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 $x$ , 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,

$\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

where $u$ is the “dummy variable”. Notice how the $du$ ’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 $2$ cm/min. At what rate is the volume increasing when the radius of the balloon is $7$ 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.

$\frac{dV}{dt}=D\times \frac{dr}{dt}$

Where $D$ is the required differential, $V$ is the volume of the balloon at any point in time and $r$ is the radius at any point in time.

Solving for $D$ gives,

$D=\frac{dV}{dr}$

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

Thus we have that

$\frac{dV}{dt}=\frac{dV}{dr}\times \frac{dr}{dt}$

We now are required to find the value of .

$\frac{dV}{dr}$

Recall that for a sphere,

$V=\frac{4}{3}\pi r^{3}$

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

$\frac{dV}{dr}=4\pi r^{2}$

Now, substituting in this value into the expression for $\frac{dV}{dt}$ gives;

$\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}$

We require the rate of change when the radius $r$ equals $7$ cm. Also, it is given that the rate of change of the radius is equal to $2$ cm/min. Hence we have that the rate of change of the volume is,

$\frac{dV}{dt}=4\pi (7)^{2}(2)=392\pi cm^{3}/min$

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 $9$ cm and the base of the triangle is increasing at a constant rate of $3$ cm/s. At what rate is the hypotenuse increasing when its length is 15 cm?

Solution 2

We firstly draw a good diagram.

Let $h$ be the hypotenuse, $b$ be the base and $a=9$ cm be the altitude of the right angled triangle.

We are given $\frac{db}{dt}$ and we are required to find $\frac{dh}{dt}$ when the hypotenuse has length $15$ cm. Using the chain rule gives,

$\frac{dh}{dt}=D\times \frac{db}{dt}$

-> $D=\frac{dh}{db}$

Hence we have that,

$\frac{dh}{dt}=\frac{dh}{db}\times \frac{db}{dt}$

We must now find $\frac{dh}{db}$ .

By Pythagoras’ theorem we have that,

$h^{2}=a^{2}+b^{2}$

$h^{2}=9^{2}+b^{2}$

$h^{2}=81+b^{2}$

Hence we have that,

$h=(81+b^{2})^{\frac{1}{2}}$

Now, to find $\frac{dh}{db}$ we must use the chain rule,

$\frac{dh}{db}=\frac{1}{2}(2b)(81+b^{2})^{-\frac{1}{2}}=\frac{b}{(81+b^{2})^{\frac{1}{2}}}$

At $h=15$ cm, we have that,

$15^{2}=9^{2}+b^{2}?b=12 (b>0)$

Hence we have that,

$\frac{dh}{db}=\frac{12}{(81+12^{2})^{\frac{1}{2}}}=\frac{12}{15}=\frac{4}{5}$

Hence we have that,

$\frac{dh}{dt}=\frac{4}{5}\times 3=\frac{12}{5} cm/s$

The length of the hypotenuse is increasing at $2.4$ 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 $8$ cm, given that the volume is increasing at rate of $12$ cm³/sec.

Solution 3

With this problem, we have the value of $\frac{dV}{dt}$ and wish to find the value of $\frac{dS}{dt}$ . Using the chain rule gives,

$\frac{dS}{dt}=\frac{dS}{dV}\times \frac{dV}{dt}$

Now, we must calculate $\frac{dS}{dV}$ . 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,

$V=\frac{4}{3}\pi r^{3} , S=4\pi r^{2}$

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

$\frac{dS}{dV}=\frac{dS}{dr}\times \frac{dr}{dV}$

Now $\frac{dS}{dr}$ is given by,

$\frac{dS}{dr}=8\pi r$

and $\frac{dV}{dr}$ is given by,

$\frac{dV}{dr}=4\pi r^{2}$

Hence we have that,

$\frac{dS}{dV}=8\pi r\times \frac{1}{4\pi r^{2}}$

-> $\frac{dS}{dV}=\frac{2}{r}$

So we have that,

$\frac{dS}{dt}=\frac{2}{r}\times \frac{dV}{dt}$

$\frac{dS}{dt}=\frac{2}{8}\times 12=3 cm^{2}/s$

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

$\frac{dS}{dt}=\frac{dS}{dr}\times \frac{dr}{dV}\times \frac{dV}{dt}$

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

$\frac{dS}{dt}=\frac{dS}{dr}\times \frac{dr}{dV}\times \frac{dV}{dt}$

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 $r$ is heated so that its area expands at a constant rate of $4$ cm²/min. At what rate does the radius increase when $r=10$ cm?

