1. A water tank has the shape of an inverted right circular cone of altitude 12 ft and
base radius of 6 ft. There is a hole in the bottom that is leaking at a rate of 0.3
gal/min. If water is being pumped into the tank at a rate of 10 gal/min,
a) Approximate the rate at which the water level is rising when the water is3 ft deep. (1 gallon = 1337 ft^2 ).
b) How long will it take for the tank to fill?
a) Rate at which the water level is rising when the water is 3 ft deep:
Let's start by setting up the volume of a cone with height (h) and base radius (r):
\[ V = \frac{1}{3} \pi r^2 h \]
Since we have an inverted cone, we'll be working with a relationship between the water level depth and the radius. Because we're given the tank's dimensions when the tank is full (altitude = 12 ft and base radius = 6 ft), we can set up a relationship between the radius and the height of the water at any given time using similar triangles:
\[ \frac{r}{h} = \frac{6 \text{ ft}}{12 \text{ ft}} \]
\[ r = \frac{h}{2} \]
If we substitute \( r = \frac{h}{2} \) into the volume formula, we get:
\[ V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h \]
\[ V = \frac{1}{3} \pi \frac{h^3}{4} \]
\[ V = \frac{\pi h^3}{12} \]
Now, let's find the rate of change of the volume with respect to time. We'll denote this as \( \frac{dV}{dt} \). The rate at which water is being added to the tank is the rate of water being pumped in minus the rate of the leak:
\[ \frac{dV}{dt} = 10 \text{ gal/min} - 0.3 \text{ gal/min} \]
\[ \frac{dV}{dt} = 9.7 \text{ gal/min} \]
We need to convert this rate to cubic feet per minute, using the conversion factor 1 gallon = 1337 ft\(^3\):
\[ \frac{dV}{dt} = 9.7 \text{ gal/min} \times 1337 \text{ ft}^3/\text{gal} \]
\[ \frac{dV}{dt} = 9.7 \times 1337 \text{ ft}^3/\text{min} \]
Next, we want to relate this rate of change of volume to the rate of change of the height (\( \frac{dh}{dt} \)). We differentiate \( V = \frac{\pi h^3}{12} \) with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} \]
\[ 9.7 \times 1337 = \frac{\pi}{4} \cdot 3h^2 \frac{dh}{dt} \]
We want to find \( \frac{dh}{dt} \) when \( h = 3 \) ft:
\[ 9.7 \times 1337 = \frac{3\pi}{4} \cdot 3^2 \frac{dh}{dt} \]
\[ \frac{dh}{dt} = \frac{9.7 \times 1337}{\frac{3\pi}{4} \cdot 9} \]
\[ \frac{dh}{dt} = \frac{9.7 \times 1337 \times 4}{3\pi \times 9} \]
Using a calculator:
\[ \frac{dh}{dt} = \frac{9.7 \times 1337 \times 4}{3 \times 3.1416 \times 9} \]
\[ \frac{dh}{dt} = \frac{9.7 \times 1337 \times 4}{84.8232} \]
\[ \frac{dh}{dt} \approx \frac{9.7 \times 1337 \times 4}{84.8232} \text{ ft/min} \]
\[ \frac{dh}{dt} \approx 0.595 \text{ ft/min} \]
So the water level is rising at approximately 0.595 feet per minute when the water is 3 ft deep.
b) Time to fill the tank:
Since we already have the volume formula for the full tank,
\[ V_{\text{full}} = \frac{\pi \cdot 6^3}{12} \]
To find out when the tank is full, we'll need to use the net flow rate that we previously found (pumping in minus leaking). Given that the tank has no water initially, we'll solve for the time it takes for the volume at the net flow rate to equal the full volume.
\[ V_{\text{full}} = \frac{dV}{dt} \cdot t \]
\[ \frac{\pi \cdot 6^3}{12} = 9.7 \times 1337 \cdot t \]
\[ t = \frac{\pi \cdot 6^3}{12 \cdot 9.7 \cdot 1337} \]
\[ t \approx \frac{3.1416 \cdot 216}{12 \cdot 9.7 \cdot 1337} \]
\[ t \approx \frac{678.584}{124528.4} \]
\[ t \approx 0.00545 \text{ min} \]
However, this number is suspiciously small and likely indicates a miscalculation during the process, as even the net input flow will not fill a large cone that quickly. Let's correct the calculation.
\[ V_{\text{full}} = \frac{\pi \cdot 6^3}{12} \]
\[ V_{\text{full}} = \frac{216\pi}{12} \]
\[ V_{\text{full}} = 18\pi \text{ ft}^3 \]
To find the exact time, we convert this volume to gallons and then divide by the net change rate:
\[ V_{\text{full}} = 18\pi \times 7.481 (1 \text{ ft}^3 \approx 7.481 \text{ gal}) \]
\[ V_{\text{full}} \approx 134.678\pi \text{ gal} \]
\[ t = \frac{V_{\text{full}}}{9.7 \text{ gal/min}} \]
\[ t \approx \frac{134.678\pi}{9.7} \text{ min} \]
\[ t \approx \frac{134.678 \times 3.1416}{9.7} \text{ min} \]
\[ t \approx 43.63 \text{ min} \]
So, it will take approximately 43.63 minutes to fill the tank (assuming the rate of filling and leaking remains constant).