Mathematics Extension 1 – Applications of Calculus to the Real World

Applications of Calculus to the Physical World

At this point we have learnt quite a bit about calculus, and wish to apply it to the physical world. This topic allows one to see the power of calculus in describing real life situations, and is the topic which shows the “applications of maths” so to say. We shall investigate a variety of topics including rates of change, growth and decay, Newton’s law of cooling, velocity and acceleration as a function of $Latex formula$, projectile motion and simple harmonic motion.

Rates of Change

By use of the chain rule, we may find the rate of change of certain variable given the rate of change of another. The questions in this section typically require the student to find the rate of change of the volume or the surface area of some polyhedron given the rate of change of some other factor. An example would be to find the rate of change of the volume of a balloon at a certain radius, given that the radius is changing at a certain rate.

To be able to solve problems in this section, recall the chain rule that states,

$Latex formula$

where $Latex formula$ is the “dummy variable”. Notice how the $Latex formula$’s cancel. We shall now consider an example to illustrate the type of questions being asked in this section.

Example 1

A spherical balloon is being inflated and its radius is increasing at the rate of $Latex formula$ cm/min. At what rate is the volume increasing when the radius of the balloon is $Latex formula$ cm?

Solution 1

So we are required to find the rate of change of the volume given the rate of change of the radius. We must firstly express the rate of change of the volume in terms of the rate of change of the radius. However, we do not know what the other expression is going to being the chain rule.

To figure out what we are required to find to solve the question, we set up the following.

$Latex formula$

Where $Latex formula$ is the required differential, $Latex formula$ is the volume of the balloon at any point in time and $Latex formula$ is the radius at any point in time.

Solving for $Latex formula$ gives,

$Latex formula$

(Of course here we treat the differentials as Leibniz quotients)

Thus we have that

$Latex formula$

We now are required to find the value of .

$Latex formula$

Recall that for a sphere,

$Latex formula$

Hence we have that upon differentiating the expression with respect to $Latex formula$,

$Latex formula$

Now, substituting in this value into the expression for $Latex formula$ gives;

$Latex formula$

We require the rate of change when the radius $Latex formula$ equals $Latex formula$ cm. Also, it is given that the rate of change of the radius is equal to $Latex formula$ cm/min. Hence we have that the rate of change of the volume is,

$Latex formula$

Notice that the units are the same in this case, however students should be wary in case they are not. Also, noticing the units being multiplied together gives an indication of the final answer required (for e.g. we have multiplied a quantity with units cm/min by a quantity being squared with units cm).

Example 2

The altitude of a right angled triangle is $Latex formula$ cm and the base of the triangle is increasing at a constant rate of $Latex formula$cm/s. At what rate is the hypotenuse increasing when its length is 15 cm?

Solution 2

We firstly draw a good diagram.

Let $Latex formula$ be the hypotenuse, $Latex formula$ be the base and $Latex formula$ cm be the altitude of the right angled triangle.

We are given $Latex formula$ and we are required to find $Latex formula$ when the hypotenuse has length $Latex formula$ cm. Using the chain rule gives,

$Latex formula$

->$Latex formula$

Hence we have that,

$Latex formula$

We must now find $Latex formula$.

By Pythagoras’ theorem we have that,

$Latex formula$

$Latex formula$

$Latex formula$

Hence we have that,

$Latex formula$

Now, to find $Latex formula$ we must use the chain rule,

$Latex formula$

At $Latex formula$ cm, we have that,

$Latex formula$

Hence we have that,

$Latex formula$

Hence we have that,

$Latex formula$

The length of the hypotenuse is increasing at $Latex formula$ cm/s.

The next example introduces a question which is more difficult than the questions shown so far.

Example 3

Find the rate of change of the surface area of a spherical balloon when the radius is $Latex formula$ cm, given that the volume is increasing at rate of $Latex formula$ cm3/sec.

