Category Archives: A Level H2 Math

Modelling the leakage of water using Differential Equations

A rectangular tank with its base horizontal is filled with water to a depth 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.                                             

JJC/2013/II/Q2 Using Differential Equations to Model the spread of Disease in a town

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 = 30 when = 0, show that 

(ii)        It is further observed that = 240 when = 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.

VJC/2013/Prelim/P1/Q1

(a)Without using a calculator, solve the inequality

(b) Deduce the range of values of x that satisfies.

Solutions: (a): x≤-6 or 1≤x<3 or x>3

                 (b): 0<x≤e-6 or e≤x<e3 or x>e3

(a)

(b)

lnx ≤ -6              

x ≤ e-6 

 1≤ lnx <3

 e≤x<e3 

lnx>3

 x>e3

 

 

 

 

Applications of integration

Differentiate y=5x(e3x) with respect to x and hence find ∫x(e3x) dx

Solution: 1/3(x)(e3x) – 1/9(e3x) + C

Y = 5x e3x

dy/dx = 5x(e3x)(3) + 5 e3x

dy/dx = 15x(e3x) + 5(e3x)

∫15x(e3x) + 5(e3x) dx = 5x(e3x)

∫15x(e3x) dx + (5(e3x))/3 = 5x(e3x)

15∫x(e3x) dx = 5x(e3x) – 5/3(e3x)

∫x(e3x) dx = 1/3(x)(e3x) – 1/9(e3x) + C