related Rates: Implicit differentiation with respect to time ($t$) to relate rates of change.
• Trigonometry: Using trigonometric ratios (specifically, tangent) to relate the radius and height of the cone.

Step-by-Step Solution

Step 1: Define Variables and Given Information

Let's define the variables:

• $V$: Volume of water in the tank (in cubic meters)
• $r$: Radius of the water surface at any time (in meters)
• $h$: Depth (height) of the water at any time (in meters)
• $t$: Time (in minutes)

We are given:

• $\frac{dV}{dt} = 5$ m$^3$/min (rate at which water is poured into the tank)
• $\tan^{-1}\left(\frac{1}{2}\right)$ is the semi-vertical angle, which means $\tan(\theta) = \frac{r}{h} = \frac{1}{2}$
• We want to find $\frac{dh}{dt}$ when $h = 10$ m

Step 2: Establish the Relationship between r and h

From the semi-vertical angle, we have: $\tan(\theta) = \frac{r}{h} = \frac{1}{2}$ This implies: $r = \frac{1}{2}h$

Step 3: Express the Volume in terms of h

Substitute the expression for $r$ in terms of $h$ into the formula for the volume of a cone: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{1}{2}h\right)^2 h = \frac{1}{3}\pi \left(\frac{1}{4}h^2\right)h = \frac{1}{12}\pi h^3$

Step 4: Differentiate with respect to time t

Differentiate both sides of the volume equation with respect to time $t$: $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{1}{12}\pi h^3\right)$ $\frac{dV}{dt} = \frac{1}{12}\pi \cdot 3h^2 \cdot \frac{dh}{dt}$ $\frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}$

Step 5: Substitute known values and solve for dh/dt

We are given $\frac{dV}{dt} = 5$ and we want to find $\frac{dh}{dt}$ when $h = 10$. Substitute these values into the equation: $5 = \frac{1}{4}\pi (10)^2 \frac{dh}{dt}$ $5 = \frac{1}{4}\pi (100) \frac{dh}{dt}$ $5 = 25\pi \frac{dh}{dt}$ $\frac{dh}{dt} = \frac{5}{25\pi} = \frac{1}{5\pi}$

Step 6: Check for Errors and Consistency

The calculations seem correct and the units are consistent.

Step 7: Express in requested format

The rate at which the level of water is rising is $\frac{1}{5\pi}$ m/min when the depth of the water is 10m.

Common Mistakes & Tips

• Units: Always pay attention to units and make sure they are consistent throughout the problem.
• Implicit Differentiation: Remember to apply the chain rule when differentiating with respect to time.
• Relating Variables: The key to solving related rates problems is to find the correct equation that relates the variables. Make sure to use any given information (like the semi-vertical angle) to establish this relationship.
• Double-check your algebra and calculus steps to avoid making careless errors.

Summary

We were given the rate at which water is poured into an inverted conical tank and asked to find the rate at which the water level is rising at a specific depth. We used the relationship between the radius and height of the cone, the formula for the volume of a cone, and implicit differentiation to relate the rates of change. Substituting the given values, we found that the rate at which the water level is rising is $\frac{1}{5\pi}$ m/min.

Final Answer The final answer is $\boxed{\frac{1}{5\pi}}$, which corresponds to option (B).