Solution 3

With this problem, we have the value of $Latex formula$ and wish to find the value of $Latex formula$. Using the chain rule gives,

$Latex formula$

Now, we must calculate $Latex formula$. However, this is not possible to do so directly, as volume is not expressed in terms of surface area. The equations we do have are,

$Latex formula$

Now, since we have that both equations contain $Latex formula$ as their variable, then it follows that we may use the chain rule again.

$Latex formula$

Now $Latex formula$ is given by,

$Latex formula$

and $Latex formula$ is given by,

$Latex formula$

Hence we have that,

$Latex formula$

->$Latex formula$

So we have that,

$Latex formula$

$Latex formula$

That is, the surface area of the balloon is increasing at 3 cm2/sec.

$Latex formula$

Notice that in this question we in fact use a “double chain” rule. Like so,

$Latex formula$

And this allowed for the calculation of the required rate.

We have so far exclusively used the chain rule to calculate rates of change. We shall now employ a technique taught in Mathematics Extension 2 called implicit differentiation, which considers every variable within the expression as a function of the expression to be differentiated with respect to.

Example 4

A circular metal plate of radius $Latex formula$ is heated so that its area expands at a constant rate of $Latex formula$ cm2/min. At what rate does the radius increase when $Latex formula$cm?

Solution 4

In this question we are given that $Latex formula$.

Now, the area of a circle is given by,

$Latex formula$

Treating $Latex formula$ as a function of time $Latex formula$ and $Latex formula$ also as a function of $Latex formula$ we may differentiate both sides with respect to $Latex formula$.

$Latex formula$

This gives,

$Latex formula$

Now, since $Latex formula$ is a function of $Latex formula$, then we may use the chain rule and hence we have that,

$Latex formula$

Since we have that $Latex formula$ is a function of $Latex formula$.

Now, it is known that $Latex formula$ is equal 4 cm2/min. At $Latex formula$ cm we have that,

$Latex formula$

Hence the radius is increasing at $Latex formula$ cm/min.

The method of implicit differentiation although tricky at first, is more efficient than using the chain rule only.

Furthur Growth and Decay

We shall now further investigate the concept of natural growth and decay. Recall the differential equation

$Latex formula$

Where $Latex formula$ is a real constant (some number determining the growth rate). This equation is a crude model for population growth and is extensively used in the Mathematics course. Recall that the solution to this differential equation is given by,

$Latex formula$

where $Latex formula$ is some real constant.

In the Mathematics Extension 1 course we shall study the differential equation,

$Latex formula$

Where $Latex formula$ and $Latex formula$ are constants. This equation also describes growth and decay but also takes into account the difference from a certain threshold. The solution to this differential equation is given by,

$Latex formula$

Where $Latex formula$ is yet again a constant. The proof of this is relatively simple.

$Latex formula$

Taking the reciprocal of all variables on LHS and RHS gives,

$Latex formula$

Integrating with respect to $Latex formula$ gives,

$Latex formula$

Where $Latex formula$ is a constant. Now, rearranging gives,

$Latex formula$

->$Latex formula$

->$Latex formula$

Now let $Latex formula$. Hence we have that,

$Latex formula$

Which is the solution to the differential equation. The initial conditions given will allow the values of $Latex formula$, $Latex formula$ and $Latex formula$ to be determined.

We shall now go through an example to illustrate the type of question that may be asked.

Example 5

The population $Latex formula$ of a town has a rate of change proportional to the difference between $Latex formula$ and 15 000. This may be expressed as

$Latex formula$

where $Latex formula$ and $Latex formula$ is the time in years.

a) Show that $Latex formula$ is a solution to the above differential equation ($Latex formula$ and $Latex formula$ are constants).

b) If the population at the beginning of 1960 was 21 000, find the population after:

i) One year

ii) Five years.

c) During which year will the population double?

d) What happens as $Latex formula$?

Solution 5

a) To show that $Latex formula$ is a solution to the differential equation, we simply differentiate the equation with respect to time and then show that the LHS is equal to RHS.

