Applications

Differential Equations

Writing Equations

Write differential equations to represent the following situations:

1. The rate of increase of the capital value, C, of an insurance policy is inversely proportional to the age, t, of the policy. 2. As time, t, passes the population, p, of a bacteria culture increases at the rate proportional to the actual population itself. 3. Animals on a farm drink from a square trough. The rate of evaporation of water, E, is proportional to the square of the length, l, of the trough.

Write differential equations to represent the following situations:

1. The rate of increase of the capital value, C, of an insurance policy is inversely proportional to the age, t, of the policy.

Write differential equations to represent the following situations:

2. As time, t, passes the population, p, of a bacteria culture increases at the rate proportional to the actual population itself.

Write differential equations to represent the following situations:

3. Animals on a farm drink from a square trough. The rate of evaporation of water, E, is proportional to the square of the length, l, of the trough.

Write differential equation to express the following (Do NOT solve):

The area of an object is decreasing at a rate inversely proportional to the square root of the side length and directly proportional to the time. At time 10s and side length 4m the area is decreasing at 0.5 m2s-1.

Write differential equations to express the following (Do NOT solve):

The area of an object is decreasing at a rate inversely proportional to the square root of the side length and directly proportional to the time. At time 10s and side length 4m the area is decreasing at 0.5 m2s-1.

Exponential Growth and Decay

Example Question

In Newtown the population of the town, N, changes at a rate proportional to the population of the town.

The population of the town in 1974 was 50 000 people and in 2004 it is 90 000 people.

Find an equation for the population, N, in terms of time t, where t is the number of years after 1974.

Step One: Set up the differential equation

In Newtown the population of the town, N, changes at a rate proportional to the population of the town.

K is the constant of proportionality

Step Two: Solve the differential equation

You must show the working- shortcuts = no marks

This is the general solution

Step Three: Find the particular solution

The population of the town in 1974 was 50 000 people and in 2004 it is 90 000 people.

Initial condition: 1974 = t = 0, N = 50000

Step Three: Find the particular solution

The population of the town in 1974 was 50 000 people and in 2004 it is 90 000 people.

Condition 2: 2004 = t = 30, N = 90000

We expect a positive value as the population is increasing

Step Three: Find the particular solution

The population of the town in 1974 was 50 000 people and in 2004 it is 90 000 people.

The rate of increases/decreases at a rate that is proportional to…

Important words to look for in the question

Learn…

Also learn…

Bursary 1989 Question 3 (b)

…then the rate of increase of N is proportional to N.

Differential equation Form of the solution

Two sets of information…

At one point, it is estimated that there were 500 ferrets in a particular area.

4 months later 600 ferrets were present in the area.

.. how many ferrets would be present 10 months after the initial

estimate was made?

Note the rounding

1990 2 (c)

…the rate of decrease of P at any time is proportional to P.

Differential equation Form of solution you must develop this

…there were 600 deer at the start of the programme

…after 2 months there were 400

Solve for k

Rate is decreasingImplies k is negative

Write down a differential equation which models the deer

population

…state an expression for P as a function of t.

.. find the number of deer that would be present at 3.5 months.

Bursary 1993 Question 4 (b)

…the area of scar tissue, S mm2, after a serious skin burn, decreases at a rate that is proportional to S itself.

Explain why S = So models the area of scar tissue at time t days after the burn, where

So and k are constants

…the area of scar tissue decreased to one-third of the

original area in 13 days

You are not given the amount of initial scar tissue

…in 26 days

Now notice this pattern…

In equal time periods,

S forms a geometric progression

If S26 represents the area of scar tissue after 26 days, express the original area So in terms of S26

Find the value of k.

k is always the ln of the ratio divided bythe time period

When will scarring have diminished to 1% of the original

area?

Newton’s Law of Cooling

• The rate at which a body changes its temperature is proportional to the difference between its temperature and the temperature of its surroundings.

Example

According to Newton’s law of cooling the temperature T of a liquid after t minutes is modelled by the differential equation:

where k and T0 are constants. If the liquid starts at 50˚C and after 5 minutes it is 40˚C, find:

Example

where k and T0 are constants. If the liquid starts at 50˚C and after 5 minutes it is 40˚C, find:

(a) The temperature T at any time t.

(b) The time at which the temperature will reach 30˚C.

Solve the equation

If the liquid starts at 50˚C and after 5 minutes it is 40˚C, find:

(a) The temperature T at any time t.

(b) The time at which the temperature will reach 30˚C.

Assume room temperature is 20˚C

Solve the equation

If the liquid starts at 50˚C and after 5 minutes it is 40˚C, find

a)The temperature T at any time t.

(b) The time at which the temperature will reach 30˚C.

Solve the equation

If the liquid starts at 50˚C and after 5 minutes it is 40˚C, find

a)The temperature T at any time t.

