## Population Differential Equations and Laplace Transform

Malthus Model
$\displaystyle \frac{dN}{dt}=BN-DN=kN$

$N$: Total population

$B$: Birth-rate per capita

$D$: Death-rate per capita

$k=B-D$

Solution to D.E.:
$\displaystyle \boxed{N(t)=\widehat{N}e^{kt}},$

where $\widehat{N}=N(0)$.

Logistic Equation
\begin{aligned} D&=sN\\ \frac{dN}{dt}&=BN-sN^2\\ \widehat{N}&=N(0)\\ N_\infty&=B/s \end{aligned}

Logistic Case 1: Increasing population ($\widehat{N})
\begin{aligned} N(t)&=\frac{B}{s+(\frac{B}{\widehat{N}}-s)e^{-Bt}}\\ &=\frac{N_\infty}{1+(\frac{N_\infty}{\widehat{N}}-1)e^{-Bt}} \end{aligned}

The second expression can be derived from the first: divide by $s$ in both the numerator and denominator.

Logistic Case 2: Decreasing population ($\widehat{N}>N_\infty$)
\begin{aligned} N(t)&=\frac{B}{s-(s-\frac{B}{\widehat{N}})e^{-Bt}}\\ &=\frac{N_\infty}{1-(1-\frac{N_\infty}{\widehat{N}})e^{-Bt}} \end{aligned}

Logistic Case 3: Constant population ($\widehat{N}=N_\infty$)
$\displaystyle N(t)=N_\infty$

Harvesting
Basic Harvesting Model: $\displaystyle \boxed{\frac{dN}{dt}=(B-sN)N-E}.$

$E$: Harvest rate (Amount harvested per unit time)

Maximum harvest rate without causing extinction: $\boxed{\dfrac{B^2}{4s}}$.

$\displaystyle \boxed{\beta_1,\beta_2=\frac{B\mp\sqrt{B^2-4Es}}{2s}}.$

$\beta_1$: Unstable equilibrium population

$\beta_2$: Stable equilibrium population

Extinction Time: $\displaystyle \boxed{T=\int_{\widehat{N}}^0\frac{dN}{N(B-sN)-E}}.$

Laplace transform of $f$
$\displaystyle F(s)=L(f)=\int_0^\infty e^{-st}f(t)\,dt$

Tip: Use this equation when the questions contains the words “show from the definition”.

Inverse transform of $F(s)$
$\displaystyle f(t)=L^{-1}(F(s))$

Linearity
\begin{aligned} L(af(t)+bg(t))&=aL(f)+bL(g)\\ L^{-1}(aF(s)+bG(s))&=aL^{-1}(F)+bL^{-1}(g) \end{aligned}

## List of common Laplace Transforms

\begin{aligned} L(e^{at})&=\frac{1}{s-a}\\ L(1)&=\frac{1}{s}\\ L(\cos wt)&=\frac{s}{s^2+w^2}\\ L(\sin wt)&=\frac{w}{s^2+w^2}\\ L(t^n)&=\frac{n!}{s^{n+1}}\\ L(f')&=sL(f)-f(0)\\ L(f'')&=s^2L(f)-sf(0)-f'(0)\\ L(f^{(n)})&=s^nL(f)-s^{n-1}f(0)\\ &\quad -s^{n-2}f'(0)-\dots-f^{(n-1)}(0)\\ L\left(\int_0^t f(\tau)\,d\tau\right)&=\frac{1}{s}L(f) \end{aligned}

$s$-shifting
If $L(f)=F(s)$, $s>a$, then $\displaystyle \boxed{L(e^{ct}f(t))=F(s-c)},$
$s-c>a$.

