Interpretation of the Derivative — Question 6

Question 6 - Solution

We are given: $$T(t) = 100 - 20e^{-0.5t}$$

(a) Rate of Change of Temperature:

Differentiate $T(t)$ with respect to $t$:

$$T'(t) = \frac{d}{dt}(100 - 20e^{-0.5t}) = -20 \cdot \frac{d}{dt}(e^{-0.5t}) = -20 \cdot (-0.5e^{-0.5t}) = 10e^{-0.5t}$$

$$\boxed{T'(t) = 10e^{-0.5t}}$$

This represents the rate of temperature change at time $t$.

(b) Rate of Change at $t = 0$:

$$T'(0) = 10e^{-0.5 \cdot 0} = 10e^{0} = 10$$

$$\boxed{T'(0) = 10 \quad \text{(°C per minute)}}$$

Interpretation: At the beginning of the reaction, the temperature is increasing at 10°C per minute.

(c) Long-Term Behavior:

As $t \to \infty$, $e^{-0.5t} \to 0$, so:

$$T'(t) = 10e^{-0.5t} \to 0$$

Conclusion: The rate of temperature change decreases over time and approaches zero. This means the temperature stabilizes and stops changing as time progresses.

$$\boxed{\lim_{t \to \infty} T'(t) = 0}$$