$Latex formula$

$Latex formula$

Now, $Latex formula$. Hence we have that,

$Latex formula$

Hence as $Latex formula$ satisfies the differential equation, then it follows that it is a solution to the differential equation.

b)

i) Firstly, we use the given information to find the required values $Latex formula$ of $Latex formula$.

At $Latex formula$, $Latex formula$. So we have that,

$Latex formula$

$Latex formula$

Now, we have that,

$Latex formula$

Since $Latex formula$

Now, at time $Latex formula$ we have that

$Latex formula$

Thus the population is approximately $Latex formula$ after one year.

ii) At time $Latex formula$,

$Latex formula$

Hence the population is approximately 31 310 after 5 years.

c) At $Latex formula$, (twice original population) we have that,

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

Hence the population will double in the year $Latex formula$

d) As $Latex formula$ the value of $Latex formula$ as $Latex formula$.

Newtons Law of Cooling

An important application of the differential equation

$Latex formula$

is to measure the rate at which an object cools or heats up under natural conditions. Letting $Latex formula$ represent the temperature in degrees Celsius we have that

$Latex formula$

Where $Latex formula$ is the surrounding temperature, and $Latex formula$ is a constant.

This simply states that the larger the temperature difference between the surrounding temperature and the objects temperature, the faster the rate at which the object changes its temperature to match the temperature of the surrounding area. We shall consider an example of a problem involving this law.

Example 6

(HSC 2003) Dr Kool wishes to find the temperature of a very hot substance using his thermometer, which only measures up to $Latex formula$. Dr Kool takes a sample of his substance and places it in a room with a surrounding air temperature of $Latex formula$, and allows it to cool.

After $Latex formula$ minutes the temperature of the substance is $Latex formula$, and after a further $Latex formula$ minutes it is $Latex formula$. If $Latex formula$ is the temperature of the substance after $Latex formula$ minutes, then Newton’s law of cooling states that $Latex formula$ satisfies the equation

$Latex formula$

where $Latex formula$ is a constant and $Latex formula$ is the surrounding air temperature.

i) Verify that $Latex formula$ satisfies the above differential equation.

ii) Show that $Latex formula$ and find the value of $Latex formula$.

iii) Hence find the initial temperature of the substance.

Solution 6

i) To verify this, we simply find $Latex formula$ and show that it is equal to $Latex formula$.

$Latex formula$

Now, $Latex formula$. Hence,

$Latex formula$

Hence $Latex formula$ satisfies the differential equation.

ii) Firstly, we know that $Latex formula$ since $Latex formula$ represents the surrounding temperature. We let $Latex formula$ be the time from which the object starts cooling. At $Latex formula$, $Latex formula$. Hence we have that;

$Latex formula$

=>$Latex formula$

At $Latex formula$ (further $Latex formula$ minutes), $Latex formula$. Hence we have;

$Latex formula$

=> $Latex formula$

Now, $Latex formula$ gives,

=>$Latex formula$

=>$Latex formula$

$Latex formula$

To find $Latex formula$ we substitute $Latex formula$ into $Latex formula$

$Latex formula$

Now, we have that

$Latex formula$

That is,

$Latex formula$

Since

$Latex formula$

Now, simplifying gives,

$Latex formula$

iii) To find the initial temperature of the substance, we let $Latex formula$. At time $Latex formula$ we have that,

$Latex formula$

Hence the temperature of the object is $Latex formula$

In the above example we could have assigned $Latex formula$ to be 6 minutes after the object had started cooling. This would allow for easy calculation of the values of $Latex formula$ and $Latex formula$.

Velocity and Acceleration

We shall now study the mathematical representation of velocity and acceleration. Velocity is the quantity which, put loosely, describes how fast an object is moving in a certain direction. Acceleration is the rate of change of this quantity. Both these quantities are “vector” quantities in that they have a magnitude (in the case of velocity how fast an object is moving) and they also have a direction.

The instantaneous velocity of an object at a point is defined to be the velocity that the object is moving with at the time in question. The average velocity of an object over a certain period is defined to be the total distance travelled over the period, divided by the time taken to travel that distance.

