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Chapter 4: Problem 93

GENERAL: Temperature A mug of beer chilled to 40 degrees, if left in a 70 -degree room, will warm to a temperature of \(T(t)=70-30 e^{-3.5 t}\) degrees in \(t\) hours. a. Find \(T(0.25)\) and \(T^{\prime}(0.25)\) and interpret your answers. b. Find \(T(1)\) and \(T^{\prime}(1)\) and interpret your answers.

Short Answer

Expert verified
After 0.25 hours, the temperature is 57.5掳F, increasing at 43.8掳F/hr. After 1 hour, it is 69.1掳F, increasing at 3.2掳F/hr.

Step by step solution

01

Understanding the Temperature Function

The temperature of the beer is given by \[T(t) = 70 - 30e^{-3.5t}\] This formula shows how temperature changes over time \(t\) as the beer warms towards room temperature.
02

Finding \(T(0.25)\)

Substitute \(t = 0.25\) into the temperature function:\[T(0.25) = 70 - 30e^{-3.5(0.25)}\]Calculate \(e^{-3.5 \times 0.25}\) and subtract from 70.\[T(0.25) \approx 70 - 30e^{-0.875} \approx 70 - 30 \times 0.41686 \approx 57.4942\] Thus, the temperature of the beer after 0.25 hours is approximately 57.5 degrees.
03

Finding the Derivative \(T^{\prime}(t)\)

Find the derivative of the function \(T(t)\). The derivative is:\[T^{\prime}(t) = \frac{d}{dt}[70 - 30e^{-3.5t}] = 30 \times 3.5 \times e^{-3.5t}\] which simplifies to: \[ T^{\prime}(t) = 105e^{-3.5t} \] This function gives the rate of change of the temperature over time.
04

Finding \(T^{\prime}(0.25)\)

Substitute \(t = 0.25\) into the derivative function:\[T^{\prime}(0.25) = 105e^{-3.5 \times 0.25} = 105 \times e^{-0.875}\]Calculate:\[T^{\prime}(0.25) \approx 105 \times 0.41686 \approx 43.7703\] This means the temperature is increasing at a rate of approximately 43.8 degrees per hour after 0.25 hours.
05

Finding \(T(1)\)

Substitute \(t = 1\) into the temperature function:\[T(1) = 70 - 30e^{-3.5 \times 1}\]Calculate:\[T(1) = 70 - 30 \times e^{-3.5} \approx 70 - 30 \times 0.0302 \approx 69.094\] Thus, the temperature of the beer after 1 hour is approximately 69.1 degrees.
06

Finding \(T^{\prime}(1)\)

Substitute \(t = 1\) into the derivative function:\[T^{\prime}(1) = 105e^{-3.5 \times 1} = 105 \times e^{-3.5}\]Calculate:\[T^{\prime}(1) \approx 105 \times 0.0302 \approx 3.171\]This means the temperature is increasing at a rate of approximately 3.2 degrees per hour after 1 hour.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions are mathematical expressions where the variable appears in the exponent. They are crucial in modeling situations involving rapid growth or decay, such as temperature change, population growth, and radioactive decay. In the context of our exercise, the function \( T(t) = 70 - 30e^{-3.5t} \) is exponential because it involves the term \( e^{-3.5t} \). This term models the rate at which the temperature of the beer approaches the ambient room temperature.
  • The constant in the base of the exponential, \( e \), is approximately equal to 2.718 and represents the natural exponential function.
  • The negative exponent \(-3.5t\) indicates decay. As time \( t \) increases, \( e^{-3.5t} \) gets smaller, slowing the rate of temperature change.
  • The constants 70 and 30 signify the room temperature and the temperature range over which the beer warms up, respectively.
By understanding the components of exponential functions, students can analyze how variables change over time in various scenarios.
Differentiation
Differentiation is a fundamental concept in calculus used to find the rate of change of a quantity. It provides us with a powerful tool for calculating how a function changes at any given point. In our exercise, we differentiate the given temperature function to find the rate at which the temperature changes over time \( t \).
  • The derivative of a function gives us the slope of the tangent line and indicates how steep the curve is at any point.
  • For the function \( T(t) = 70 - 30e^{-3.5t} \), the derivative is \( T^{\prime}(t) = 105e^{-3.5t} \).
  • This derivative shows that as \( t \) increases, the rate of temperature increase decreases because \( e^{-3.5t} \) becomes smaller.
By working through differentiation, students gain insight into how quickly or slowly changes occur within a system.
Temperature Models
Temperature models like the one in our problem help predict how temperatures change over time under specific conditions. The model \( T(t) = 70 - 30e^{-3.5t} \) illustrates how a beverage left in a warmer environment reaches thermal equilibrium, which is a common real-world scenario.
  • The initial temperature of the beer is 40 degrees, and it attempts to reach the room's ambient temperature of 70 degrees.
  • The term \( 70 - 30e^{-3.5t} \) indicates how the temperature difference decreases over time, showing the beer gradually warming to room temperature.
  • By evaluating this function at different time points, we can understand both the temperature of the beer and the rate of temperature change, helping to predict when it will reach certain temperatures.
Developing a temperature model aids students in interpreting real-life thermal processes, improving their understanding of heat transfer dynamics.

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