Solution 4

In this question we are given that $\frac{dA}{dt}=4 cm^{2}/min$ .

Now, the area of a circle is given by,

$A=\pi r^{2}$

Treating $r$ as a function of time $(t)$ and $A$ also as a function of $t$ we may differentiate both sides with respect to $t$ .

$\frac{d}{dt}(A)=\frac{d}{dt}(\pi r^{2})$

This gives,

$\frac{dA}{dt}=\pi \frac{d}{dt}(r^{2})$

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

$\frac{dA}{dt}=\pi (2r\frac{dr}{dt})$

Since we have that $r$ is a function of $t$ .

Now, it is known that $\frac{dA}{dt}$ is equal 4 cm²/min. At $r=10$ cm we have that,

$\frac{dr}{dt}=\frac{4}{20\pi }=\frac{1}{5\pi } cm/min$

Hence the radius is increasing at $1/5\pi$ 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

$\frac{dP}{dt}=kP$

Where $k$ 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,

$P(t)=Ae^{kt}$

where $A$ is some real constant.

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

$\frac{dP}{dt}=k(P-P_{0})$

Where $P_{0}$ and $k$ 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,

$P(t)=P_{0}+Ae^{kt}$

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

$\frac{dP}{dt}=k(P-P_{0})$

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

$\frac{dt}{dP}=\frac{1}{k(P-P_{0})}$

Integrating with respect to $P$ gives,

$t=\frac{1}{k}\ln (P-P_{0})+C$

Where $C$ is a constant. Now, rearranging gives,

$\ln (P-P_{0})=kt-kC$

-> $P-P_{0}=e^{kt-kC}$

-> $P=P_{0}+(e^{-kC})e^{kt}$

Now let $e^{-kC}=A$ . Hence we have that,

$P(t)=P_{0}+Ae^{kt}$

Which is the solution to the differential equation. The initial conditions given will allow the values of $A$ , $k$ and $P_{0}$ to be determined.

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

Example 5

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

$\frac{dP}{dt}=k(P-15 000)$

where $k=0.2$ and $t$ is the time in years.

a) Show that $P=15000+Ae^{kt}$ is a solution to the above differential equation ( $A$ and $k$ 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 $t\to \infty$ ?

Solution 5

a) To show that $P=15000+Ae^{kt}$ 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.

$P=15000+Ae^{kt}$

$? \frac{dP}{dt}=Ake^{kt}$

Now, $Ae^{kt}=P-15000$ . Hence we have that,

$\frac{dP}{dt}=k(Ae^{kt})=k(P-15 000)$

Hence as $P=15000+Ae^{kt}$ satisfies the differential equation, then it follows that it is a solution to the differential equation.

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

At $t=0$ , $P=21000$ . So we have that,

$21000=15000+Ae^{k(0)}$

$A=6000$

Now, we have that,

$P(t)=15 000+6000 e^{0.2t}$

Since $k=0.2$

Now, at time $t=1$ we have that

$P(1)=15 000+6000e^{0.2}\cong 22 328$

Thus the population is approximately $22 328$ after one year.

ii) At time $t=5$ ,

$P(5)=15000+6000e^{0.2\times 5}\cong 31 310$

Hence the population is approximately 31 310 after 5 years.

c) At $P=42 000$ , (twice original population) we have that,

$42 000=15 000+6000 e^{0.2t}$

$=> e^{0.2t}=\frac{9}{2}$

$0.2t=\ln (\frac{9}{2})$

$=> t=5\ln (4.5)\cong 7.52 years$

Hence the population will double in the year $1960+7=1967$

d) As $t\to \infty$ the value of $P(t)\to \infty$ as $e^{0.2t}\to \infty$ .

