Problem 65 - Calculus Early Transcendentals: Pearson New International Edition

Question

Water leaks out of a 200 -gallon storage tank (initially full) at the rate \(V^{\prime}(t)=20-t\), where \(t\) is measured in hours and \(V\) in gallons. How much water leaked out between 10 and 20 hours? How long will it take the tank to drain completely?

Step-by-Step Solution

Answer

50 gallons leaked out; the tank drains completely in 20 hours.

Step 1: Understanding the Leakage Rate

The rate of water leakage from the tank is given by the function \( V'(t) = 20 - t \). This equation tells us the rate at which water is leaking from the tank in gallons per hour.

Step 2: Find the Total Water Leaked Between 10 and 20 Hours

To find how much water leaked between 10 and 20 hours, we need to integrate the rate function \( V'(t) \) from 10 to 20. This involves calculating the definite integral:\[\int_{10}^{20} (20 - t) \, dt. \]This will give us the total amount of water leaked in this time interval.

Step 3: Calculate the Definite Integral

Compute the integral:\[\int_{10}^{20} (20 - t) \, dt = \left[ 20t - \frac{t^2}{2} \right]_{10}^{20}.\]Evaluate this from 10 to 20:\[= \left( 20(20) - \frac{20^2}{2} \right) - \left( 20(10) - \frac{10^2}{2} \right)\]which simplifies to:\[= (400 - 200) - (200 - 50) = 200 - 150 = 50.\]Thus, 50 gallons of water leaked out between the 10th and 20th hour.

Step 4: Determine When the Tank Drains Completely

To find when the tank will drain entirely, set the original volume of water equal to the total outflow over time. Integrate from 0 to \( t \) to find the total water that has leaked out:\[V(t) = \int (20 - t) \, dt = 20t - \frac{t^2}{2}.\]Set \( V(t) = 200 \) (initial volume of water):\[20t - \frac{t^2}{2} = 200.\]

Step 5: Solve for the Time When Volume is Zero

Solve the equation \( 20t - \frac{t^2}{2} = 200 \). Rearranging terms gives:\[-t^2 + 40t - 400 = 0. \]Multiply through by \(-1\):\[t^2 - 40t + 400 = 0.\]Use the quadratic formula: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]For \( a = 1, b = -40, c = 400 \), calculate:\[t = \frac{-(-40) \pm \sqrt{1600 - 1600}}{2} = \frac{40 \pm 0}{2} = 20.\]Thus, it will take 20 hours for the tank to drain completely.

Key Concepts

Definite Integral

The concept of a definite integral is pivotal in calculus, particularly in scenarios where we seek to determine the total accumulation of a quantity over a time interval. In this problem, we have a water tank with a leaking rate described by the function \( V'(t) = 20 - t \). To find out how much water has leaked between 10 and 20 hours, we deploy the definite integral: \[\int_{10}^{20} (20 - t) \, dt\]This integral calculates the total amount of water leaked during this timeframe by summing up the infinitesimal quantities of water leaving the tank continuously. Think of the definite integral as summarizing tiny slices of change happening every instant between 10 and 20 hours.

When calculated, the definite integral evaluates to 50, thereby indicating 50 gallons of water leaked during this period. This cumulative approach is why definite integrals are fundamental in understanding total change over a specific interval.

Rate of Change

Understanding the rate of change is crucial as it reflects how one quantity shifts relative to another, often time. In this exercise, the rate of water leaking from the tank is depicted by the derivative function \( V'(t) = 20 - t \). This function informs us of the water leakage speed at any moment \( t \). For example, at time \( t = 0 \), the rate of leakage is \( 20 \) gallons per hour, but as \( t \) increases, the rate decreases linearly.

• The rate is positive when \( t < 20 \), indicating water is actively leaking out.
• After \( t = 20 \), the rate would become negative, meaning the tank would have theoretically reached its empty state, assuming no physical constraints.

Being able to interpret a rate of change function like this can provide great insights into dynamic processes, helping predict future states of systems based on current trends.

Quadratic Equation

Quadratic equations are essential algebraic expressions represented typically as \( ax^2 + bx + c = 0 \). They often emerge in calculus problems, particularly when dealing with motion or change over time. In this exercise, after integrating the rate function to find when the tank will be empty, we derived the quadratic equation:\[t^2 - 40t + 400 = 0\]This equation arose by setting the accumulated outflow \( V(t) \) equal to the tank's initial capacity of 200 gallons. Solving this using the quadratic formula:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \( a = 1, b = -40, c = 400 \), gives us the time \( t = 20 \). This calculation tells us it will take 20 hours for the total tank to drain, verifying our practical rate of leakage against the initial conditions.

Quadratic equations are powerful in revealing when particular conditions (like tank emptiness) are met, highlighting their importance beyond pure algebra.

Volume Calculation

Volume calculations are critical in many applied mathematics problems, particularly those involving fluids, as seen in this calculus exercise. Here, the volume makes up the essential initial condition and desired outcome of our calculation:

• The tank starts fully filled with a volume of 200 gallons.
• Our task was to determine how much volume is left or has leaked at particular intervals and when it completely empties.

Utilizing calculus, we integrate the rate of change of the water's volume to discover resulting volumes over time. The integral provides us with accumulated leaked volume at any point, enabling us to predict when the tank will be empty based on initial data.

Thus, volume calculation in calculus ties neatly into determining cumulative results of dynamic processes, assuring efficiency in predictive and analytical tasks.