Let $Latex formula$ be the displacement (vector representation of distance) from the point $Latex formula$, which we shall call the origin from here on in. We have that the (instantaneous) velocity $Latex formula$ at any time $Latex formula$ is given by,

$Latex formula$

Since acceleration is defined to be the rate of change of the velocity of an object, then it follows that the acceleration $Latex formula$ is given by,

$Latex formula$

There are several notations used for velocity and acceleration. The main notations used for velocity are $Latex formula$ and $Latex formula$. The main notations used for acceleration are $Latex formula$ and $Latex formula$. We shall look at an example to illustrate the type of question that may be asked.

Example 7

Find an expression for the

a) velocity

b) displacement

given that the acceleration, $Latex formula$ at a time $Latex formula$, is given by $Latex formula$, and the particle is stationary at the origin at time $Latex formula$.

Solution 7

a) Firstly we have that $Latex formula$ where $Latex formula$ is the acceleration. Now, the acceleration is given by,

$Latex formula$

We may now integrate both sides with respect to $Latex formula$ to obtain an expression for the velocity $Latex formula$. Doing so gives,

$Latex formula$

=>$Latex formula$

At time $Latex formula$, $Latex formula$. Hence we have that,

$Latex formula$

$Latex formula$

b) We now have an expression for the velocity which is defined to be $Latex formula$. Using this information gives,

$Latex formula$

Upon integrating both sides with respect to $Latex formula$ we have,

$Latex formula$

Now, at time $Latex formula$ we have that $Latex formula$.

=>$Latex formula$

Thus we have that

$Latex formula$

Now, so far we have defined the acceleration in terms of the rate of change of time. We can now define the acceleration in terms of the change of displacement. That is, we have;

$Latex formula$

The proof is quite simple. We firstly note that;

$Latex formula$

$Latex formula$

$Latex formula$

Now,

$Latex formula$

Hence we have that,

$Latex formula$

$Latex formula$

Students must now be able to differentiate between cases where the expression for the acceleration is integrated with respect to time $Latex formula$, and when the integral is to be taken with respect to displacement $Latex formula$. Observe the example below.

Example 8

Find an expression for the velocity of a particle, given that the acceleration is given by $Latex formula$ $Latex formula$, and that the particle is stationary at $Latex formula$.

Solution 8

We have that,

$Latex formula$

Now, note that the acceleration is in terms of the displacement $Latex formula$, and hence must be integrated with respect to this variable.

$Latex formula$

Integrating with respect to $Latex formula$ gives,

$Latex formula$

=>$Latex formula$

At $Latex formula$, $Latex formula$. Hence we have that,

$Latex formula$

Hence we have that,

$Latex formula$

That is,

$Latex formula$

Now, since $Latex formula$ at $Latex formula$ and $Latex formula$, then it follows that $Latex formula$ for $Latex formula$. That is,

$Latex formula$

Projectile Motion

A projectile is any object whose path of motion (known as the trajectory) is only influenced by gravity; a good example would be a thrown rock. The motion of an ideal projectile is a parabola, although in reality there is a slight deviation from a parabola owing to the air resistance encountered (this is studied in some depth in the Mathematics Extension 2 course). The motion of a projectile may be broken up into two separate and independent motions. This notion of separating the motion of a projectile into two independent motions is important, and allows for the representation of the Cartesian equation of the parabola in terms of two parametric equations that are in terms of time. We shall now study these.

Firstly, note that we represent the motion of a projectile on a set of Cartesian axes and use the origin $Latex formula$ as the point of projection, apart from exceptional cases. Also, note that two perpendicular lines will not have any component in each other – that is to say that you cannot find any directional constituent of one line in the other line. Since we may represent vectors by lines, it thus follows that perpendicular vectors also do not have any component in each other.

Now, we may resolve the motion of the projectile into two perpendicular motions. Consider a particle projected at $Latex formula$ m/s at an angle of $Latex formula$ degrees.