Newtons Law of Cooling

An important application of the differential equation

$\frac{dP}{dt}=k(P-P_{0})$

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

$\frac{dT}{dt}=k(T-S)$

Where $S$ is the surrounding temperature, and $k$ 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 $100degC$ . Dr Kool takes a sample of his substance and places it in a room with a surrounding air temperature of $20degC$ , and allows it to cool.

After $6$ minutes the temperature of the substance is $80degC$ , and after a further $2$ minutes it is $50degC$ . If $T(t)$ is the temperature of the substance after $t$ minutes, then Newton’s law of cooling states that $T$ satisfies the equation

$\frac{dT}{dt}=k(T-A)$

where $k$ is a constant and $A$ is the surrounding air temperature.

i) Verify that $T(t)=A+Be^{kt}$ satisfies the above differential equation.

ii) Show that $k=-(ln2)/2$ and find the value of $B$ .

iii) Hence find the initial temperature of the substance.

Solution 6

i) To verify this, we simply find $dT/dt$ and show that it is equal to $k(T-A)$ .

$\frac{dT}{dt}=\frac{d}{dt}(A+Be^{kt})=Bke^{kt}$

Now, $Be^{kt}=T-A$ . Hence,

$\frac{dT}{dt}=k(Be^{kt})=k(T-A)$

Hence $T(t)=A+Be^{kt}$ satisfies the differential equation.

ii) Firstly, we know that $A=20degC$ since $A$ represents the surrounding temperature. We let $t=0$ be the time from which the object starts cooling. At $t=6$ , $T=80degC$ . Hence we have that;

$80=20+Be^{6k}$

=> $Be^{6k}=60 (1)$

At $t=8$ (further $2$ minutes), $T=50$ . Hence we have;

$50=20+Be^{8k}$

=> $Be^{8k}=30 (2)$

Now, $(2)/(1)$ gives,

=> $^{2k}=\frac{1}{2}$

=> $k=\frac{1}{2}\ln (\frac{1}{2})$

$k=-\frac{(\ln 2)}{2}$

To find $B$ we substitute $k$ into $(1)$

$Be^{-3\ln 2}=60$

Now, we have that

$B(e^{\ln 2})^{-3}=60$

That is,

$B(2)^{-3}=60$

Since

$e^{\ln x}=x$

Now, simplifying gives,

$B=480$

iii) To find the initial temperature of the substance, we let $t=0$ . At time $t=0$ we have that,

$T(0)=20+480e^{0}=500degC$

Hence the temperature of the object is $500degC$

In the above example we could have assigned $t=0$ to be 6 minutes after the object had started cooling. This would allow for easy calculation of the values of $k$ and $B$ .

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

$v=\frac{dx}{dt}$

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

$a=\frac{dv}{dt}=\frac{d}{dt}(\frac{dx}{dt})=\frac{d^{2}x}{dt^{2}}$

There are several notations used for velocity and acceleration. The main notations used for velocity are $v , \dot{x}$ and $dx/dt$ . The main notations used for acceleration are $a , \ddot{x} , dv/dt$ and $d^{2}x/dt^{2}$ . 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, $a$ at a time $t$ , is given by $a=t+1$ , and the particle is stationary at the origin at time $t$ .

Solution 7

a) Firstly we have that $a=t+1$ where $a$ is the acceleration. Now, the acceleration is given by,

$\frac{dv}{dt}=t+1$

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

$v=\int{(t+1) dt}$

=> $v=\frac{t^{2}}{2}+t+C_{1}$

At time $t=0$ , $v=0$ . Hence we have that,

$C_{1}=0$

$v=\frac{1}{2}t^{2}+t$

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

$\frac{dx}{dt}=\frac{1}{2}t^{2}+t$

Upon integrating both sides with respect to $t$ we have,

$x=\frac{1}{6}t^{3}+\frac{1}{2}t^{2}+C_{2}$

Now, at time $t=0$ we have that $x=0$ .

