Water runs into a conical tank at the rate of 9 ft³/min. The tank stands point down and has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the... Water runs into a conical tank at the rate of 9 ft³/min. The tank stands point down and has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the water is 6 ft deep?
Understand the Problem
The question is asking how fast the water level is rising in a conical tank when the water is 6 feet deep, given the rate at which water is entering the tank and the dimensions of the tank.
Answer
The water level is rising at a rate of $\frac{1}{\pi} \text{ ft/min}$.
Answer for screen readers
The water level is rising at a rate of $\frac{1}{\pi} \text{ ft/min}$ when the water is 6 ft deep.
Steps to Solve
- Identify the given information
The given values are:
- Rate of change of volume of water, $\frac{dV}{dt} = 9 \text{ ft}^3/\text{min}$
- Height of the tank, $h_{\text{tank}} = 10 \text{ ft}$
- Radius of the tank, $r_{\text{tank}} = 5 \text{ ft}$
- Water depth at the moment of interest, $h = 6 \text{ ft}$
- Relate radius and height of water in the cone
Since the tank is conical, at any height $h$, the radius $r$ of the water can be found using similar triangles: $$ \frac{r}{h} = \frac{5}{10} \Rightarrow r = \frac{h}{2} $$
- Volume of water in the conical tank
The volume $V$ of water in a cone is given by: $$ V = \frac{1}{3} \pi r^2 h $$ Substituting $r = \frac{h}{2}$ into the volume equation, we get: $$ V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \frac{h^2}{4} h = \frac{\pi}{12} h^3 $$
- Differentiate volume with respect to time
To find the rate of change of the height with respect to time, we differentiate the volume equation: $$ \frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt} $$
- Substitute known values and solve for $\frac{dh}{dt}$
At the moment when $h = 6 \text{ ft}$:
- Substitute $V = 9 \text{ ft}^3/\text{min}$ and $h = 6 \text{ ft}$ into the differentiated equation: $$ 9 = \frac{\pi}{4} (6^2) \frac{dh}{dt} $$
- This simplifies to: $$ 9 = \frac{\pi}{4} (36) \frac{dh}{dt} $$
- Isolate $\frac{dh}{dt}$ and compute
To find $\frac{dh}{dt}$, isolate it: $$ \frac{dh}{dt} = \frac{9 \cdot 4}{36\pi} = \frac{36}{36\pi} = \frac{1}{\pi} \text{ ft/min} $$
The water level is rising at a rate of $\frac{1}{\pi} \text{ ft/min}$ when the water is 6 ft deep.
More Information
This problem involves applying the concept of related rates in calculus and using the formula for the volume of a cone. The geometry of the cone and the relationship between the dimensions are critical to solving the problem.
Tips
- Forgetting to relate the radius and height using similar triangles.
- Misapplying the volume formula by not adjusting for the dimensions of the specific cone.
- Confusing the rates of change, forgetting which variables change with respect to time.
