Problem 57 - Calculus and its applications

Question

Suppose the temperature in degrees Celsius over a 10 -hr period is given by $$ f(t)=-t^{2}+5 t+40, \quad 0 \leq t \leq 10 $$ a) Find the average temperature. b) Find the minimum temperature. c) Find the maximum temperature.

Step-by-Step Solution

Answer

a) Average temperature is 31.67°C. b) Minimum temperature is -10°C. c) Maximum temperature is 46.25°C.

Step 1: Formula for Average Temperature

The average temperature over a time interval from \( t = a \) to \( t = b \) is given by the formula \( \text{Average} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt \). In this case, the function is given as \( f(t) = -t^2 + 5t + 40 \), and the interval is \( 0 \leq t \leq 10 \).

Step 2: Set Up the Integral for Average Temperature

We need to evaluate the integral \( \int_{0}^{10} (-t^2 + 5t + 40) \, dt \). This integral will help find the total temperature over the 10-hour period, whose average can then be calculated.

Step 3: Integrate the Function

Integrate each term separately: \( \int (-t^2) \, dt = -\frac{t^3}{3} \), \( \int 5t \, dt = \frac{5t^2}{2} \), and \( \int 40 \, dt = 40t \). The integrated function is \(-\frac{t^3}{3} + \frac{5t^2}{2} + 40t \).

Step 4: Evaluate the Definite Integral

Evaluate the integrated function from \( t = 0 \) to \( t = 10 \): \[ \left[-\frac{t^3}{3} + \frac{5t^2}{2} + 40t \right]_{0}^{10} = \left(-\frac{10^3}{3} + \frac{5 \times 10^2}{2} + 40 \times 10 \right) - \left(0\right) \].

Step 5: Calculate the Average Temperature

Compute the values: \(-\frac{1000}{3} + 250 + 400 = -\frac{1000}{3} + 650 \). Convert \(-\frac{1000}{3}\) to decimal, which is approximately \(-333.33\). Thus, \(-333.33 + 650 = 316.67\). Divide by the interval length (10): \( \frac{316.67}{10} = 31.67 \).

Step 6: Find Critical Points for Minimum and Maximum Temperature

Calculate the derivative of \( f(t) \): \( f'(t) = -2t + 5 \). Set the derivative to zero: \( -2t + 5 = 0 \) which gives \( t = 2.5 \). This is a critical point.

Step 7: Evaluate Temperature at Critical Point and Endpoints

Evaluate \( f(t) \) at the critical point \( t = 2.5 \) and endpoints \( t = 0 \) and \( t = 10 \): \( f(0) = 40 \), \( f(10) = -10 \), \( f(2.5) = -2.5^2 + 5(2.5) + 40 = 46.25 \).

Step 8: Determine Minimum and Maximum Temperature

From the calculated values, the minimum temperature is \( f(10) = -10 \), and the maximum temperature is \( f(2.5) = 46.25 \).

Key Concepts

Average Temperature Calculation

Calculating average temperature is an excellent way to understand how temperatures vary over time. It's especially useful in calculus, where the average value of a function over an interval is sought. To find this average temperature over a period, like a 10-hour interval, you use a specific formula. The formula for the average temperature is: \[ \text{Average} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt \]Here, \(a\) and \(b\) are the boundaries of the time period, and \(f(t)\) represents the temperature function.To break this down with an example:

• The function given is \(-t^2 + 5t + 40\), which describes how the temperature changes over time.
• The interval is from \(t = 0\) to \(t = 10\), representing a 10-hour period.

By using the definite integral, which calculates the total temperature over these 10 hours, and dividing by the length of the interval, we determine the average temperature during this time.

Definite Integral Evaluation

Evaluating definite integrals is a core part of calculus, often used to find the total accumulation of a quantity. Here, we focus on how it helps compute the total temperature over time.To evaluate an integral for the temperature function \(f(t) = -t^2 + 5t + 40\), we start by breaking it into simpler parts. Each term in the function is integrated separately:

• \( \int (-t^2) \, dt = -\frac{t^3}{3} \)
• \( \int 5t \, dt = \frac{5t^2}{2} \)
• \( \int 40 \, dt = 40t \)

Once integrated, the function becomes:\[-\frac{t^3}{3} + \frac{5t^2}{2} + 40t\]Next, this integrated function is evaluated between the limits \(t = 0\) and \(t = 10\). This gives us the total temperature, helping us understand the overall behavior of the function over the specified time interval.

Critical Points in Calculus

Critical points are vital in determining the behavior of functions, such as finding minimum and maximum values. In temperature problems, they reveal when temperatures spike or drop significantly.To locate critical points, find where the derivative of the function equals zero. For \(f(t) = -t^2 + 5t + 40\), the derivative is:\[f'(t) = -2t + 5\]Set the derivative to zero to find the critical point.

• \( -2t + 5 = 0 \)
• This simplifies to \( t = 2.5 \)

At this critical point and at the endpoints of the interval (\(t = 0\) and \(t = 10\)), we evaluate the function:

• \(f(0) = 40\)
• \(f(2.5) = 46.25\)
• \(f(10) = -10\)

The highest temperature in this range, 46.25°C, occurs at \(t = 2.5\) and is the maximum value, while the lowest temperature, -10°C, at \(t = 10\), is the minimum value. Critical points help elucidate these extremes.