=> $C_{2}=0$

Thus we have that

$x=\frac{1}{6}t^{3}+\frac{1}{2}t^{2}$

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;

$a=\frac{d}{dx}(\frac{1}{2}v^{2})$

The proof is quite simple. We firstly note that;

$a=\frac{dv}{dt}$

$a=\frac{dx}{dt}\times \frac{dv}{dx}$

$a=v\frac{dv}{dx}$

Now,

$v=\frac{d}{dv}(\frac{1}{2}v^{2})$

Hence we have that,

$a=\frac{d(\frac{1}{2}v^{2})}{dv}\times \frac{dv}{dx}$

$a=\frac{d}{dx}(\frac{1}{2}v^{2})$

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

Example 8

Find an expression for the velocity of a particle, given that the acceleration is given by $a=x+$ $\frac{1}{x}$ , and that the particle is stationary at $x=1$ .

Solution 8

We have that,

$a=x+\frac{1}{x}$

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

$\frac{d}{dx}(\frac{1}{2}v^{2})=x+\frac{1}{x}$

Integrating with respect to $x$ gives,

$\frac{1}{2}v^{2}=\int{(x+\frac{1}{x})dx}$

=> $\frac{1}{2}v^{2}=\frac{1}{2}x^{2}+\ln x+C$

At $x=1$ , $v=0$ . Hence we have that,

$C=-\frac{1}{2}$

Hence we have that,

$\frac{1}{2}v^{2}=\frac{1}{2}x^{2}+\ln x-\frac{1}{2}$

That is,

$v^{2}=x^{2}+2\ln x-1$

Now, since $a>0$ at $x=1$ and $v=0$ , then it follows that $v\ge 0$ for $x\ge 1$ . That is,

$v(x)=\sqrt{x^{2}+2\ln x-1}$

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 $O$ 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 $V$ m/s at an angle of $\theta$ degrees.

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

Now it is obvious that that

$u_{x}=V\cos \theta u_{y}=V\sin \theta$

by using simple trigonometric ratios. Now, in the $y$ 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 $x$ 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 $\ddot{x}$ notation mentioned beforehand, we have that;

$\ddot{x}=0 , \ddot{y}=-g$

Now, we shall define $t=0$ to be at the point of projection. Upon integrating with respect to $t$ and using the initial conditions, we obtain the equations,

$\dot{x}=V\cos \theta , \dot{y}=-gt+V\sin \theta$

Now, integrating again and using the initial conditions gives,

$x=Vt\cos \theta , y=-\frac{1}{2}gt^{2}+Vt\sin \theta$

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

$y=x\tan \theta -(\frac{g\sec ^{2}\theta }{2v^{2}})x^{2}$

To calculate the range (the horizontal distance travelled) by projectile, we simply find the value of $t$ for which $y=0$ and substitute this into the expression for $x$ .

$R=\frac{2V^{2}\sin \theta \cos \theta }{g}=\frac{V^{2}\sin 2\theta }{g}$

Now, obviously $R$ takes its maximum value when $\sin 2\theta$ is equal to $1$ because this sinusoidal factor alternates between $1$ and $-1$ for any value of $V$ . So,

$\sin 2\theta =1 ? 2\theta =90deg ? \theta =45deg$

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 $t$ when $\dot{y}=0$ . (When the vertical velocity is equal to zero, the particle reaches its maximum height). After finding the value of $t$ , simply substitute the value of $t$ into the formula for $y$ 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 $t$ . To do so, simply substitute the value of $t$ into the expressions for $\dot{y}$ and $\dot{x}$ and using the exact opposite process used above, combine the two component vectors into one resultant vector. Doing so gives the speed as $v_{t}=\sqrt{\dot{x}^{2}+\dot{y}^{2}}$ . Now, one must also find the angle of deviation from the horizontal. Drawing an accurate diagram, and considering the angle $\theta$ as drawn in the diagram, we can show using simple trigonometry that,

