Solving Real-Life Problems Using Differential Equations Revision Notes for Leaving Cert Applied Maths

Solving Real-Life Problems Using Differential Equations

Understanding acceleration in differential equations

When solving real-life motion problems using differential equations, we need to understand how acceleration relates to other quantities. Acceleration is fundamentally the rate at which velocity changes with time.

The basic definition of acceleration gives us our first key relationship. We can express acceleration in two different ways depending on what variables we want to connect:

Linking velocity and time: When we know how acceleration depends on velocity and want to find how velocity changes with time, we use a = dv/dt
Linking velocity and displacement: When we want to connect velocity changes with distance travelled, we use the chain rule to derive a = v(dv/ds)

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The second formula comes from applying the chain rule: $a = \frac{dv}{dt} = \frac{dv}{ds} \times \frac{ds}{dt} = v\frac{dv}{ds}$

This gives us flexibility in choosing the most appropriate approach for each problem.

Worked example: Particle with velocity-dependent acceleration

Let's examine how to solve a problem where a particle starts from rest and experiences acceleration equal to $v^2 + 100$ , where $v$ is the particle's velocity.

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Worked Example: Particle with acceleration $a = v^2 + 100$

Problem setup: A particle starts from rest at point P and moves in a straight line with acceleration $a = v^2 + 100$ . Find the time taken to reach 20 m/s.

Solution approach:

First, we sketch the situation to establish our positive direction. Since we're finding time and have acceleration in terms of velocity, we use $a = \frac{dv}{dt}$ .

Setting up the differential equation: $\frac{dv}{dt} = v^2 + 100$

We separate variables to get all velocity terms on one side and time terms on the other: $\frac{1}{v^2 + 100} dv = dt$

Integrating both sides: $\int \frac{1}{v^2 + 100} dv = \int dt$

The left side integrates to $\frac{1}{10}\tan^{-1}\left(\frac{v}{10}\right)$ , giving us: $\frac{1}{10}\tan^{-1}\left(\frac{v}{10}\right) = t + C$

Using initial conditions (at $t = 0$ , $v = 0$ ), we find $C = 0$ .

Worked example: Particle with retardation

Now let's look at a more complex example involving retardation (negative acceleration).

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Worked Example: Particle with retardation $\frac{v^3}{25}$

Problem setup: A particle moves in a straight line and experiences retardation of $\frac{v^3}{25}$ , where $v$ is the speed.

The negative sign indicates deceleration. For the first part, we need to link velocity and displacement, so we use $a = v\frac{dv}{ds}$ .

Setting up the equation: $v\frac{dv}{ds} = -\frac{v^3}{25}$

Dividing by $v^3$ and rearranging: $\frac{1}{v^2}dv = -\frac{1}{25}ds$

Integrating both sides and applying initial conditions gives us the relationship between velocity and displacement.

For the second part, linking velocity and time, we use $a = \frac{dv}{dt}$ and follow a similar separation of variables approach.

Deriving the equations of motion

We can use differential equations to derive the familiar kinematic equations from first principles.

Starting with the definition $a = \frac{dv}{dt}$ and integrating: $\int dv = \int a , dt$

This gives us $v = at + C$ . Using initial condition $v = u$ when $t = 0$ , we get $C = u$ .

Therefore: $v = u + at$

Similarly, starting from $v = \frac{ds}{dt}$ and substituting $v = u + at$ : $ds = (u + at)dt$

Integrating: $s = ut + \frac{1}{2}at^2 + C$

Using initial condition $s = 0$ when $t = 0$ gives $C = 0$ .

Therefore: $s = ut + \frac{1}{2}at^2$

For the third equation, we use $a = v\frac{dv}{ds}$ : $\int v , dv = \int a , ds$

This gives: $\frac{v^2}{2} = as + C$

Using initial conditions and rearranging: $v^2 = u^2 + 2as$

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These three equations of motion that we use in kinematics can all be derived from the fundamental definition of acceleration using calculus. This shows the power of differential equations in connecting basic definitions to practical formulas.

Power problems

In mechanical systems, power relates to the tractive force and velocity through the equation Power = Tv, where $T$ is the tractive force and $v$ is velocity.

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Worked Example: Car engine at constant power

A car of mass 1200kg starts from rest with engine working at constant power of 1500W. Find the acceleration in terms of velocity.

Solution: Power = $Fv$ , so $F = \frac{1500}{v}$

Using Newton's second law: $F = ma = 1200a$

Therefore: $\frac{1500}{v} = 1200a$

Solving for acceleration: $a = \frac{5}{4v}$

This gives us a differential equation that we can solve using the methods shown earlier.

Newton's law of cooling

Differential equations also apply to non-mechanical problems. Newton's law of warming states that the rate of temperature change is proportional to the temperature difference between a body and its surroundings.

If $\theta$ is the temperature difference between a body and its surroundings, then: $\frac{d\theta}{dt} = k\theta$

This separates to give: $\frac{1}{\theta}d\theta = k , dt$

Integrating: $\ln \theta = kt + C$

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Worked Example: Body warming problem

A body warms from 2°C to 8°C in 6 minutes when the surrounding temperature is constant at 25°C.

Solution approach:

Set up the differential equation using $\theta$ as the temperature difference
Separate variables and integrate
Use the given conditions to find the constant $k$
Apply the equation to find the final temperature

The temperature difference initially is $\theta = 25 - 2 = 23$ °C. After 6 minutes, $\theta = 25 - 8 = 17$ °C.

Using these conditions in our integrated equation allows us to find $k$ and predict future temperatures.

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Problem-solving strategy

When approaching differential equation problems:

Sketch the situation to establish directions and understand the physical setup
Choose the appropriate acceleration formula based on what variables you need to connect
Set up the differential equation using the given information
Separate variables to get like terms on each side
Integrate both sides carefully
Apply initial conditions to find constants of integration
Substitute final conditions to find the required answer

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Key Points to Remember:

Use a = dv/dt when linking velocity and time, and a = v(dv/ds) when linking velocity and displacement
Always establish a positive direction and stick to it throughout the problem
Set your calculator to radian mode when dealing with inverse trigonometric functions
Negative acceleration indicates retardation or deceleration
Newton's law of cooling applies to many real-world temperature change problems

Solving Real-Life Problems Using Differential Equations (Leaving Cert Applied Maths): Revision Notes