Form a differential equation from the information given:
If something is proportional, add a constant of proportionality, k
If rate is decreasing, add a negative sign
Separate variables, bring dx and dt on opposite sides
Integrate both sides to form an equation
Use the conditions given to find c and/or k
Examples
A biologist is investigating the spread of a weed in a particular region. At time t weeks, the area covered by the weed is A m². The biologist claims that the rate of increase of A is proportional to √(2A – 5)
Write down a differential equation with the given info
dA/dt ∝ √(2A – 5)
dA/dt = k√(2A – 5)
At the start of the investigation, the area covered by the weed was 7 m². 10 weeks later, the area covered was 27 m². Find the are covered 20 weeks after the start of the investigation.
Using dA/dt above
Separate variables: dA1/√(2A – 5) = kdt
Integrate both sides: kt + c = (2A – 5)½
When t = 0:
A = 7
∴ c = 3
∴ kt + 3 = (2A – 5)½
When t = 10:
10k + 3 = (2(27) – 5)½
10k = √(49) – 3
∴ k = 0.4
Therefore the differential equation is 0.4t + 3 = (2A – 5)½
Now substitute 20 as t and then find A:
0.4(20) + 3 = (2A – 5)½
11 = (2A – 5)½
121 = 2A – 5
∴ A = 63 m²
Liquid is flowing into a small tank which has a leak. Initially the tank empty, and t minutes later, the volume of the liquid in the tank is V cm³. The liquid is flowing into the tank at a constant rate of 80 cm³ per minute.
Because of the leak, liquid is being lost from the tank at a rate which, at any instant, is equal to kV cm³ per minute, where k is a positive constant
Write down a differential equation describing this situation and solve it to show that V = 1/k (80 – 80e-kt)
Represent the given information as a derivative: dV/dt = 80 – kV
Proceed to solve the differential equation:
dt/dV = 1/80 – kV
dt = dv/80 – kV
∫dt = ∫dV/80 – kV
t + c = – 1/k ln|80 – kV|
Use the given information; when t = 0, V = 0: ∴c = 1/k ln(80)
Substitute c into the equation:
t – 1/k ln(80) = – 1/k ln|80 – kV|
t = 1/k ln(80) – 1/k ln|80 – kV|
t = 1/k ln(80/80 – kV)
kt = ln(80/80 – kV)
ekt = 80/80 – kV
80 – kV = 80/ekt
kV = 80 – 80e-kt
∴ V = 1/k (80 – 80e-kt)
V = 500 when t = 15, show k = 4 – 4e-15k/25. Find k using iterations, initially k = 0.1
500 = 1/k (80 – 80e-15k)
k = (80 – 80e-kt)/500
k = 4 – 4e-15k/25
Doing the mishwaar iterations gives k = 0.14
Work out the volume of the liquid at t = 20 and state what happens to the volume after a long time
Using k above: V = 1/0.14 (80 – 80e-15k)
Substitute t = 20 in V above: ∴ V = 537 cm³
The volume of the liquid in the tank after a long time approaches the maximum volume:
∴ V = 1/0.14 (80)
∴ V = 571 cm³
A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi-vertical angle is 60°, as shown in the diagram below.
At time t = 0, the tank is full and the depth of the water is H. At this instant, a tap at C is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional √h, where h is the depth of the water at time t. The tank becomes empty when t = 60.
Show that h and t satisfy a differential equation of the form dh/dt = -Ah–3/2, where A is a positive constant
Represent the information given as an equation:
V = 1/3 πr²h
r = tan 60 × h = h√3
∴ V = 1/3 π(h√3)²h = πh³
dV/dt ∝ -√h = -kh½
Find the rate of change of h:
dh/dt = dV/dt ÷ dV/dh
dh/dt = –kh½/3πh² = – k/3π h-3/2
Solve the differential equation given in (a) and obtain an expression for t in terms of h and H