Tip: Use this when doing Laplace Transform of a function with an exponential factor $e^{ct}$. Note that the reverse direction can sometimes be used as well: $\displaystyle L^{-1}[F(s-c)]=e^{ct}f(t).$

$t$-shifting
If $L(f(t))=F(s)$, then $\displaystyle \boxed{L(f(t-a)u(t-a))=e^{-as}F(s)}.$

Tip: Frequently, we use the reverse direction $\displaystyle L^{-1}[e^{-as}F(s)]=f(t-a)u(t-a).$

Delta function
$\delta(t)$: infinitely tall and narrow spike at $t=0$.

$\delta(t-a)$: infinitely tall and narrow spike at $t=a$.

$\boxed{L[\delta(t-a)]=e^{-as}}$

Two properties of delta function
\begin{aligned} \int_0^\infty\delta(t-a)\,dt&=1\\ \int_0^\infty \delta(t-a)g(t)\,dt&=g(a) \end{aligned}
for $a\geq 0$.

Tip: Use delta function when the keywords “suddenly”, “burst”, etc. appear.

Unit step function
$\displaystyle u(t-a)=\begin{cases} 0, &ta. \end{cases}$

For $0, $\displaystyle u(t-a)-u(t-b)=\begin{cases} 0, &tb. \end{cases}$

Tip: Use unit step function for questions that require a force to “switch on / switch off” at certain times.

$\displaystyle \boxed{L(u(t-a))=\frac{e^{-as}}{s}}$

## Linear First Order ODE, Bernoulli Equations and Applications

Linear First Order ODE
DE of the form: $y'+P(x)y=Q(x)$.

Integrating factor: $R(x)=e^{\int P(x)\,dx}$.
\begin{aligned} R'&=RP\\ Ry'+RPy&=RQ\\ (Ry)'&=RQ\\ Ry&=\int RQ\,dx \end{aligned}
$\displaystyle \boxed{y=\frac{\int RQ\,dx}{R}}$
(Remember to have a constant $C$ when integrating the numerator $\int RQ\,dx$.)

Integration by parts
$\displaystyle \boxed{\int uv'\,dx=uv-\int u'v\,dx}$

Acronym: LIATE (Log, Inverse Trig., Algebraic, Trig., Exponential), where L is the best choice for $u$. (This is only a rough guideline.)

Bernoulli Equations
DE of the form: $y'+p(x)y=q(x)y^n$.

$y^{-n}y'+y^{1-n}p(x)=q(x)$

Set $\boxed{y^{1-n}=z}$.

Then $(1-n)y^{-n}y'=z'$. The given DE becomes
$\displaystyle \boxed{z'+(1-n)p(x)z=(1-n)q(x)}.$

Fundamental Theorem of Calculus (FTC)
Part 1: $\displaystyle \frac{d}{dx}\int_a^x f(t)\,dt=f(x)$

Part 2: $\displaystyle \int_a^b F'(t)\,dt=F(b)-F(a)$

Hyperbolic Functions
\begin{aligned} \sinh x&=\frac{e^x-e^{-x}}{2}\\ \cosh x&=\frac{e^x+e^{-x}}{2}\\ \cosh^2 x-\sinh^2 x&=1\\ \end{aligned}
\begin{aligned} \frac{d}{dx}\sinh x&=\cosh x\\ \frac{d}{dx}\cosh x&=\sinh x\\ \frac{d}{dx}\sinh^{-1}x&=\frac{1}{\sqrt{x^2+1}}\\ \frac{d}{dx}\cosh^{-1}x&=\frac{1}{\sqrt{x^2-1}} \end{aligned}
$\displaystyle \int \tanh(ax)\,dx=\frac{1}{a}\ln(\cosh(ax))+C.$

Uranium-Thorium Dating
Starting Equations:
$\displaystyle \begin{cases} \frac{dU}{dt}=-k_U U\implies U=U_0e^{-k_Ut}\\ \frac{dT}{dt}=k_UU-k_TT. \end{cases}$

$\frac{dT}{dt}+k_T T=k_U U_0e^{-k_Ut}$

$R=e^{\int k_T}=e^{k_T t}$.

