Question

Newton's law of cooling is given by: \(\theta=\theta_{0} \mathrm{e}^{-k t}\), where the excess of temperature at zero time is \(\theta_{0}^{\circ} \mathrm{C}\) and at time \(t\) seconds is \(\theta^{\circ} \mathrm{C}\). Determine the rate of change of temperature after \(50 \mathrm{~s}\), given that \(\theta_{0}=15^{\circ} \mathrm{C}\) and \(k=-0.02\)

Short answer

The rate of change of temperature after 50 seconds is approximately 0.81549°C/s.

Step-by-step solution

Understand the Problem

We need to find the rate of change of the temperature \( \theta \) at \( t = 50 \, s \). This is the derivative \( \frac{d\theta}{dt} \) of the function \( \theta(t) = \theta_0 e^{-kt} \). We are given \( \theta_0 = 15^{\circ}C \) and \( k = -0.02 \).

Differentiate the Function

The formula is \( \theta(t) = \theta_0 e^{-kt} \). To find the rate of change, we differentiate this with respect to \( t \), giving the derivative: \( \frac{d\theta}{dt} = -k \theta_0 e^{-kt} \).

Substitute Known Values

Substitute \( \theta_0 = 15 \) and \( k = -0.02 \) into the derivative: \[ \frac{d\theta}{dt} = -(-0.02) \times 15 \times e^{0.02 t} = 0.02 \times 15 \times e^{0.02 t} \].

Calculate at t = 50 seconds

Substitute \( t = 50 \) into the expression obtained in the previous step: \( \frac{d\theta}{dt} = 0.02 \times 15 \times e^{0.02 \times 50} \). Simplify it further: \( \frac{d\theta}{dt} = 0.02 \times 15 \times e^{1} \).

Evaluate Expression

Calculate the numerical value: \( e^{1} \approx 2.7183 \), so \( \frac{d\theta}{dt} = 0.02 \times 15 \times 2.7183 \). Thus, \( \frac{d\theta}{dt} \approx 0.02 \times 15 \times 2.7183 = 0.81549 \).

Conclusion on Rate of Change

The rate of change of temperature at \( t = 50 \) seconds is approximately \( 0.81549^{\circ}C/s \).

Key concepts

Understanding Temperature Rate of Change

Newton's Law of Cooling helps us understand how an object's temperature changes over time, especially how quickly it cools down to match its surrounding environment.
This is expressed by the formula \( \theta(t) = \theta_0 e^{-kt} \). Here, \( \theta(t) \) is the temperature at time \( t \), \( \theta_0 \) is the initial temperature difference, and \( k \) is a cooling constant.
The rate of change of temperature, represented as \( \frac{d\theta}{dt} \), tells us how fast the temperature is changing at any given moment. In practical terms, it's like saying how fast your coffee is cooling down after being poured.
To find the rate of change, you need to calculate the derivative of the temperature function, which involves solving a mathematical expression that includes constants and variables.
Why is it important to know the rate of temperature change? By knowing this, scientists and engineers can make predictions and consider safety measures for various processes that involve temperature regulation, such as food cooling, material manufacturing, and even weather predictions.

Exponential Functions and Their Role

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent, like \( e^{-kt} \) in our problem. They are fascinating and widely used in scientific fields because they model growth or decay processes.
When you have a negative exponent, as in \( e^{-kt} \), the function models exponential decay. This is exactly what's happening in Newton's Law of Cooling, where the temperature difference decreases over time.
Exponential decay is common in real life. It's used to explain processes like radioactive decay, population decline, and of course, cooling of objects. It's important to understand the concept of half-life, which refers to the time it takes for a quantity to reduce to half its initial value—that concept is inherently tied to exponential functions.
In simpler terms, anytime you're looking at something that decreases rapidly initially and then slows down over time, you're likely looking at an exponential decay process. Making sense of this can help you better understand temperature changes and other natural processes that follow similar patterns.

Differentiation in Action

Differentiation is a key concept in calculus, dealing with how a function's output changes as its input changes. In our situation, differentiation helps us find the rate of temperature change from the temperature-time function.
The derivative of a function gives you the rate at which one variable changes with respect to another. For example, if you differentiate \( \theta(t) = \theta_0 e^{-kt} \), you get \( \frac{d\theta}{dt} = -k \theta_0 e^{-kt} \). Here, it tells us how quickly the temperature decreases over time.
When solving real-world problems like our temperature cooling scenario, differentiation provides insights by breaking down complex changes into simpler, more manageable rates of change. This is true across many fields, from physics to economics.
Always remember: finding the derivative is about understanding change. Whether you're dealing with temperatures cooling down or stocks fluctuating, derivatives show the instantaneous rate of how things change, giving invaluable information for decision-making, analysis, and prediction.