(b) The time at which the temperature will reach 30˚C.

Solve the equation

If the liquid starts at 50˚C and after 5 minutes it is 40˚C, find

a)The temperature T at any time t.

(b) The time at which the temperature will reach 30˚C.

Solve the equation

If the liquid starts at 50˚C and after 5 minutes it is 40˚C, find

a)The temperature T at any time t.

(b) The time at which the temperature will reach 30˚C.

Kinematics

Belinda was lying on a viaduct peering down at the water far below. She held a stone out at arms length and let it go. The sound of the stone hitting the water reached her 4 seconds later.The stone falls under gravity with an acceleration of 9.8 m s-2, and sound travels at 33 m s-1. 1. How far does the stone fall in t seconds?2. How far does the sound travel in t seconds?3. How many seconds after it is dropped does the stone hit

the water?4. How high is the viaduct?

Belinda was lying on a viaduct peering down at the water far below. She held a stone out at arms length and let it go. The sound of the stone hitting the water reached her 4 seconds later.The stone falls under gravity with an acceleration of 9.8 m s-2, and sound travels at 33 m s-1. 1. How far does the stone fall in t seconds?2. How far does the sound travel in t seconds?3. How many seconds after it is dropped does the stone hit

the water?4. How high is the viaduct?

Belinda was lying on a viaduct peering down at the water far below. She held a stone out at arms length and let it go. The sound of the stone hitting the water reached her 4 seconds later.The stone falls under gravity with an acceleration of 9.8 m s-2, and sound travels at 33 m s-1. 1. How far does the stone fall in t seconds? 2. How far does the sound travel in t seconds?3. How many seconds after it is dropped does the stone hit

the water?4. How high is the viaduct?

Belinda was lying on a viaduct peering down at the water far below. She held a stone out at arms length and let it go. The sound of the stone hitting the water reached her 4 seconds later.The stone falls under gravity with an acceleration of 9.8 m s-2, and sound travels at 33 m s-1. 1. How far does the stone fall in t seconds? 2. How far does the sound travel in t seconds?3. How many seconds after it is dropped does the stone hit

the water?4. How high is the viaduct?

Belinda was lying on a viaduct peering down at the water far below. She held a stone out at arms length and let it go. The sound of the stone hitting the water reached her 4 seconds later.The stone falls under gravity with an acceleration of 9.8 m s-2, and sound travels at 33 m s-1. 1. How far does the stone fall in t seconds? 2. How far does the sound travel in t seconds?3. How many seconds after it is dropped does the stone hit

the water?4. How high is the viaduct?

Belinda was lying on a viaduct peering down at the water far below. She held a stone out at arms length and let it go. The sound of the stone hitting the water reached her 4 seconds later.The stone falls under gravity with an acceleration of 9.8 m s-2, and sound travels at 33 m s-1. 1. How far does the stone fall in t seconds? 2. How far does the sound travel in t seconds?3. How many seconds after it is dropped does the stone hit

the water?4. How high is the viaduct?

2012A cylindrical tank is full of water. A small hole is drilled in the bottom of the tank so that the water leaks out. The rate at which the volume V of water in the tank changes is given by:

where h is the height of the water surface above the hole and k is a positive constant.The volume of water in the tank is given by V = Ah where A is the cross-sectional area of the tank.(i) Form a differential equation for the rate at which the height of the water in the tank is changing.

2012A cylindrical tank is full of water. A small hole is drilled in the bottom of the tank so that the water leaks out. The rate at which the volume V of water in the tank changes is given by:

where h is the height of the water surface above the hole and k is a positive constant.The volume of water in the tank is given by V = Ah where A is the cross-sectional area of the tank.(i) Form a differential equation for the rate at which the height of the water in the tank is changing.

2012

• The cross-sectional area of the tank is 0.5 m2.• The height of the water in the tank before it

starts leaking out is 144 cm. • After 160 hours, the height of the water in the

tank is 100 cm. • Find out how long it will take for the tank to

empty.

• The cross-sectional area of the tank is 0.5 m2.• The height of the water in the tank before it

starts leaking out is 144 cm. • After 160 hours, the height of the water in the

tank is 100 cm. • Find out how long it will take for the tank to

empty.

• The cross-sectional area of the tank is 0.5 m2.

• The height of the water in the tank before it starts leaking out is 144 cm.

• After 160 hours, the height of the water in the tank is 100 cm.

• Find out how long it will take for the tank to empty.

t = 0 , h = 144 gives c = 24•

• The cross-sectional area of the tank is 0.5 m2.

• The height of the water in the tank before it starts leaking out is 144 cm.

• After 160 hours, the height of the water in the tank is 100 cm.

• Find out how long it will take for the tank to empty.

• t = 160 , h = 100

•

960 hours.

‘Salt’ questions