$\displaystyle \boxed{T(t)=\frac{k_U}{k_T-k_U}U_0(e^{-k_Ut}-e^{-k_Tt})}$

$\displaystyle \boxed{\frac{T}{U}=\frac{k_U}{k_T-k_U}[1-e^{(k_U-k_T)t}]}$

## Trigonometric Identities and Differential Equations

Trigonometric Identities
\begin{aligned} \sin(A\pm B)&=\sin A\cos B\pm\cos A\sin B\\ \cos(A\pm B)&=\cos A\cos B\mp\sin A\sin B\\ \tan(A\pm B)&=\frac{\tan A\pm \tan B}{1\mp \tan A\tan B} \end{aligned}
\begin{aligned} \sin A+\sin B&=2\sin\frac{1}{2}(A+B)\cos\frac{1}{2}(A-B)\\ \sin A-\sin B&=2\cos\frac{1}{2}(A+B)\sin\frac{1}{2}(A-B)\\ \cos A+\cos B&=2\cos\frac{1}{2}(A+B)\cos\frac{1}{2}(A-B)\\ \cos A-\cos B&=-2\sin\frac{1}{2}(A+B)\sin\frac{1}{2}(A-B) \end{aligned}

Reduction to Separable Form
Equations of the form
$y'=g(\frac{y}{x}).$
Set $\frac{y}{x}=v$. Then $y=vx$ and $y'=v+xv'$. Thus the original equation becomes $v+xv'=g(v)$ which is separable.

Linear Change of Variable
$y'=f(ax+by+c)$ can be solved by setting $u=ax+by+c$.

Newton’s Law of Cooling
Rate of change of temperature of an object ($\frac{dT}{dt}$) is proportional to the difference between its own temperature ($T$) and the ambient temperature ($T_\text{env}$). That is, $\frac{dT}{dt}=-k(T-T_\text{env})$.

## Maths Skills to be a Doctor

Doctor and Lawyer are the top two favourite careers in Singapore. Do doctors need to use Maths? Read the below to find out.

Even if Maths is not directly needed, the logical thinking skills learnt in Mathematics will definitely be of great use. 🙂

I am not a medical doctor, but my two younger siblings are medical students, and the Mathematical knowledge and thinking skills have definitely helped them in their medical studies.

Functional numeracy is as essential to an aspiring medical professional as functional literacy. As a physician, perhaps the most important mathematical skills you will need are:

1. Basic mathematical knowledge sufficient to calculate drug doses, concentrations, etc.

2. An understanding of the core statistical concepts most commonly represented in the medical literature.

3. Knowledge of algebra to understand calculations of acid–base status, etc.

4. Ability to appreciate whether or not results are mathematically plausible.    (Nusbaum, 2006)

The careful logical reasoning that is necessary for the study of mathematics is an essential element of clinical reasoning. Although you do not need higher mathematics to get through medical school, you will need the ability to manipulate numbers, including fractions, ratios, powers of 10 and logarithms. You will also need a basic understanding of probability, graphs and simple algebra. You will need to rearrange equations and convert between units of measure.

It’s often unclear from your interactions with a doctor how much math she is using in order to treat you. While not all doctors have to use math as directly and frequently as engineers do, all of them must understand the complex mathematical equations that inform different medical treatments in order to administer treatments correctly.

## Dosages and Half-Life

One of the most common ways in which doctors use mathematics is in the determination of medicine prescriptions and dosages. Doctors not only have to use basic arithmetic to calculate what dosage of a particular drug will be effective for your height and body type over a specific period of time, they will also have to be aware of the medicine’s cycle through the body and how the dosage of one drug compares with the dosage of a similar type of drug. Sometimes doctors have to use calculus to figure out the right dosage of a drug. Calculus is the study of how changing variables affect a system. In the human body, the kidney processes medicine. However, people’s kidneys are at varying levels of health. Doctors can designate the kidney as a changing function in a calculus equation known as the Cockroft-Gault equation. This equation uses the level of creatine in a patient’s blood to find the level of the kidney’s functioning, which allows the doctor to determine the appropriate dose.

