Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

A cylindrical barrel with radius 3 feet is lled with oil. The barrel stands upright and oil comes out a faucet at the base of the tank. (a) How are the volume of oil in the tank and the height of oil in the tank related? (b) What is the rate of change of volume with respect to height? Is it constant, or does it depend upon the height? Does your answer make sense to you? (c) As oil leaves the tank, both the volume and the height of oil in the tank change with respect to time. i. When the height is decreasing at a rate of \(0.5\) feet per hour, how fast is the volume of oil in the tank changing? ii. When oil is leaving the tank at a rate of 3 cubic feet per hour, how fast is the height of oil in the tank changing?

Step-by-Step Solution

Answer

a) The volume of the oil is directly proportional to its height. b) The rate of change of volume with respect to height is constant and equals \(9\pi\) cubic feet per foot. c.i) When the height is dropping at 0.5 feet per hour, the volume decreases at a rate of \(-4.5\pi\) cubic feet per hour. c.ii) When oil is leaving the barrel at 3 cubic feet per hour, the height of oil is decreasing at a rate of \(1/(3\pi)\) feet per hour.

Step 1: Relating Volume and Height

Let's denote the radius \(r\) and height \(h\) of the oil in the tank. The volume \(V\) of a cylinder is given by \(\pi r^2 h\). Since the radius is constant (3 ft), the volume of oil is \(V = \pi (3)^2 h = 9\pi h\). The volume of oil is directly proportional to its height.

Step 2: Computing Volume-Hight Change Rate

Taking derivative of both sides with respect to \(h\), we get \(\frac {dV}{dh} = 9\pi)\). This implies the rate of change of volume w.r.t height is constant and equals \(9\pi\). This means that for each foot of height change, the volume changes by \(9\pi\) cubic feet.

Step 3: Relating Volume Change to Time

Now we want to find out how fast the volume is changing when the height decreases by \(0.5\) feet per hour. We can simply substitute this value into the derivative we obtained in the previous step, i.e., \( \frac {dV}{dt} = \frac {dV}{dh} \frac {dh}{dt}\) = \( (9\pi) * (-0.5) \) = \(-4.5\pi\) cubic feet per hour.

Step 4: Relating Height Change to Volume discharge

Lastly, we want to know how fast the height is changing when the volume is decreasing by \(3\) cubic feet per hour. So, rearrange the formula to yield \( \frac {dh}{dt} = \frac {dV}{dt} / \frac {dV}{dh} \), then substitute the given value, i.e. \( \frac {dh}{dt} = 3 / 9\pi = 1/(3\pi) ft/hr \).

Key Concepts

Volume of a Cylinder

In calculus, understanding the volume of a cylinder is foundational in solving problems involving cylindrical shapes. A cylinder is characterized by a circular base and a specific height. The volume \( V \) of a cylinder can be mathematically expressed by the formula \( V = \pi r^2 h \), where \( r \) is the radius of the circular base, and \( h \) is the height of the cylinder.
In our exercise, the cylindrical barrel has a fixed radius \( r = 3 \) feet. Thus, the volume formula simplifies to \( V = 9\pi h \).
This simplification shows that the volume of oil is directly proportional to the height of oil in the tank. In practical terms, this means as the height increases, the volume also increases linearly, assuming the cylinder remains upright and the radius unchanged. This relationship is essential for understanding how changes in height affect the overall volume without recalibrating for other factors.

Derivative with Respect to Height

The derivative concept in calculus helps us understand how one quantity changes with respect to another. In the case of the volume of the cylinder, we're interested in how the volume \( V \) changes as the height \( h \) changes. This is expressed via the derivative \( \frac{dV}{dh} \).
Since the volume equation for the cylinder is \( V = 9\pi h \), taking the derivative with respect to height \( h \) yields \( \frac{dV}{dh} = 9\pi \).
This derivative is constant, indicating that the rate at which the volume changes with respect to the height is consistent, regardless of how much oil is in the tank. Thus, every 1-foot increase in height results in an increase of \( 9\pi \) cubic feet of oil; the relationship does not change with height. Understanding this helps verify that our initial rate assumptions are mathematically sound and independent of current conditions in the tank.

Rate of Volume Change with Respect to Time

When considering how the volume of oil changes over time as oil flows out of the tank, we introduce the concept of rates with respect to time. Using calculus, we can determine how fast the volume changes concerning time \( t \) by employing \( \frac{dV}{dt} \).
Using the chain rule, \( \frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} \). In the exercise, it suggests scenarios such as oil height decreasing at \( 0.5 \) feet per hour. Substituting into our rate equations, \( \frac{dV}{dt} = (9\pi) \cdot (-0.5) = -4.5\pi \) cubic feet per hour. This negative sign highlights a decrease in volume, meaning oil is leaving the tank.

• When the volume decreases at \( 3 \) cubic feet per hour, you find the rate of height change using \( \frac{dh}{dt} = \frac{dV}{dt} / \frac{dV}{dh} = \frac{3}{9\pi} = \frac{1}{3\pi} \) feet per hour.

These calculations underscore an important concept: the interplay of volume, height, and the passage of time, crucial for discerning the dynamic behavior of the system as it evolves.