Problem 74 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

AIR PURIFICATION During testing of a certain brand of air purifier, the amount of smoke remaining \(t\) min after the start of the test was $$ \begin{aligned} A(t)=&-0.00006 t^{5}+0.00468 t^{4}-0.1316 t^{3} \\ &+1.915 t^{2}-17.63 t+100 \end{aligned} $$ percent of the original amount. Compute \(A^{\prime}(10)\) and \(A^{\prime \prime}(10)\) and interpret your results. Source: Consumer Reports

Step-by-Step Solution

Answer

The first derivative, \(A'(10) \approx -0.67\), indicates that at 10 minutes after engaging the air purifier, the amount of smoke is decreasing at a rate of approximately 0.67% per minute. The second derivative, \(A''(10) \approx -2.79\), shows that the rate at which the smoke is decreasing is accelerating at a value of approximately 2.79% per square minute at 10 minutes after the air purifier is turned on.

Step 1: Find the first derivative of the given function

To find the first derivative, we'll apply differentiation rules to the given function: $$ \begin{aligned} A(t) = &-0.00006 t^{5} + 0.00468 t^{4} - 0.1316 t^{3} + 1.915 t^{2} - 17.63 t + 100 \\ A'(t) = &\frac{d}{dt}(-0.00006 t^{5} + 0.00468 t^{4} - 0.1316 t^{3} + 1.915 t^{2} - 17.63 t + 100) \\ \end{aligned} $$ Applying differentiation rules: $$ A'(t) = -5(0.00006) t^{4} + 4(0.00468) t^{3} - 3(0.1316) t^{2} + 2(1.915) t - 17.63 $$ Simplify: $$ A'(t) = -0.0003 t^{4} + 0.01872 t^{3} - 0.3948 t^{2} + 3.83 t - 17.63 $$

Step 2: Evaluate the first derivative at \(t = 10\)

Now that we have the first derivative, we can plug in \(t = 10\) to find the rate of change at that time: $$ \begin{aligned} A'(10) = &-0.0003 (10)^{4} + 0.01872 (10)^{3} - 0.3948 (10)^{2} + 3.83(10) - 17.63 \\ \end{aligned} $$ Evaluate: $$ A'(10) \approx -0.67 $$

Step 3: Interpret the meaning of the result from step 2

In this case, \(A'(10)\) represents the rate of change of the smoke remaining at \(t = 10\). Since \(A'(10) \approx -0.67\), the negative value means that the amount of smoke is decreasing at a rate of approximately 0.67% per minute at 10 minutes after engaging the air purifier.

Step 4: Find the second derivative of the given function

Now let's find the second derivative of the function, which represents the rate of change of the first derivative. Start by differentiating \(A'(t)\): $$ A''(t) = \frac{d}{dt}(-0.0003 t^{4} + 0.01872 t^{3} - 0.3948 t^{2} + 3.83 t - 17.63) $$ Applying differentiation rules: $$ A''(t) = -4(0.0003) t^{3} + 3(0.01872) t^{2} - 2(0.3948) t + 3.83 $$ Simplify: $$ A''(t) = -0.0012 t^{3} + 0.05616 t^{2} - 0.7896 t + 3.83 $$

Step 5: Evaluate the second derivative at \(t = 10\)

Now, we can plug in \(t = 10\) into the second derivative to evaluate the acceleration at that time: $$ \begin{aligned} A''(10) = &-0.0012 (10)^{3} + 0.05616 (10)^{2} - 0.7896(10) + 3.83 \\ \end{aligned} $$ Evaluate: $$ A''(10) \approx -2.79 $$

Step 6: Interpret the meaning of the result from step 5

The second derivative, \(A''(10)\), indicates the acceleration of the rate of change of the smoke remaining in the environment. Since \(A''(10) \approx -2.79\), the negative value means that the rate at which the smoke is decreasing is also speeding up or accelerating at a value of approximately 2.79% per square minute at 10 minutes after the air purifier is turned on.

Key Concepts

Differentiation

Differentiation is one of the most fundamental concepts in calculus. It is the process through which we determine the rate at which a function is changing at any given point. This is essential in many real-world scenarios, such as understanding how quantities evolve over time. For instance, in the context of the air purifier, differentiation allows us to compute how quickly the smoke level decreases over time.

• Understanding Differentiation: At its core, differentiation calculates the slope of the tangent line to the curve of a function at any point. This slope is a representation of the rate of change.
• Notation: The derivative of a function is denoted by various symbols, such as \(A'(t)\) or \(\frac{dA}{dt}\).
• Rules: Differentiation follows several rules, like the power rule, to compute derivatives efficiently. For example, if you have \(t^n\), the derivative is \(nt^{n-1}\).

With these insights, you can not only compute derivatives but also interpret them within contexts like reducing smoke particles efficiently through purification devices.

First Derivative

The first derivative of a function, often simply called the "derivative," is the result of differentiation and shows the rate of change at any point on the function's graph. In the context of the air purifier test, the first derivative \(A'(t)\) tells us how fast the percentage of smoke is decreasing over time.

• Interpretation: Physically, \(A'(10) \approx -0.67\) indicates that at \(t = 10\) minutes, the smoke level is reducing at a rate of 0.67% per minute.
• Significance of Negative Values: The negative value of \(A'(10)\) signifies a decrease in the amount of smoke, which means the purifier is effectively reducing smoke. Positive values would imply an increase, which is generally less desirable in this context.
• Importance: Knowing the first derivative helps in understanding the immediate effect and efficiency of the air purifier at any specific minute during the operation.

The first derivative is a simple yet powerful tool to analyze and interpret dynamic systems.

Second Derivative

The second derivative of a function provides insight into the curvature or concavity of the function's graph. In physical terms, it can represent the acceleration or deceleration of a particular rate of change. For the air purifier's efficiency, the second derivative \(A''(t)\) informs us about the change in the rate at which the smoke level decreases.

• Mathematical Role: Mathematically, the second derivative is the derivative of the first derivative, symbolized as \(A''(t)\), and it details how the first derivative \(A'(t)\) is changing.
• Interpretation of Calculations: The result \(A''(10) \approx -2.79\) suggests that the rate of decrease in smoke is itself increasing. This means the purifier is speeding up its smoke removal efficiency at 10 minutes.
• Importance of Sign: A negative second derivative signifies that the rate at which smoke decreases is speeding up, enhancing the purifier's performance effectiveness over time.

The second derivative provides a deeper insight into the behavior of changes, ensuring a complete understanding of the effects of technological tools like air purifiers.

Rate of Change

The term "rate of change" refers to how a quantity is varying over time. In the study of calculus, especially in differentiating functions as seen in the air purifier's function, it becomes a crucial concept. Here, it is vital to determine how the smoke level changes with respect to time.

• Instantaneous Rate of Change: This is what the first derivative, like \(A'(t)\), provides. It tells us about the precise rate of change at an exact moment.
• Acceleration of Rate: The second derivative \(A''(t)\) describes how this rate is changing. A change in rate can be just as important as the rate itself, especially in optimizing performance.
• Application: In practical terms, understanding the rate of change can lead to improved device settings or enhancements, maximizing overall efficiency in processes like purification.

By analyzing both the rate of change and its acceleration, we can optimize systems for better performance and ensure that technological solutions are being applied effectively.