Problem 3 - Active Calculus

Question

Water is leaking out of an inverted conical tank at a rate of \(8600.0 \mathrm{~cm}^{3} / \mathrm{min}\) at the same time that water is being pumped into the tank at a constant rate. The tank has height \(12.0 \mathrm{~m}\) and the the diameter at the top is \(4.0 \mathrm{~m}\). If the water level is rising at a rate of \(24.0 \mathrm{~cm} / \mathrm{min}\) when the height of the water is \(5.0 \mathrm{~m}\), find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Step-by-Step Solution

Answer

The rate at which water is being pumped into the tank is approximately 532198.78 cm³/min.

Step 1 - Convert Units

First, convert all units to be consistent. Given that height and diameter are in meters, convert them to centimeters. Thus, the height is 1200 cm and the diameter is 400 cm. The radius of the tank is then 200 cm.

Step 2 - Relate the radius and height

Write the relationship between the radius and height of the water. Since the tank is a cone, the radius and height are proportional: \[ \frac{r}{h} = \frac{200}{1200} = \frac{1}{6} \]. Then, for any height h of the water, the radius r of the water is given by \[ r = \frac{h}{6} \].

Step 3 - Express Volume in Terms of Height

The volume V of a cone is given by: \[ V = \frac{1}{3} \pi r^2 h \]. Substitute r from the previous step: \[ V = \frac{1}{3} \pi \left( \frac{h}{6} \right)^2 h = \frac{1}{3} \pi \frac{h^3}{36} = \frac{\pi h^3}{108} \].

Step 4 - Differentiate Volume with Respect to Time

Differentiate the volume equation with respect to time t: \[ \frac{dV}{dt} = \frac{\pi}{108} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{36} \cdot \frac{dh}{dt} \].

Step 5 - Plug in Given Values

Given that the height of the water is 500 cm and the height is rising at 24 cm/min, substitute these values: \[ \frac{dV}{dt} = \frac{\pi \cdot 500^2}{36} \cdot 24 \].

Step 6 - Calculate the Volume Rate

Calculate the rate of change of the volume: \[ \frac{dV}{dt} = \frac{\pi \cdot 250000}{36} \cdot 24 = \frac{\pi \cdot 250000 \cdot 24}{36} \approx 523598.78 \mathrm{~cm}^3/ \mathrm{min} \].

Step 7 - Account for Leaking Water

Since water is leaking out at a rate of 8600 cm³/min and the volume rate calculated represents the net rate of volume increase, the rate at which water is being pumped in, P, is the sum of the net rate and the leaking rate: \[ P = \frac{dV}{dt} + 8600 \approx 523598.78 + 8600 = 532198.78 \mathrm{~cm}^3/ \mathrm{min} \].

Key Concepts

Volume of a Cone

Understanding the volume of a cone is essential in this problem. The formula to calculate the volume of a cone is given by \[\frac{1}{3} \pi r^2 h\]. Here, \[r\] is the radius of the base of the cone and \[h\] is the height of the cone. This formula derives from the general formula for the volume of a geometric solid, where the base area is multiplied by the height and divided by three. For our exercise, the radius is obtained in terms of the height of the water in the tank to help simplify the calculations. Since the problem involves the height and not directly the radius, putting the volume formula in terms of height becomes crucial.

Unit Conversion

Before solving any problem involving measurements, ensure the units are consistent. For our problem, we have measurements in both meters and centimeters. To avoid errors, convert all measurements to a single unit system. Here, we convert everything to centimeters:

• The height of the tank from 12.0 meters to 1200 centimeters.
• The diameter from 4.0 meters to 400 centimeters, giving a radius of 200 centimeters.

This uniformity simplifies the calculations and prevents confusion while differentiating or substituting values.

Differentiation with Respect to Time

Differentiating a function with respect to time is crucial when dealing with rates of change. In our problem, the volume of water in the tank is changing over time due to leaking and pumping actions. By differentiating the volume of the cone equation \[\frac{\razor}{108} \pi h^3\] with respect to time (denoted as t), we get the rate of change of the volume. We use the chain rule of differentiation since the height \[h\] is itself a function of time
:

• First, differentiate the volume equation, treating h as a function of time: \[ \frac{dV}{dt} = \frac{\razor h^2}{36} \frac{dh}{dt} \].
• This equation tells us that the rate of change of the volume depends on the current height of water and the rate at which that height changes.

Substituting the given values for height and the rate of height increase allows us to find the net rate of volume change.

Chain Rule in Calculus

The chain rule is a fundamental method in calculus for differentiating composite functions. In our exercise, it helps in linking the rates of different variables. The volume \[V\] is a function of the height \[h\], which itself is a function of time \[t\]. Thus, to find \[ \frac{dV}{dt} \], we must apply the chain rule. This involves:

• First finding the derivative of volume with respect to height: \[ \frac{d}{dh} ( \frac{\razor h^3 }{108} ) = \frac{\razor h^2}{36} \].
• Then multiplying this by the derivative of height with respect to time: \[ \frac{dh}{dt} \].

Altogether, the differentiation step is represented as \[ \frac{dV}{dt} = \frac{\razor h^2}{36} \frac{dh}{dt} \]. This chain rule application shows how the derivative of a function of another function is handled, making it easier to solve related rates problems like the one we are tackling.