$\theta =\tan ^{-1}\vert \frac{\dot{y}}{\dot{x}}\vert$

and hence that the angle of deviation is $\tan ^{-1}\vert \frac{\dot{y}}{\dot{x}}\vert$ from the horizontal. (Whether it is below or above the horizontal is a matter of the sign of $\dot{y}$ . In his case $\dot{y}<0$ 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, $\theta$ , is allowed to vary. The speed of the water as it leaves the hose, $v$ 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

$x=vt\cos \theta$

$y=vt\sin \theta -\frac{1}{2}gt^{2}$

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

a) Show that the water returns to ground level at a distance $\frac{v^{2}\sin 2\theta }{g}$ 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 $40$ metres from the base of the wall. It is known that when the angle $\theta$ is $15deg$ , the water just reached the base of the wall.

b) Show that $v^{2}=80g$ .

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

$y=x\tan \theta -\frac{x^{2}\sec ^{2}\theta }{160}$

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

$\tan ^{2}\theta -4\tan \theta +3=0$

e) Find all the values of $\theta$ for which the water hits the front of the wall.

Solution 9

a) To show this we must find the value of $t$ for which $y=0$ and then substitute into the expression for $x$ .

When $y=0$ ,

$vt\sin \theta -\frac{1}{2}gt^{2}=0$

=> $t=0 , v\sin \theta -\frac{1}{2}gt=0$

However, we do not require $t=0$ . Hence,

$t=\frac{2v\sin \theta }{g}$

So, substituting into the expression for $x(t)$ gives,

$x=\frac{2v^{2}\sin \theta \cos \theta }{g}=\frac{v^{2}(2\sin \theta \cos \theta )}{g}=\frac{v^{2}\sin 2\theta }{g}$

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

$R=\frac{v^{2}\sin 2\theta }{g}$

b) When the angle $\theta$ is $15deg$ , it follows that $R=40$ . Since at this angle, the water hits the base of the wall.

$? 40=\frac{v^{2}\sin 2(15)}{g}$

$40=\frac{v^{2}}{2g}$

$v^{2}=80g$

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

$t=\frac{x}{v\cos \theta }$

Hence,

$y=v(\frac{x}{v\cos \theta })\sin \theta -\frac{1}{2}g (\frac{x}{v\cos \theta })^{2}$

$y=x\tan \theta -\frac{gx^{2}}{2v^{2}\cos ^{2}\theta }$

Now, we have that $v^{2}=80g$ . Hence,

$y=x\tan \theta -\frac{gx^{2}}{2(80g)\cos ^{2}\theta }$

$? y=x\tan \theta -\frac{x^{2}\sec ^{2}\theta }{160}$

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

$20=40\tan \theta -\frac{40^{2}\sec ^{2}\theta }{160}$

$? 2=4\tan \theta -\sec ^{2}\theta$

Now, we have that $\sec ^{2}\theta =1+\tan ^{2}\theta$ . Using this gives,

$2=4\tan \theta -(1+\tan ^{2}\theta )$

$? \tan ^{2}\theta -4\tan \theta +3=0$

e) For the water to hit the front of the wall, we must have that $y>0$ and $y<20$ at $x=40$ .

So at $x=40$ ,

$0<y(x)<20$

$0<40\tan \theta -10(1+\tan ^{2}\theta )<20$

That is,

$\tan ^{2}\theta -4\tan \theta +1<0 (1)$

and

$\tan ^{2}\theta -4\tan \theta +3>0 (2)$

After separating out into two separate inequalities.

Now, we firstly solve the corresponding equalityto $(1)$ , to be able to solve the inequality.

$\tan ^{2}\theta -4\tan \theta +1=0$

$\tan \theta =\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}$

So we have that,

$(\tan \theta -2-\sqrt{3})(\tan \theta -2+\sqrt{3})<0$

That is,

$\tan \theta >2-\sqrt{3} , \tan \theta <2+\sqrt{3}$

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

$\theta >15deg , \theta <75deg$

Now we apply process onto $(2)$ .

$\tan \theta =\frac{4\pm \sqrt{16-12}}{2}=\frac{4\pm 2}{2}=1 , 3$

So we have that,

$(\tan \theta -3)(\tan \theta -1)>0$

That is,

$\tan \theta >3 , \tan \theta <1$

That is,

$\theta >71.57deg , \theta <45deg$

Hence we have that for $15deg<\theta <45deg$ and $71.57deg<\theta <75deg$ , the water hits the wall.

