Problem 55 - Brief Applied Calculus

Question

The temperature in an industrial pasteurization tank is \(f(x)=\) \(x^{2}-8 x+110\) degrees centigrade after \(x\) minutes (for \(0 \leq x \leq 12\) ). a. Find \(f^{\prime}(x)\) by using the definition of the derivative. b. Use your answer to part (a) to find the instantaneous rate of change of the temperature after 2 minutes. Be sure to interpret the sign of your answer. c. Use your answer to part (a) to find the instantaneous rate of change after 5 minutes.

Step-by-Step Solution

Answer

a. \( f^{\prime}(x) = 2x - 8 \); b. -4 (temperature decreases); c. 2 (temperature increases).

Step 1: Recall the Definition of a Derivative

The derivative of a function \( f(x) \) at a point \( x = a \) is given by the limit: \[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\] This represents the instantaneous rate of change of \( f(x) \) at \( x = a \).

Step 2: Substitute into the Derivative Formula

Given the function \( f(x) = x^2 - 8x + 110 \), substitute \( f(a+h) \) and \( f(a) \) into the derivative formula:\[f^{\prime}(x) = \lim_{h \to 0} \frac{(x+h)^2 - 8(x+h) + 110 - (x^2 - 8x + 110)}{h}\] Simplify the numerator.

Step 3: Simplify the Expression

Expand and simplify the expression inside the limit:\[(x+h)^2 = x^2 + 2xh + h^2\]\[-8(x+h) = -8x - 8h\]Then simplify the whole expression:\[f^{\prime}(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 8x - 8h + 110 - x^2 + 8x - 110}{h}\]Combine like terms:\[f^{\prime}(x) = \lim_{h \to 0} \frac{2xh + h^2 - 8h}{h}\]

Step 4: Factor and Evaluate the Limit

Factor out \( h \) from the simplified expression:\[f^{\prime}(x) = \lim_{h \to 0} \frac{h(2x + h - 8)}{h}\]Cancel \( h \) in the numerator and denominator and evaluate the limit:\[f^{\prime}(x) = \lim_{h \to 0} (2x + h - 8) = 2x - 8\] Thus, the derivative \( f^{\prime}(x) = 2x - 8 \).

Step 5: Calculate Instantaneous Rate at 2 Minutes

Substitute \( x = 2 \) into the derivative:\[f^{\prime}(2) = 2(2) - 8 = 4 - 8 = -4\]This indicates that the temperature is decreasing at a rate of 4 degrees per minute after 2 minutes.

Step 6: Calculate Instantaneous Rate at 5 Minutes

Substitute \( x = 5 \) into the derivative:\[f^{\prime}(5) = 2(5) - 8 = 10 - 8 = 2\]This indicates that the temperature is increasing at a rate of 2 degrees per minute after 5 minutes.

Key Concepts

Instantaneous Rate of Change

The concept of instantaneous rate of change is essential in calculus to understand how a variable changes at a specific point in time. When we talk about the rate at which something changes, we're referring to how fast it's moving—something very useful for measuring changes like temperature in a pasteurization tank, as in our problem.

In a mathematical sense, the instantaneous rate of change of a function at a particular point is equivalent to the slope of the tangent line at that point. Think of it as capturing the essence of how the temperature in the tank evolves at an exact minute mark, rather than over a stretch of time.

For example, in the exercise, we found that the rate of temperature change after 2 minutes is \(-4\) degrees per minute, indicating a cooling process. After 5 minutes, the rate is \(2\) degrees per minute, highlighting a warming trend. Interpreting the sign in these answers is crucial:

• A negative rate means the value is decreasing, as seen at 2 minutes.
• A positive rate means the value is increasing, as observed at 5 minutes.

Understanding this helps in predicting and controlling processes like pasteurization more effectively.

Limit Definition of Derivative

The limit definition of a derivative is fundamental in calculus, acting as the bridge to understanding how to compute derivatives. It allows us to formally capture the notion of the instantaneous rate of change. Here's how it works:

For a function \(f(x)\), the derivative at a point \(x = a\) is defined by the limit:\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\]This represents the slope of the tangent at the point \(a\). The idea is to look at how the function \(f(x)\) changes as we make very small changes \(h\) around the point \(a\).

In our pasteurization problem, to find \(f'(x)\) for \(f(x) = x^2 - 8x + 110\), we substitute into this definition, leading us through steps such as expanding \((x+h)^2\) and simplifying the expression. At the end of simplification, you arrive at a formula for the derivative: \(f'(x) = 2x - 8\).

This derived formula now easily allows us to evaluate the temperature's rate of change at any given moment \(x\), showcasing the power of derivatives in practical applications.

Temperature Change Rate

When dealing with processes that involve changing temperatures, like in a pasteurization tank, understanding the rate at which temperature changes is crucial. This is effectively captured by the derivative of the temperature function.

Using the derivative \(f'(x)\), we can determine at any minute \(x\) how quickly the temperature is rising or falling. From the problem, this derivative \(f'(x) = 2x - 8\) gives us a straightforward indication:

• Positive values of \(f'(x)\) signify temperature increases, pointing to heating.
• Negative values indicate cooling, where the temperature decreases over time.

For instance, at 2 minutes, \(f'(2)\) yields a negative 4, thus indicating cooling. At 5 minutes, \(f'(5)\) is positive 2, suggesting a warming phase. This analysis helps to effectively manage and optimize industrial processes where controlled temperature is crucial.

By examining the rates at specific time points, operators can make informed decisions to regulate temperature, ensuring the process stays within desired parameters for safety and efficiency.