We may draw the following vector diagram with the arrow representing the velocity vector.

Now it is obvious that that

$Latex formula$

by using simple trigonometric ratios. Now, in the $Latex formula$ direction there exists only one force causing acceleration in the downwards direction which we is defined to be negative due to the set up of the Cartesian axes. Also, there is no acceleration in the $Latex formula$ direction since the only force acting on the object is gravity (recall that forces and accelerations are in proportion to one another). Finding the equations for the acceleration using the $Latex formula$ notation mentioned beforehand, we have that;

 $Latex formula$

Now, we shall define $Latex formula$ to be at the point of projection. Upon integrating with respect to $Latex formula$ and using the initial conditions, we obtain the equations,

 $Latex formula$

Now, integrating again and using the initial conditions gives,

 $Latex formula$

These two equations are in essence parametric equations with $Latex formula$ (time) as the parameter. We may eliminate $Latex formula$ to obtain a cartesian equation for the parabola. To obtain the cartesian equation, simply find $Latex formula$ in terms of $Latex formula$, then substitute into the equation for $Latex formula$. We then have that,

 $Latex formula$

To calculate the range (the horizontal distance travelled) by projectile, we simply find the value of $Latex formula$ for which $Latex formula$ and substitute this into the expression for $Latex formula$.

 $Latex formula$

Now, obviously $Latex formula$ takes its maximum value when $Latex formula$ is equal to $Latex formula$ because this sinusoidal factor alternates between $Latex formula$ and $Latex formula$ for any value of $Latex formula$. So,

$Latex formula$

Now there also exist formulae for the maximum height and so on, but these are not required knowledge. Rather an understanding of how to obtain the formulae is required. To obtain the maximum height you may find the value of $Latex formula$ when $Latex formula$. (When the vertical velocity is equal to zero, the particle reaches its maximum height). After finding the value of $Latex formula$, simply substitute the value of $Latex formula$ into the formula for $Latex formula$ to obtain the maximum height.

Now, some questions may ask the student to find the velocity of the projectile upon impact or at any time $Latex formula$. To do so, simply substitute the value of $Latex formula$ into the expressions for $Latex formula$ and $Latex formula$ and using the exact opposite process used above, combine the two component vectors into one resultant vector. Doing so gives the speed as $Latex formula$. Now, one must also find the angle of deviation from the horizontal. Drawing an accurate diagram, and considering the angle $Latex formula$ as drawn in the diagram, we can show using simple trigonometry that,

 $Latex formula$

and hence that the angle of deviation is $Latex formula$ from the horizontal. (Whether it is below or above the horizontal is a matter of the sign of $Latex formula$. In his case $Latex formula$ and hence we have that the angle is below the horizontal).

We shall now consider some examples of projectile motion questions.

Example 9

(HSC 2004) A fire hose is at ground level on a horizontal plane. Water is projected from the hose. The angle of projection, $Latex formula$, is allowed to vary. The speed of the water as it leaves the hose, $Latex formula$ metres per second remains constant. You may assume that if the origin is taken to be the point of projection, the path of the water is given by the parametric equations

$Latex formula$

$Latex formula$

where $Latex formula$ ms-2 is the acceleration due to gravity. (Do NOT prove this).

a) Show that the water returns to ground level at a distance $Latex formula$ metres from the point of projection.

The fire hose is no aimed at a 20 metre high thin wall from a point of projection at ground level $Latex formula$ metres from the base of the wall. It is known that when the angle $Latex formula$ is $Latex formula$, the water just reached the base of the wall.

b) Show that $Latex formula$.

c) Show that the cartesian equation of the path of the water is given by

$Latex formula$

d) Show that the water just clears the top of the wall if

$Latex formula$

e) Find all the values of $Latex formula$ for which the water hits the front of the wall.

Solution 9

a) To show this we must find the value of $Latex formula$ for which $Latex formula$ and then substitute into the expression for $Latex formula$.