Example 10

(HSC 1995) A cap $C$ is lying outside a softball field, $r$ metres from the fence $F$ , which is $h$ metres high. The fence is $R$ metres from the point $O$ , and the point $P$ is $h$ metres above $O$ . Axes are based at $O$ as shown.

At time $t=0$ , a ball is hit from $P$ at a speed $V$ metres per second at an angle $\alpha$ to the horizontal, towards the cap.

a) The equations of motion of the ball are

$\ddot{x}=0 , \ddot{y}=-g$

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

$x=Vt\cos \alpha$

$y=Vt\sin \alpha -\frac{1}{2}gt^{2}+h$

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

$y=h+x\tan \alpha -x^{2}\frac{g}{2V^{2}\cos ^{2}\alpha }$

c) The ball clears the fence. Show that

$V^{2}\ge \frac{gR}{2\sin \alpha \cos \alpha }$

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

$\tan \alpha \ge \frac{Rh}{(R+r)r}$

e) Suppose that the ball clears the fence, and that $V\le 50$ , $g=10$ , $R=80$ and $h=1$ . What is the closest point to the fence where the ball can land?

Solution 10

a) We firstly deal with the equation $\ddot{y}=-g$ . Integrating with respect to $t$ gives;

$\ddot{y}=-g$

=> $\dot{y}=-gt+C_{1}$

At $t=0$ , $\dot{y}=V\sin \alpha$ (initial vertical velocity). Hence we have,

$V\sin \alpha =C_{1}$

Hence we have,

$\dot{y}=-gt+V\sin \alpha$

Integrating again with respect to $t$ gives,

$y=-\frac{1}{2}gt^{2}+Vt\sin \alpha +C_{2}$

At $t=0$ , $y=h$ . Hence we have,

$h=C_{2}$

Hence we have,

$y(t)=Vt\sin \alpha -\frac{1}{2}gt^{2}+h$

Now, applying the same process to $\ddot{x}=0$ gives,

$\ddot{x}=0$

=> $\dot{x}=C_{3}$

At $t=0, \dot{x}=V\cos \alpha$ (initial horizontal velocity).

Hence,

$\dot{x}=V\cos \alpha$

Integrating with respect to $t$ again gives,

$x=Vt\cos \alpha +C_{4}$

At $t=0$ , $x=0$ . Hence we have,

$C_{4}=0$

=> $x(t)=Vt\cos \alpha$

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

$t=\frac{x}{V\cos \alpha }$

Substituting this into the expression for $y$ gives,

$y=V(\frac{x}{V\cos \alpha })\sin \alpha -\frac{1}{2}g(\frac{x}{V\cos \alpha })^{2}+h$

$y=x\tan \alpha -\frac{gx^{2}}{2V^{2}\cos ^{2}\alpha }+h$

i.e.

$y(x)=h+x\tan \alpha -x^{2}\frac{g}{2V^{2}\cos ^{2}\alpha }$

c) If the ball clears the fence then we have that at $x=R$ , $y\ge h$ (Since the height of the wall is $h$ ). Using the Cartesian equation,

$y(R)=h+R\tan \alpha -R^{2}\frac{g}{2V^{2}\cos ^{2}\alpha }$

And we have that,

$y(R)\ge h$

Hence,

$h+R\tan \alpha -R^{2}\frac{g}{2V^{2}\cos ^{2}\alpha }\ge h$

$\tan \alpha \ge \frac{Rg}{2V^{2}\cos ^{2}\alpha }$

=> $V^{2}\ge \frac{gR}{2\cos ^{2}\alpha \tan \alpha }$

$V^{2}\ge \frac{gR}{2\sin \alpha \cos \alpha }$

d) Now, if the ball hits the cap, then we have that at $x=R+r$ , $y=0$ . Upon substituting into the cartesian equation we obtain,

