▶️ Answer/Explanation
Solution
(a)
\( C'(5) \approx \frac{C(7) – C(3)}{7 – 3} = \frac{69 – 85}{4} = -4 \).
Answer: \(-4\) degrees Celsius per minute.
(b)
Left Riemann sum: \( (3 – 0) \cdot C(0) + (7 – 3) \cdot C(3) + (12 – 7) \cdot C(7) = 3 \cdot 100 + 4 \cdot 85 + 5 \cdot 69 = 985 \).
\( \frac{1}{12} \cdot 985 \approx 82.083 \), the average temperature over \( [0, 12] \).
Answer: 985, average temperature \(\approx 82.083\) degrees Celsius.
(c)
\( C(20) = C(12) + \int_{12}^{20} C'(t) \, dt = 55 + \int_{12}^{20} -\frac{24.55 e^{0.01t}}{t} \, dt \).
\( \int_{12}^{20} \frac{e^{0.01t}}{t} \, dt \approx 0.597 \), so \( C(20) = 55 – 24.55 \cdot 0.597 \approx 40.329 \).
Answer: 40.329 degrees Celsius.
(d)
\( C”(t) = \frac{0.2455 e^{0.01t} (100 – t)}{t^2} \). Since \( 100 – t > 0 \), \( C”(t) > 0 \), so \( C'(t) \) is increasing.
Answer: Changing at an increasing rate.