When $Latex formula$,

$Latex formula$

=>$Latex formula$

However, we do not require $Latex formula$. Hence,

$Latex formula$

So, substituting into the expression for $Latex formula$ gives,

$Latex formula$

Hence we have that the point at which it lands is given by,

$Latex formula$

b) When the angle $Latex formula$ is $Latex formula$, it follows that $Latex formula$. Since at this angle, the water hits the base of the wall.

$Latex formula$

$Latex formula$

$Latex formula$

c) To obtain the cartesian equation, we simply eliminate $Latex formula$ from the parametric equations. We have that,

$Latex formula$

Hence,

$Latex formula$

$Latex formula$

Now, we have that $Latex formula$. Hence,

$Latex formula$

$Latex formula$

d) The water just clears the top of the wall if at $Latex formula$, we have that $Latex formula$. (Since the wall is 40 metres away and is 20 metres tall). Substituting into the cartesian equation gives,

$Latex formula$

$Latex formula$

Now, we have that $Latex formula$. Using this gives,

$Latex formula$

$Latex formula$

e) For the water to hit the front of the wall, we must have that $Latex formula$ and $Latex formula$ at $Latex formula$.

So at $Latex formula$,

$Latex formula$

$Latex formula$

That is,

$Latex formula$

and

$Latex formula$

After separating out into two separate inequalities.

Now, we firstly solve the corresponding equalityto $Latex formula$, to be able to solve the inequality.

$Latex formula$

$Latex formula$

So we have that,

$Latex formula$

That is,

$Latex formula$

That is, upon taking the inverse tangent of both sides,

$Latex formula$

Now we apply process onto $Latex formula$.

$Latex formula$

So we have that,

$Latex formula$

That is,

$Latex formula$

That is,

$Latex formula$

Hence we have that for $Latex formula$ and $Latex formula$, the water hits the wall.

Example 10

(HSC 1995) A cap $Latex formula$ is lying outside a softball field, $Latex formula$ metres from the fence $Latex formula$, which is $Latex formula$ metres high. The fence is $Latex formula$ metres from the point $Latex formula$, and the point $Latex formula$ is $Latex formula$ metres above $Latex formula$. Axes are based at $Latex formula$ as shown.

At time $Latex formula$, a ball is hit from $Latex formula$ at a speed $Latex formula$ metres per second at an angle $Latex formula$ to the horizontal, towards the cap.

a) The equations of motion of the ball are

$Latex formula$

Using calculus, show that the position of the ball at time $Latex formula$ is given by

$Latex formula$

$Latex formula$

b) Hence show that the trajectory of the ball is given by,

$Latex formula$

c) The ball clears the fence. Show that

$Latex formula$

d) After clearing the fence, the ball hits the cap $Latex formula$. Show that,

$Latex formula$

e) Suppose that the ball clears the fence, and that $Latex formula$, $Latex formula$, $Latex formula$ and $Latex formula$. What is the closest point to the fence where the ball can land?

Solution 10

a) We firstly deal with the equation $Latex formula$. Integrating with respect to $Latex formula$ gives;

$Latex formula$

=>$Latex formula$

At $Latex formula$, $Latex formula$ (initial vertical velocity). Hence we have,

$Latex formula$

Hence we have,

$Latex formula$

Integrating again with respect to $Latex formula$ gives,

$Latex formula$

At $Latex formula$, $Latex formula$. Hence we have,

$Latex formula$

Hence we have,

$Latex formula$

Now, applying the same process to $Latex formula$ gives,

$Latex formula$

=>$Latex formula$

At $Latex formula$ (initial horizontal velocity).

Hence,

$Latex formula$

Integrating with respect to $Latex formula$ again gives,

$Latex formula$

At $Latex formula$, $Latex formula$. Hence we have,

$Latex formula$

=>$Latex formula$

b) In this question we simply need to find the cartesian equation, which clearly describes the trajectory of the ball. So we have,

$Latex formula$

Substituting this into the expression for $Latex formula$ gives,

$Latex formula$

$Latex formula$

i.e.