$0=h+(R+r)\tan \alpha -\frac{(R+r)^{2}g}{2V^{2}\cos ^{2}\alpha }$

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

So, since we have that,

$V^{2}\ge \frac{gR}{2\sin \alpha \cos \alpha }$

Then it follows that,

$\frac{1}{V^{2}}\le \frac{2\sin \alpha \cos \alpha }{gR}$

That is,

$-\frac{(R+r)^{2}g}{2V^{2}\cos ^{2}\alpha }\ge -\frac{2(R+r)^{2}\sin \alpha \cos \alpha }{2R\cos ^{2}\alpha }=-\frac{(R+r)^{2}\sin \alpha }{R\cos \alpha }=-\frac{1}{R}(R+r)^{2}\tan \alpha$

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

$h+(R+r)\tan \alpha -\frac{(R+r)^{2}g}{2V^{2}\cos ^{2}\alpha }\ge h+(R+r)\tan \alpha -\frac{1}{R}(R+r)^{2}\tan \alpha$

But the LHS is equal to $0$ . Hence we have,

$0\ge Rh+R(R+r)\tan \alpha -(R+r)^{2}\tan \alpha$

$\lbrace (R+r)^{2}-(R+r)\rbrace \tan \alpha \ge Rh[latex] Now, removing a factor of [latex](R+r)$ from LHS gives,

$(R+r)(R+r-R)\tan \alpha \ge Rh$

$\tan \alpha \ge \frac{Rh}{(R+r)r}$

e) We have from (f) that,

$\tan \alpha \ge \frac{Rh}{(R+r)r}$

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

Substituting the values and obtaining a quadratic inequality $r$ gives,

$r^{2}+80r-\frac{80}{\tan \alpha }\ge 0$

Now, firstly we solve the equality corresponding inequality.

$r^{2}+80r-\frac{80}{\tan \alpha }=0$

$? r=\frac{-80\pm \sqrt{80^{2}+\frac{4\times 80}{\tan \alpha }}}{2}$

$? r=\frac{-80\pm 8\sqrt{100+\frac{5}{\tan \alpha }}}{2}$

$r=-40\pm 4\sqrt{100+\frac{5}{\tan \alpha }}$

So we have,

$(r+40+4\sqrt{100+\frac{5}{\tan \alpha }})(r+40-4\sqrt{100+\frac{5}{\tan \alpha }})\ge 0$

i.e.

$r\le -40-4\sqrt{100+\frac{5}{\tan \alpha }}, r\ge -40+4\sqrt{100+\frac{5}{\tan \alpha }}$

Since $r>0$ (As the ball passes over the fence) then we have that

$r\ge 4\sqrt{100+\frac{5}{\tan \alpha }}-40$

$\min (r(\alpha ))=4\sqrt{100+\frac{5}{\tan \alpha }}-40$

So we can see that $r$ takes its minimum value when $\tan \alpha$ takes its maximum value (since at this point $5/\tan \alpha$ takes its minimum value and hence the square root value is decreased, decreasing the difference and hence the value of $r$ ).

Now to find the maximum value of $\tan \alpha$ we have to use the last piece of information left. This is the fact that $V\ge 50$ . So we have that,

$V^{2}\ge \frac{gR}{2\sin \alpha \cos \alpha }$

That is,

$50^{2}\ge V^{2}\ge \frac{80(10)}{\sin 2\alpha }$

That is,

$50^{2}\ge V^{2}\ge \frac{80(10)}{\sin 2\alpha }$

Hence we have,

$\sin 2\alpha \ge \frac{8}{25}$

Now, a quick sketch of the function $y=\sin 2\alpha$ will show that,

$\sin 2\alpha \ge \frac{8}{25} ? \sin ^{-1}(\frac{8}{25})\le 2\alpha \le \pi -\sin ^{-1}(\frac{8}{25})$

(Note that $2\alpha$ may be obtuse since $\alpha$ is still acute in this case as well i.e since $0\le \alpha \le \frac{\pi }{2}$ then it follows that $0\le 2\alpha \le \pi$ )

