Solve the following differential equations using the substitutions given.
A rectangular tank with its base horizontal is filled with water to a depth h at time t = 0. Water leaks out of the tank from a small hole in the base at a rate proportional to the square root of the depth of the water. If the depth of water is ½h at time T, find the further time, in terms of T, it will take before the tank is empty.
A researcher is investigating the spread of a certain disease in a town with a population of 3000 people. The researcher suggests that I, the number of people infected by the disease at time t days satisfies the differential equation , where k is a positive constant.
(i) Given that I = 30 when t = 0, show that .
(ii) It is further observed that I = 240 when t = 7 find the time it takes for 90% of the population to be infected by the disease.
(iii) State, in the context of this question, one assumption needed to model the spread of the disease in the town by the given differential equation.
(iii) Possible Answer:
Assume that the total population of the town is 3000 during the spread of the disease.
Or: Assume that a person infected by the disease will remain infected by the disease.
Or: Assume that everyone in the town has an equal chance of being infected by the disease.
The diagram shows a river,500m wide between the straight parallel banks where AD = 500m and DB = 2000. A man swims across the river from A to C at 0.5 m/s and then he runs along the banks from C to B at 0.8 m/s
(a)If he swims in the direction AC making an angle of Ɵ and reaches C in t seconds, show that t = 1000cosec Ɵ
(b)If T(in seconds) is the total time taken from A to B, express T in terms of Ɵ
(c)Find the minimum value of T in minutes and seconds, correct to the nearest second.
Using trigo ratio,
sin Ɵ = 500/AC
AC = 500/sin Ɵ
= 500cosec Ɵ
0.5 = 500cosec Ɵ/t
T = 1000cosec Ɵ
BC = 2000 – DC
BC = 2000 – 500/tan Ɵ
TBC = BC/s = 2500 – 625/tan Ɵ
Total time, T = tAC + tBC
= 1000cosec Ɵ + 2500 – 625/tan Ɵ