Problem 61 - Brief Applied Calculus

Question

ENVIRONMENTAL SCIENCE: Water Purification If the cost of purifying a gallon of water to a purity of \(x\) percent is \(C(x)=\frac{100}{100-x}\) cents \(\quad\) for \(50 \leq x<100\) a. Find the instantaneous rate of change of the cost with respect to purity. b. Evaluate this rate of change for a purity of \(95 \%\) and interpret your answer. c. Evaluate this rate of change for a purity of \(98 \%\) and interpret your answer.

Step-by-Step Solution

Answer

a. \( C'(x) = \frac{100}{(100-x)^2} \). b. At 95%, the rate is 4 cents/percent; costs rise steadily. c. At 98%, the rate is 25 cents/percent; costs rise sharply.

Step 1: Differentiate the Cost Function

The first step is to find the derivative of the cost function \(C(x) = \frac{100}{100 - x}\) with respect to \(x\). This will give us the instantaneous rate of change. Using the quotient rule, the derivative \(C'(x)\) is calculated as follows:\[ C'(x) = \frac{d}{dx} \left( \frac{100}{100-x} \right) = \frac{0 \cdot (100-x) - 100 \cdot (-1)}{(100-x)^2} = \frac{100}{(100 - x)^2} \]

Step 2: Evaluate the Rate of Change at 95% Purity

Now, substitute \(x = 95\) into the derivative to find the rate of change at 95% purity:\[ C'(95) = \frac{100}{(100 - 95)^2} = \frac{100}{5^2} = \frac{100}{25} = 4 \text{ cents per percent} \]This means the cost of increasing the purity from 95% by 1% is 4 cents.

Step 3: Interpret the Rate of Change at 95%

The instantaneous rate of change at 95% indicates that for every 1% increase in purity around 95%, the cost increases by 4 cents. This suggests that as purity increases, costs rise significantly.

Step 4: Evaluate the Rate of Change at 98% Purity

Substitute \(x = 98\) into the derivative to find the rate of change at 98% purity:\[ C'(98) = \frac{100}{(100 - 98)^2} = \frac{100}{2^2} = \frac{100}{4} = 25 \text{ cents per percent} \]The cost of increasing purity from 98% by 1% is 25 cents.

Step 5: Interpret the Rate of Change at 98%

The instantaneous rate of change at 98% indicates that for every additional 1% increase in purity around 98%, the cost increases by 25 cents. This shows a much steeper cost increase compared to 95%, demonstrating how expensive it becomes to achieve higher purity levels.

Key Concepts

Instantaneous Rate of Change

The instantaneous rate of change represents how a quantity changes immediately as you make a small change in another variable. Imagine it like checking your speedometer while driving; it shows your speed at that exact moment.
The mathematical tool we use to determine instantaneous rate of change is differentiation. In the context of the exercise, we're interested in how the cost of purifying water changes as we try to make the water even purer.To find this rate, we differentiate the cost function, which is given as \(C(x) = \frac{100}{100-x}\) cents per gallon. By using calculus, specifically the quotient rule, we find the derivative of this function, denoted \(C'(x)\). This derivative provides us with the rate at which the cost changes with respect to changes in purity level \(x\). The formula is:

• \(C'(x) = \frac{100}{(100-x)^2}\)

This expression will tell us exactly how many more cents it costs to increase the purity by 1% at any given purity level.

Cost Function Differentiation

Differentiating the cost function helps us understand the relationship between the cost and level of water purity. Think of differentiation as a tool that zooms in to see how changes are unfolding, similar to a detective looking for clues.In this scenario, the cost function is \(C(x) = \frac{100}{100-x}\). The differentiation shows us the sensitivity of the cost to changes in purity. If you increase the purity level by a tiny amount, the derivative \(C'(x)\) will tell you how much the cost will change.By substituting specific values for \(x\), like 95% or 98% purity, into the derivative, we find:

• At 95% purity, \(C'(95) = 4\) cents per percent, meaning an additional 1% purity costs 4 cents more.
• At 98% purity, \(C'(98) = 25\) cents per percent. This makes increasing purity at this level much more expensive.

Differentiation reveals these cost dynamics, highlighting the challenges and expenses of striving for purer water.

Water Purification Economics

Understanding the economics behind water purification involves comprehending how costs escalate as the desired purity level approaches near-perfect quality. This economic insight is crucial because every extra percentage point of purity, especially at higher levels, incurs a more significant increase in costs.The exercise shows by calculating \(C'(95)\) and \(C'(98)\), that it's much cheaper to purify water from 95% to 96% compared to from 98% to 99%. This steep cost increase is due to the greater precision and resources needed when achieving high purity levels.

• At 95%, a 1% increase costs an additional 4 cents.
• At 98%, the same increase costs 25 cents.

This economic principle is not only crucial for environmental science but is also applicable to various industrial processes where purity and quality are pivotal. Such insights help industries budget effectively for purification processes and make informed decisions to balance cost against necessary purity levels.