Answer to Question #124113 in Calculus for Michael

Question

a) A bowl of water has a temperature of 50◦C. It is put into a refrigerator where the temperature is 5◦C. After 0.5 hours, the water is stirred and its temperature is measured to be 20◦C. It is then left to cool further, take t in Newton’s law of cooling to be measured in minutes. Use Newton’s law of cooling to predict when the temperature will be 10◦C
b)) Find the volume of the solid generated by revolving the region bounded by the curves y = x^2 and y = 4x − x^2 about the line y = 6

Answer 1

a) As $\frac{dT}{dt}=-k(T-a)$ then $\ln(T-a)|_{T_1}^{T_2}=-kt_1$ where $T_1=50^0C, T_2=20^0C,$

$t_1=30 min,$ then $k=-\frac{1}{t_1}\ln\frac{T_2-a}{T_1-a},$ then $t=-\frac{\ln(T-a)|_{T_1}^{T_3}}{k}$ where $t_3=10^0C$ then

$t=2t_1=60min$

b)$V=\pi\int_{0}^{2}(6-y_1)^2dx-\pi\int_{0}^2(6-y_2)^2dx$ where $y_1=x^2,y_2=4x-x^2$ then

$V=\pi\int_{0}^2((6-x^2)^2-(6-4x+x^2)^2)dx=\pi\sdot21\frac{1}{3}$