Tangent Lines and Rates of Change — Question 7

Question 7 - Solution

Given: $$T(t) = 25 + 60e^{-0.4t}$$

(a) Average rate of change on $[1, 4]$: $$T(1) = 25 + 60e^{-0.4} \approx 25 + 60(0.6703) \approx 65.22$$ $$T(4) = 25 + 60e^{-1.6} \approx 25 + 60(0.2019) \approx 37.11$$ $$\text{Average rate} = \frac{37.11 - 65.22}{4 - 1} \approx \frac{-28.11}{3} \approx -9.37 \text{ °C/min}$$

(b) Instantaneous rate of change at $t = 1$: $$T'(t) = 60(-0.4)e^{-0.4t} = -24e^{-0.4t}$$ $$T'(1) = -24e^{-0.4} \approx -24(0.6703) \approx -16.09 \text{ °C/min}$$

(c) Interpretation: At $t = 1$, the object is cooling at a faster rate than the average cooling over the interval $[1, 4]$, consistent with exponential decay.