$Latex formula$

c) If the ball clears the fence then we have that at $Latex formula$, $Latex formula$ (Since the height of the wall is $Latex formula$). Using the Cartesian equation,

$Latex formula$

And we have that,

$Latex formula$

Hence,

$Latex formula$

$Latex formula$

=>$Latex formula$

$Latex formula$

d) Now, if the ball hits the cap, then we have that at $Latex formula$, $Latex formula$. Upon substituting into the cartesian equation we obtain,

$Latex formula$

Now, to obtain the inequality, we must use the inequality in the previous question.

So, since we have that,

$Latex formula$

Then it follows that,

$Latex formula$

That is,

$Latex formula$

Upon multiplying by the missing factors and simplifying. Hence we have that,

$Latex formula$

But the LHS is equal to $Latex formula$. Hence we have,

$Latex formula$

$Latex formula$ from LHS gives,

$Latex formula$

$Latex formula$

e) We have from (f) that,

$Latex formula$

We now let $Latex formula$ represent the distance the ball lands from the wall after passing it.

Substituting the values and obtaining a quadratic inequality $Latex formula$ gives,

$Latex formula$

Now, firstly we solve the equality corresponding inequality.

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

So we have,

$Latex formula$

i.e.

$Latex formula$

Since $Latex formula$ (As the ball passes over the fence) then we have that

$Latex formula$

$Latex formula$

So we can see that $Latex formula$ takes its minimum value when $Latex formula$ takes its maximum value (since at this point $Latex formula$ takes its minimum value and hence the square root value is decreased, decreasing the difference and hence the value of $Latex formula$).

Now to find the maximum value of $Latex formula$ we have to use the last piece of information left. This is the fact that $Latex formula$. So we have that,

$Latex formula$

That is,

$Latex formula$

That is,

$Latex formula$

Hence we have,

$Latex formula$

Now, a quick sketch of the function $Latex formula$ will show that,

$Latex formula$

(Note that $Latex formula$ may be obtuse since $Latex formula$ is still acute in this case as well i.e since $Latex formula$ then it follows that $Latex formula$)

This means,

$Latex formula$

Now, the function $Latex formula$ takes its maximum value on the maximum value of $Latex formula$ for $Latex formula$. So the maximum value of $Latex formula$ is:

$Latex formula$

Hence we have that the minimum value of $Latex formula$ (which occurs when $Latex formula$ is a maximum) is,

$Latex formula$

$Latex formula$

Hence the minimum distance from the fence upon clearing it is $Latex formula$ cm.

Simple Harmonic Motion

Objects that undergo simple harmonic motion oscillate back and forth about some equilibrium position. Examples include;

• The swinging of a pendulum
• A cork bobbing up and down in a wave
• A mass moving up and down when attached to a spring

Objects that undergo Simple Harmonic Motion (SHM) have one defining feature; the rate at which they accelerate is in proportion to the distance from the equilibrium position and directed towards it. That is, for an object undergoing Simple Harmonic Motion, we have that;

$Latex formula$

Where $Latex formula$ is a positive real constant. We now let $Latex formula$ for the sake of simplifying later work. So we have the equation;

$Latex formula$

Solving this differential equation gives,

 $Latex formula$

where,

$Latex formula$ is a constant called the amplitude.
$Latex formula$is a constant called the angular frequency.
$Latex formula$ is a constant called the initial phase of the motion (This determines the initial position of the particle).

Note that the above formula is required learning and should be memorised by students. We shall now consider an example to illustrate the use of the above formula.

Example 11

A particle moves with simple harmonic motion according to the formula

$Latex formula$

If the particle is initially at latex]x=0[/latex] moving with a velocity of $Latex formula$$Latex formula$ in the negative direction, find;

a) The values of $Latex formula$ and $Latex formula$.

b) The equation of the position at a time $Latex formula$.

c) The position of the particle after 4 seconds.

