Question: A pot of rice pudding is removed from the stovetop at 7 a.m. and allowed to cool for two hours before being transferred to the refrigerator. During the first stage of cooling, the temperature \(u\) of the pudding (measured in \(\left.{ }^{\circ} \mathrm{C}\right)\) follows the curve \(u=23+(74-23) e^{-k t}\), where \(k\) is a positive constant and \(t\) measures the number of hours since 7 a.m. A thermometer inserted into the pudding at 9 a.m. reads \(54^{\circ} \mathrm{C}\).

1. Solve for the decay constant \(k\). (No calculus needed!)
2. How fast is the temperature dropping at 7:30 a.m.? (Hint: this rate is measured in degrees per hour.)
3. Show that \(\frac{\mathrm{d} u}{\mathrm{~d} t}=k(23-u)\), and check that this formula produces the same answer as the approach you took for part (b).

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