- 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
- dt = dh/-Ah–3/2
- ∫Adt = ∫dh/-h-3/2
- At + c = –2/5 h5/2
- Use the information given to find the unknowns
- when t = 0:
- -A(0) + c = 2/5 (H)5/2
- ∴ c = 2/5 H5/2
- when t = 60:
- -A(60) + c = 0
- c = 60A
- ∴ A = 1/150 H5/2
- Thus the initial equation becomes:
- – 1/150 H5/2 + 2/5 H5/2 = 2/5 h5/2
- H5/2 (-t/150 + 2/5) = 2/5 h5/2
- –t/150 + 2/5 = 2h5/2/5H5/2
- t/150 = 2/5 – 2h5/2/5H5/2
- t = 150(2/5 – 2h5/2/5H5/2)
- t = 60 – 60h5/25H-5/2
- t = 60 (1 – (h/H)5/2)
