Problem 66 - Calculus Early Transcendentals: Pearson New International Edition

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

Approximately 0.995 gallons leaks in the first hour, 0.914 gallons in the tenth hour, and the tank is drained in about 110 hours.

Step 1: Understanding the Problem

We are given a rate function \( V'(t) = 1 - \frac{t}{110} \) which represents the rate at which oil leaks from the tank. The tank initially holds 55 gallons. We need to determine how much oil leaks out during the first hour, during the tenth hour, and when the tank will be completely drained.

Step 2: Calculating Oil Loss During the First Hour

To find out how much oil leaks out during the first hour, integrate the rate function from \( t = 0 \) to \( t = 1 \). \[ \int_0^1 (1 - \frac{t}{110}) \, dt \]The antiderivative is: \[ \left[ t - \frac{t^2}{220} \right]_0^1 = \left( 1 - \frac{1}{220} \right) - 0 = \frac{219}{220} \]Thus, approximately 0.995 gallons leaks out in the first hour.

Step 3: Calculating Oil Loss During the Tenth Hour

Similarly, to find how much oil leaks during the tenth hour (from \( t = 9 \) to \( t = 10 \)), integrate:\[ \int_9^{10} (1 - \frac{t}{110}) \, dt \]The antiderivative evaluates to:\[ \left[ t - \frac{t^2}{220} \right]_9^{10} = \left( 10 - \frac{100}{220} \right) - \left( 9 - \frac{81}{220} \right) = 1 - \frac{19}{220} = \frac{201}{220} \]Thus, approximately 0.914 gallons leaks out during the tenth hour.

Step 4: Finding When the Tank is Empty

To find when the tank is completely drained, set up the integral from \( t = 0 \) to \( t = T \) such that the total is 55 gallons:\[ \int_0^T (1 - \frac{t}{110}) \, dt = 55 \]This integral evaluates to:\[ \left[ t - \frac{t^2}{220} \right]_0^T = T - \frac{T^2}{220} \]So, solve:\[ T - \frac{T^2}{220} = 55 \]This results in a quadratic equation \( T^2 - 220T + 12100 = 0 \). Solving this using the quadratic formula, \( T = \frac{220 \pm \sqrt{220^2 - 4 \times 1 \times 12100}}{2} \), we find \( T \approx 110 \) (discarding the negative root).Thus, the tank is completely drained at approximately 110 hours.

Key Concepts

Rate of Change

The concept of rate of change is crucial in calculus, especially when dealing with dynamic phenomena. In this exercise, the rate of change function is given by \( V'(t) = 1 - \frac{t}{110} \). This function tells us how quickly or slowly oil is leaking from the tank at any given time \( t \). A positive rate implies addition, while a negative rate indicates reduction.

Understanding the rate of change:

• The function starts at \( V'(0) = 1 \), meaning initially, oil is leaking at a constant rate of 1 gallon per hour.
• As time progresses, the rate decreases (due to \(-\frac{t}{110}\)) until it ultimately reaches zero when \( t = 110 \).
• This descending rate means the leakage slows down uniformly over time.

The rate of change gives us immediate insight into the instantaneous behavior of the system, allowing us to predict how quickly a resource (like oil) will be depleted over time.

Definite Integrals

Definite integrals play a pivotal role in determining the total change over a specified interval. In this problem, definite integrals help us calculate the amount of oil lost during specific time periods. By integrating the rate of change function over

• \( t = 0 \) to \( t = 1 \), we find the oil lost in the first hour.
• Similarly, integrating from \( t = 9 \) to \( t = 10 \) helps us determine the leakage during the tenth hour.

Through integration, the area under the curve of the rate function over these intervals provides the total quantity affected, diminishing the need for complex summation methods.

The mathematical representation is:

• Definite integral of the rate function, \( \int_a^b (1 - \frac{t}{110}) \, dt \), calculates the total oil lost over \( [a, b] \) hours.

The antiderivative is determined and evaluated at the bounds to find the exact oil loss. Definite integrals effectively automate the accumulation process, providing precise measures of quantities within dynamic processes.

Quadratic Equations

Quadratic equations often arise in situations involving acceleration or growth, and, in this case, when finding total outcomes over time. When determining when the tank is entirely drained, we derived the quadratic equation:\( T^2 - 220T + 12100 = 0 \).

Solving Quadratic Equations:

• We use the quadratic formula, \( T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -220 \), and \( c = 12100 \).
• The formula helps find possible time \( T \) when the tank is drained.
• Out of the results, \( T = 110 \) makes sense physically, as time can't be negative.

Quadratic equations assist in predicting end conditions within systems that follow a quadratic pattern, allowing us to find precise points of system change over extended time periods.

Applications of Integration

Integration has numerous applications across various fields, including physics, biology, and environmental science. In real-world contexts, like this word problem, integration not only helps in solving for rates or totals, but it also assists in resource management strategies.

Common Applications in Word Problems:

• Determining total loss or gain over time, as seen with the oil leak.
• Predicting resource depletion, helping in planning by estimating when refills or stops are needed.
• Optimization in scenarios requiring maximum efficiency over a span of operations (like timed machinery or dosing in pharmaceuticals).

Integration thus extends beyond mere calculations; it provides the framework to interpret and manage quantities where gradual change affects outcomes significantly. Whether for logistical planning or scientific analysis, mastering integration is a powerful tool for solving diverse calculus word problems.