Solution 11

a) At $Latex formula$, $Latex formula$. Hence we have that,

$Latex formula$

$Latex formula$

Now, we have that $Latex formula$ at $Latex formula$. We must differentiate the expression for $Latex formula$ with respect to time. Doing so gives,

$Latex formula$

$Latex formula$

At $Latex formula$, $Latex formula$. Hence we have,

$Latex formula$

$Latex formula$

b) Now, we have that,

$Latex formula$

c) Now, we require the position of the particle at time $Latex formula$. Substituting in gives,

$Latex formula$

Hence the particle is $Latex formula$ m to the right of the origin (equilibrium position).

Notes

• To set the value of $Latex formula$ to zero, we simply set the time $Latex formula$ to the point at which $Latex formula$.
• All measurements of angles are in radians.
• The expression $Latex formula$ within the cosine term is called the phase of the motion, and is measured in radians.

Formulae

We now look at some important formulae regarding simple harmonic motion.

Period

The period of oscillation $Latex formula$ (or just period) is the time taken for one complete oscillation or cycle. This is given by,

$Latex formula$

Frequency

The frequency is the number of oscillations per second and is measured in Hertz (Hz). The frequency $Latex formula$ is given by,

$Latex formula$

Velocity

Velocity is simply obtained by differentiating the expression for position with respect to time $Latex formula$. However to obtain a time independentexpression for the velocity, we simple use the formula,

$Latex formula$

Where $Latex formula$ is the position. Proof of this is left to the student.

Acceleration

The definition of simple harmonic motion expressed mathematically is

$Latex formula$

To prove that a particle is moving in simple harmonic motion we simply prove the above.

That is; A particle moving with an acceleration proportional to the distance from the origin (equilibrium position), directed towards the origin is said to be moving with simple harmonic motion.

Differing expressions for Position

The position of the particle at a time $Latex formula$ has multiple functional representations. The three main functional representations are,

$Latex formula$

$Latex formula$

$Latex formula$

Where $Latex formula$ and $Latex formula$ are constants.

We shall now consider some examples to illustrate questions that may involve simple harmonic motion.

Example 12

A particle is moving in simple harmonic motion. Its displacement $Latex formula$ at time $Latex formula$ is given by;

$Latex formula$

a) Find the period of the motion.

b) Find the maximum acceleration of the particle.

c) Find the speed of the particle when $Latex formula$.

Solution 12

a) We have that $Latex formula$ by comparison to the general form. Hence we have that the period $Latex formula$ is given by,

$Latex formula$

b) The maximum acceleration of the particle occurs at the maximum negative displacement from the equilibrium position since a particle undergoing simple harmonic motion has acceleration given by, $Latex formula$. As the maximum negative displacement is given by the negated amplitude, we have that the maximum acceleration is $Latex formula$

c) Recall the formula $Latex formula$ which is independent of $Latex formula$. We have that at $Latex formula$,

$Latex formula$

Hence we have that speed (magnitude of velocity) is given by $Latex formula$ at $Latex formula$.

Example 13

A particle whose displacement is $Latex formula$, moves in simple harmonic motion such that $Latex formula$. At time $Latex formula$, $Latex formula$ and $Latex formula$.

a) Show that, for all positions of the particle,

$Latex formula$

b) What is the particles greatest displacement?

c) Find $Latex formula$ as a function of $Latex formula$. You may assume the general form for $Latex formula$.

Solution 13

a) In this question we must solve the differential equation given.

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

At $Latex formula$, $Latex formula$ hence we have that,

$Latex formula$

Hence we have,

$Latex formula$

$Latex formula$

(Since at $Latex formula$, $Latex formula$)

$Latex formula$

b) By comparison with the equation $Latex formula$ we have that,

$Latex formula$

Hence the particles greatest displacement is $Latex formula$.

c) So upon comparison we have that $Latex formula$ and $Latex formula$. Thus assuming the general form,

$Latex formula$

Now, at $Latex formula$. Hence we have that,

$Latex formula$

Hence we have:

$Latex formula$