Problem 62 - Calculus

Question

Oil is leaking at the rate of \(V^{\prime}(t)=1-t / 110\) from a storage tank that is initially full of 55 gallons. How much leaks out during the first hour? During the tenth hour? How long until the entire tank is drained?

Step-by-Step Solution

Answer

In the first hour, 0.995 gallons leak; in the tenth hour, 0.1 gallons leak. The tank is completely drained at approximately 110 hours.

Step 1: Determine the rate of oil leaking

Given the rate of oil leaking is described by the function \( V'(t) = 1 - \frac{t}{110} \), where \( t \) is in hours and \( V(t) \) is the volume in gallons leaked at time \( t \).

Step 2: Calculate the amount leaked in the first hour

To find out how much oil leaks out during the first hour, integrate the rate function from 0 to 1. The integral is \( \int_0^1 \left( 1 - \frac{t}{110} \right) dt \). This evaluates to \[ t - \frac{t^2}{220} \Bigg|_0^1 = \left( 1 - \frac{1^2}{220} \right) - \left( 0 - \frac{0^2}{220} \right) = 1 - \frac{1}{220} = \frac{219}{220}. \] Thus, \( \frac{219}{220} \) gallons leaks out in the first hour.

Step 3: Calculate the amount leaked during the tenth hour

To find out how much oil leaks between the ninth and tenth hour, integrate the rate function from 9 to 10. The integral is \( \int_9^{10} \left( 1 - \frac{t}{110} \right) dt \). This evaluates to \[ t - \frac{t^2}{220} \Bigg|_9^{10} = \left( 10 - \frac{10^2}{220} \right) - \left( 9 - \frac{9^2}{220} \right) = \left( 10 - \frac{100}{220} \right) - \left( 9 - \frac{81}{220} \right) = \frac{121}{220} - \frac{99}{220} = \frac{22}{220} = \frac{1}{10}. \] Thus, 0.1 gallons leaks during the tenth hour.

Step 4: Determine when the entire tank is drained

The tank initially contains 55 gallons. To find when it is completely drained, set the integral from 0 to \( t \) equal to 55 and solve for \( t \): \( \int_0^t \left( 1 - \frac{t}{110} \right) dt = 55 \). This evaluates to \[ t - \frac{t^2}{220} \Bigg|_0^t = t - \frac{t^2}{220} = 55. \] Solving \( t - \frac{t^2}{220} = 55 \) results in a quadratic equation \( 220t - t^2 = 12100 \), or \( t^2 - 220t + 12100 = 0 \). Solving this using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1, b = -220, c = 12100 \), we find \( t \approx 110 \). Thus, the tank is completely drained at approximately 110 hours.

Key Concepts

Integration

In calculus, integration is a fundamental concept used to find areas, volumes, and total accumulations. When you integrate a function, you are essentially finding the total change or accumulation described by that function over a certain interval.
In the context of the oil leaking problem, the function given by the rate of oil leaking is described as \(V'(t) = 1 - \frac{t}{110}\). This function indicates how fast the oil is leaking out of the tank over time.
To find out how much oil leaks out during a specific time period, such as the first hour, we must integrate the rate function from the start of this interval (0 hours) to the end (1 hour):

• The integration of \(\int_0^1 \left( 1 - \frac{t}{110} \right) dt\) calculates the total oil leakage in gallons over this time frame.
• Doing the math, we find that \(\frac{219}{220}\) gallons leaks out during the first hour.

Through integration, we take an instantaneous rate (like our \(V'(t)\) function) and expand it across an interval to measure how much occurs cumulatively—be it distance, volume, or, in this case, leaked oil.

Rate of change

The concept of rate of change is central in understanding how a quantity varies over time. It is essentially the derivative of a function in calculus terms. A derivative tells us how a function value changes as its input changes.
In this oil leaking problem, the rate of change is given by the function \(V'(t) = 1 - \frac{t}{110}\). This describes how quickly the oil volume decreases as time progresses.
Here are some key points:

• The rate of change starts from an initial value of 1 gallon per hour when \(t = 0\).
• As time increases, the rate at which oil leaks decreases linearly due to the term \(- \frac{t}{110}\).

This diminishing rate signifies that initially, oil leaks out faster, but as time passes, the leaking slows. Recognizing how rates of change impact the real world can significantly aid in anticipating how processes will behave over time, such as predicting when a resource will deplete.

Quadratic equation

Quadratic equations appear frequently in problems involving motion or changes over time, as they can describe the relationship between different variables. A quadratic equation is typically written as \(ax^2 + bx + c = 0\). Solving these often involves tools like factoring, completing the square, or the quadratic formula.
In solving our leaking oil scenario, we reached a quadratic equation when finding out how long it takes for the tank to fully drain:

• The integral of the rate function equated to the total initial volume of oil gives us \(220t - t^2 = 12100\).
• This simplifies to \(t^2 - 220t + 12100 = 0\), where the solution gives the time \(t\) when all 55 gallons of oil are gone.

Utilizing the quadratic formula: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a = 1\), \(b = -220\), and \(c = 12100\), leads us to find \(t \approx 110\). This signifies that the tank will be fully drained at roughly 110 hours, illustrating how quadratic equations can help predict long-term outcomes in dynamic processes.