This means,

$\frac{1}{2}\sin ^{-1}(\frac{8}{25})\le \alpha \le \frac{\pi }{2}-\frac{1}{2}\sin ^{-1}(\frac{8}{25})$

Now, the function $f(\alpha )=\tan \alpha$ takes its maximum value on the maximum value of $\alpha$ for $0\le \alpha \le \frac{\pi }{2}$ . So the maximum value of $\tan \alpha$ is:

$max (\tan \alpha )=\tan (\frac{\pi }{2}-\frac{1}{2}\sin ^{-1}(\frac{8}{25}))$

Hence we have that the minimum value of $r$ (which occurs when $\tan \alpha$ is a maximum) is,

$\min (r)=4\sqrt{100+\frac{5}{\tan (\frac{\pi }{2}-\frac{1}{2}\sin ^{-1}(\frac{8}{25}))}}-40$

$\min (r)=0.163\ldots m\cong 16 cm$

Hence the minimum distance from the fence upon clearing it is $16$ 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;

$\frac{d^{2}x}{dt^{2}}=-kx$

Where $k$ is a positive real constant. We now let $k=n^{2}$ for the sake of simplifying later work. So we have the equation;

$\frac{d^{2}x}{dt^{2}}=-n^{2}x$

Solving this differential equation gives,

$x(t)=A\cos (nt+\alpha )$

where,

$A$ is a constant called the amplitude.
$n$ is a constant called the angular frequency.
$\alpha$ 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

$x=A\cos (3t+\alpha )$

If the particle is initially at latex]x=0[/latex] moving with a velocity of $4$ $ms^{-1}$ in the negative direction, find;

a) The values of $A$ and $\alpha$ .

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

c) The position of the particle after 4 seconds.

Solution 11

a) At $t=0$ , $x=A$ . Hence we have that,

$0=A\cos (0+\alpha )?\cos \alpha =0$

$\alpha =\frac{\pi }{2}$

Now, we have that $\dot{x}=-4$ at $t=0$ . We must differentiate the expression for $x$ with respect to time. Doing so gives,

$x=A\cos (3t+\frac{\pi }{2})$

$\dot{x}=-3A\sin (3t+\frac{\pi }{2})$

At $t=0$ , $\dot{x}=-4$ . Hence we have,

$-4=-3A\sin \frac{\pi }{2}$

$A=4/3$

b) Now, we have that,

$x(t)=\frac{4}{3}\cos (3t+\frac{\pi }{2})$

c) Now, we require the position of the particle at time $t=4$ . Substituting in gives,

$x(4)=\frac{4}{3}\cos (12+\frac{\pi }{2})\cong 0.715 m$

Hence the particle is $0.72$ m to the right of the origin (equilibrium position).

Notes

To set the value of $\alpha$ to zero, we simply set the time $t=0$ to the point at which $x=A$ .
All measurements of angles are in radians.
The expression $nt+\alpha$ 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 $(T)$ (or just period) is the time taken for one complete oscillation or cycle. This is given by,

$T=\frac{2\pi }{n}$

Frequency

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

$f=\frac{1}{T}=\frac{n}{2\pi }$

Velocity

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

$v^{2}=n^{2}(A^{2}-x^{2})$

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

Acceleration

The definition of simple harmonic motion expressed mathematically is

$\ddot{x}=-n^{2}x$

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 $t$ has multiple functional representations. The three main functional representations are,

$x(t)=A\cos (nt+\alpha )$

$x(t)=A\sin (nt+\beta )$

$x(t)=B\sin (nt)+C\cos (nt)$

Where $B$ and $C$ 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 $x$ at time $t$ is given by;

$x=3\sin (2t+5)$

a) Find the period of the motion.

b) Find the maximum acceleration of the particle.

c) Find the speed of the particle when $x=2$ .

Solution 12

a) We have that $n=2$ by comparison to the general form. Hence we have that the period $T$ is given by,

$T=\frac{2\pi }{2}=\pi$

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, $\ddot{x}=-n^{2}x$ . As the maximum negative displacement is given by the negated amplitude, we have that the maximum acceleration is $\max ( \ddot{x} )=-2^{2}(-3)=12$