## Cancer Treatment

When a doctor administers radiation therapy to a cancer patient, the radiation beams have to cross each other at specific angles, so that they harm the cancerous tumor without harming the surrounding healthy tissue. The precise numbers for these angles must be calculated mathematically. Cancer tends to respond to any drug by mutating so that its DNA is no longer affected by that drug. Oncologists and medical scientists have decided to target cancerous tumors with many different kinds of drugs at once so that the cancer is unable to respond adequately. They use complex mathematical models that plot the speed and timing of the cancer’s different mutations to figure out what combinations and dosages of different drugs should be used.

## Medical Images and Tests

Doctors in medical imaging use two-dimensional images of a patient’s body taken from thousands of angles to create a three-dimensional image for analysis. Determining what angles should be used and how they will fit together requires mathematics. Medical researchers who study disease will analyze the mathematical dimensions of these images. Neurologists who run EEGs on patients to measure their brain waves must add and subtract different voltages and use Fourier transforms to filter out signal static. Fourier transforms are used to alter functions in calculus.

## Treatment Research

Medical scientists working with cardiologists use differential equations to describe blood flow dynamics. They also build sophisticated computer models to find the ideal size of an artificial aorta and where to place it in an infant pending a heart transplant. Doctors have to read medical journals to keep up on the latest scientific findings for the benefit of their patients. In addition to describing the calculus used to model health conditions, medical journal studies also make heavy use of statistics and probability to describe the health conditions of whole populations and the likelihood that different treatments will be effective.

## H2 Maths Tuition: Foot of Perpendicular (from point to plane) (Part II)

This is a continuation from H2 Maths Tuition: Foot of Perpendicular (from point to line) (Part I).

## Foot of Perpendicular (from point to plane)

From point (B) to Plane ( $p$)

## Equation (I):

Where does F lie?

F lies on the plane  $p$.

$\overrightarrow{\mathit{OF}}\cdot \mathbf{n}=d$

## Equation (II):

Perpendicular

$\overrightarrow{\mathit{BF}}=k\mathbf{n}$

$\overrightarrow{\mathit{OF}}-\overrightarrow{\mathit{OB}}=k\mathbf{n}$

$\overrightarrow{\mathit{OF}}=k\mathbf{n}+\overrightarrow{OB}$

## Final Step

Substitute Equation (II) into Equation (I) and solve for k.

## Example

[VJC 2010 P1Q8i]

The planes $\Pi _{1}$ and $\Pi _{2}$ have equations $\mathbf{r\cdot(i+j-k)}=6$ and $\mathbf{r\cdot(2i-4j+k)}=-12$ respectively. The point $A$  has position vector  $\mathbf{{9i-7j+5k}}$ .

(i) Find the position vector of the foot of perpendicular from  $A$ to $\Pi _{2}$ .

## Solution

Let the foot of perpendicular be F.

### Equation (I)

$\overrightarrow{\mathit{OF}}\cdot \left(\begin{matrix}2\\-4\\1\end{matrix}\right)=-12$

### Equation (II)

$\overrightarrow{\mathit{OF}}=k\left(\begin{matrix}2\\-4\\1\end{matrix}\right)+\left(\begin{matrix}9\\-7\\5\end{matrix}\right)=\left(\begin{matrix}2k+9\\-4k-7\\k+5\end{matrix}\right)$

Subst. (II) into (I)

$2(2k+9)-4(-4k-7)+(k+5)=-12$

Solve for k,  $k=-3$ .

$\overrightarrow{\mathit{OF}}=\left(\begin{matrix}3\\5\\2\end{matrix}\right)$

## H2 Maths Tuition

If you are looking for Maths Tuition, contact Mr Wu at:

Email: mathtuition88@gmail.com