To find the rate of change of the temperature $T(t)$ at time $t = 8$, we need to compute the derivative of the function $T(t) = 21e^{-0.4t}$, which represents the temperature of the water.

Step 1: Differentiate $T(t)$

We can apply the chain rule to differentiate the function.

$T(t) = 21e^{-0.4t}$

The derivative of $e^{x}$ is $e^{x}$, and then we multiply by the derivative of the exponent $-0.4t$, which is $-0.4$. Therefore, the derivative $T'(t)$ is:

$T'(t) = 21 \times (-0.4)e^{-0.4t}$

Simplifying:

$T'(t) = -8.4e^{-0.4t}$

Step 2: Evaluate the derivative at $t = 8$

Now, we substitute $t = 8$ into the derivative expression to find the rate of change of temperature at that specific time:

$T'(8) = -8.4e^{-0.4(8)} = -8.4e^{-3.2}$

We can compute $e^{-3.2} \approx 0.0407622$.

So:

$T'(8) \approx -8.4 \times 0.0407622 \approx -0.3414$

Final Answer:

The rate of change of the temperature at $t = 8$ is approximately $-0.3414 \, \text{°C/min}$.

Would you like further details or have any questions about this process?

Follow-up Questions:

1. How does the rate of change of temperature affect the cooling process over time?
2. What is the significance of the negative sign in the rate of change?
3. How can the temperature of the water be modeled as a function of time using other methods, such as linear approximations?
4. How does changing the constant in the exponent (e.g., using $-0.5t$ instead of $-0.4t$) affect the rate of change?
5. What is the concept of exponential decay in the context of cooling objects?

Tip:

When finding the rate of change, remember that it provides insight into how quickly a quantity changes over time, especially in real-world situations like cooling or heating processes.