Question

Water is drained from a swimming pool at a rate given by \(R(t)=100 e^{-0.05 t}\) gal/hr. If the drain is left open indefinitely, how much water is drained from the pool?

Short answer

Answer: 2000 gallons.

Step-by-step solution

Identify the draining rate function and the integration limits

The draining rate function is given by \(R(t)=100e^{-0.05 t}\) gallons per hour. We want to find the total amount of water drained, so we will integrate this function with respect to time from 0 to infinity.

Set up the integral

To find the total amount of water drained from the pool, we need to integrate the draining rate function with respect to time from 0 to infinity. That is, we need to calculate: $$ \int_0^{\infty}100e^{-0.05 t} dt $$

Evaluate the integral

Using the substitution method, let's set \(u=-0.05t\). Then, \(\frac{du}{dt}=-0.05\), so \(dt=-\frac{1}{0.05}du=-20du\). Now, we will change the limits of integration according to our substitution: $$ t=0 \Rightarrow u=-0.05(0)=0 \quad \text{and} \quad t \to \infty \Rightarrow u \to -\infty $$ Now, substitute \(u\) and \(dt\) into our integral and adjust the limits of integration: $$ \int_{-0}^{-\infty}100e^{u}(-20)du $$ Notice that we have a negative limit on the right, so let's reverse the limits and remove the minus sign on the \((-20)\): $$ -1\int_{-\infty}^{0}100e^{u}(20)du $$ Now, pull out a constant \(-20(100)\) in front of the integral and we have: $$ -2000\int_{-\infty}^{0}e^{u}du $$

Solve for the integral

Now, we can find the antiderivative of \(e^u\) which is itself \(e^u\). Then, evaluate this antiderivative from \(-\infty\) to \(0\): $$ -2000(e^{u}\Big|_{-\infty}^{0}) $$ $$ -2000(e^0-e^{-\infty})=-2000(1-0)=-2000 $$

Interpret the result

The total amount of water drained from the swimming pool when the drain is left open indefinitely is 2000 gallons.

Key concepts

Exponential Functions

Exponential functions are mathematical expressions that involve exponents, typically represented in the form \( a^x \), where \( a \) is a constant known as the base, and \( x \) is the exponent. In calculus, exponential functions often appear in the form \( e^x \), where \( e \) is Euler’s number, approximately equal to 2.71828. This special base is fundamental due to its unique properties which simplify differentiation and integration processes.

In the context of our problem, the draining rate function of the pool is given as an exponential function, \( R(t) = 100e^{-0.05t} \). Here, \( 100 \) is a coefficient, and the exponent includes \( t \), the time variable, with a constant rate of change \(-0.05\).

Exponential functions are significant because they model real-world phenomena, such as population growth, radioactive decay, and in this case, the rate at which water is drained from the pool. Understanding how to handle these functions is essential for various applications in science and engineering.

Definite Integrals

Definite integrals are used in calculus to find the total accumulation of a quantity, represented by the area under a curve between two points. The definite integral is written as \( \int_a^b f(x) \, dx \), where \( f(x) \) is the function of interest, and \( a \) and \( b \) are the limits of integration.

In our swimming pool problem, we use the definite integral to determine the total volume of water drained over time, represented as \( \int_0^{\infty} 100e^{-0.05t} dt \). The limits are from 0 to infinity, indicating the process of draining begins at time \( t = 0 \) and continues indefinitely.

Solving this definite integral can show the accumulated amount of water drained, providing insight into how much water the pool loses over an unlimited timeframe. This method is invaluable for calculating total changes where variables continuously accumulate.

Improper Integrals

Improper integrals arise when integrating over an infinite interval or when the integrand has an infinite discontinuity within the interval of integration. They are written similarly to definite integrals but require special techniques to evaluate.

The integral in our exercise, \( \int_0^{\infty} 100e^{-0.05t} dt \), is classified as an improper integral because the upper limit of integration is infinity. To evaluate such integrals, one typically takes the limit of a definite integral, letting the upper boundary approach infinity.

In practice, to evaluate an integral from 0 to infinity, you substitute infinity with a variable, say \( M \), then compute \( \lim_{M \to \infty} \int_0^M 100e^{-0.05t} dt \). This process helps determine whether the area under the curve converges to a finite value, indicating the total volume of water drained in this scenario is finite despite the infinite time horizon.

Substitution Method

The substitution method is a powerful technique for evaluating integrals, especially when dealing with exponential and trigonometric functions. This method simplifies complicated integrals by replacing a part of the integrand with a new variable, making the integral easier to solve.

In our problem, we use substitution to tackle the improper integral \( \int_0^{\infty} 100e^{-0.05t} dt \). We set \( u = -0.05t \), and consequently, the differential \( dt \) transforms to \( -20du \). This changes the integral into: \[ -2000 \int_{-\infty}^{0} e^u \, du \]

By completing this substitution, the integral becomes a standard form of the exponential function \( e^u \), which simplifies to \( e^u \) itself upon integration. Substitution is crucial in calculus for rearranging complex integrals into more manageable forms, aiding in their solution.