Lecture 1. About Differential Calculus

by Reinaldo Baretti Machín

Reference :

1. Mathematical Techniques for Biology and Medicine (Dover Publications- Paperback)

by William Simon ; http://www.amazon.com/Mathematical-Techniques-Biology-Medicine-William/dp/0486652475/ref=sr_1_1?ie=UTF8&s=books&qid=1223731765&sr=8-1

Examples and problem from:

Chapter 1 . Reference 1. The problem numbers refer to the textbook listing.

Problem 15

Let R~rate of generation of a substance ~ moles/time

V~ volume ,

C ~ concentration ~ moles/volume

F ~ rate of a liquid flow ~volume/time

Verify that the differential equation

dC/dt= R/V –(F/V) C (1)

(DE) is dimensionally correct.

Substitute the dimensions and obtain

moles/(volume-time) ~ moles/(volume-time) –(volume/time)*(1/volume)*moles/volume)

****

2. Plot log 10 ( y =exp(α x) ) vs -2 ≤ x ≤ 2

for selected values - 2 ≤ α ≤ 2

A FORTRAN code was used

c plot log 10 (y=exp(alfa*x) )

data xi,xf, nstep / -2.,2. , 30/

data alfa /2.0/

dx=(xf-xi)/float(nstep)

do 10 i=0,nstep

x=xi+dx*float(i)

y=exp(alfa*x)

print 100 ,x, log10(y)

10 continue

100 format(1x,'x,log10(y)=',2(3x,e10.3))

stop

end

*****

Derivative of the exponential function.

Let y = A exp( f(x) ) ,where A is a constant ant f(x) is some

function of the independent variable x .

By definition exp(x) is an infinite series,

y=exp( f(x) ) = 1 + f(x) + f(x)²/2 + … f(x)ⁿ /n!

dy/dx= 0 + df/dx + f(x) (df/dx) + 3 (f²/3!)(df/dx) +

n (f^n-1/n!)(df/dx)

= (1 + f(x) + f(x)²/2 + … f(x)ⁿ /n! ) (df/dx)

Hence

d exp( f(x)) /dx = exp( f(x) ) (df/dx)

Let y₁ = A exp( - λt ) be a decaying exponential while

y₂ = B exp(+ λt ) is agrowing exponential. The constant

λ has dimension of ~ 1 /time . For example .25 /sec ,

.5/hour , .02 /year etc/

Their first derivatives are

dy₁/dt = - λ A exp( - λt ) = - λ y₁ and

dy₂/dt = + λ A exp( + λt ) = λ y₂ .

Example of exponential growth

Suppose the world population grows at the rate of

.02/year. This means that if ∆t= 1 year

∆y/(y∆t) ={ y(t+1 year) –y(t) }/[∆t y(t)]

= .02 /year ≡ λ.

In the limit dy/dt = λ y

Let y(t) = A exp (.02 t) with A=1.E9 at t=0 ( the year zero is the 1900).

Plot of y vs t .(MATLAB code)

%population growth

lambda=.02 ; A=1.e9;

t=[0:5:110];

y=A*exp(lambda*t);

year=t+1900;

plot(year,y),xlabel('year'),ylabel('Population')

The doubling time is a function of λ.

Writing y₂ / y₁ =2= A exp(λ t₂ ) / A exp(λ t₁ ) = exp(λ( t₂ – t₁))

Take the natural log and get

ln(2) = λ( t₂ – t₁) or (t₂ – t₁)_doubling = 0.693 / λ

In the present example there is a time scale

T_scale = 1/ λ= (.02 /year )^-1 = 50 years .

The world population doubles every .693(50) ≈ 35 years .

Example of exponential decay

Let y(t) = A exp( -λt) , λ ~1/time , T_scale=1/ λ.

The half life is the interval ∆t_1/2 for which y diminishes to half the y value at the origin of the chosen interval.

Let y(0) = A , y (t_1/2) = A/2 = A exp(-λt_1/2 ) .Thus

exp(((-λt_1/2) =1/2 . Take the natural log

-λt_1/2 = ln(1/2) = - ln(2) .

The half life is t_1/2 = .693/λ = .693 T_scale , which is the same expression for the doubling time in the case of exponential growth.

Problem 11 ( Ref. 1)

The concentration of a substance decays as shown in the table.

Assume y = A exp(-λt ) and find A , λ and the time scale.

time (sec)	y(conc)	ln(y)
1	0.0096	-4.64
2	0.0048	-5.34
3	0.0024	-6.03
4	0.0012	-6.72

From inspection it is clear that the half life is one second. Thus

λ = .693 /t_1/2 = 0.693 /sec . Also y(1) =.0096 = A exp(-.693)

thus A = 1.92E-2 . A will have the dimensions of A which are not specified here.

Now we treat the same problem by fitting the data with a least square software.

The following plot shows the natural log of y vs time (t).

ln y = ln A – λt . This equation is a straight line of the form

say , u = b – m x . The slope is – m = - λ and the vertical axis intercept

b= ln A.

The line should be called u = -3.95 -.693 x .The y in the plot IS NOT or y.

Anyway we have that m= λ=0.693/ second and ln (A)=-3.95 ,

A= exp(-3.95)=1.92E-2 . The half life t_1/2 = 0.693/ λ=1.0 second

Plot of y=1.92E-2 exp( -.693 t)

Matlab code

% exponential decay

lambda=.693 ; A=1.92E-2;

t=[0:.5:5];

y=A*exp(-lambda*t);

plot(t,y),xlabel('t~sec'),ylabel('Y')

*****

Prob 3 (ref 1)

Given U(x) = W(x) V(x) ; V(x) = S(x) W(x)

W(x=2)=3 ; V(2) = 4 ; ( dW/dx)_x=2=4 ; (dV/dx)_x=2= 2

find dU/dx and dS/dx at x=2 .

Answer:

dU/dx= (dW/dx) V + W (dV/dx) ;using the given values ,

= (4)(4) +(3)(2) = 22 .

dV/dx = (dS/dx)W + S (dW/dx) =(dS/dx)W + (V/W)*(dW/dx)

2= (dS/dx) (3) +(4/3)4

dS/dx = -10/9

Problem 4. Let y(3)=4 , y(3.1)=4.3 what is ∆y/∆x ?

dy/dx ≈ ∆y/∆x = (4.3-4)/(3.1-3) = .3/.1 = 3

No dimensions are given here for y or x but in general they have different units and the slope ∆y/∆x will have units . For if instead of x we have time

so that ∆x → ∆t ~ sec or minutes or hours and ∆y ~ units of length

then ∆y/∆x = 3 ( units of length/unit of time )

Problem 5. Given dy(3)/dx= 2 and y(3)=12 find an approximate value

for y(3.1). Let ∆x = 3.1-3 = 0.1 and suppose the derivative does not change appreciably in the range 3.0 ≤ x≤ 3.1 .

Then ∆y/∆x ≈ dy/dx= 2

y₂ - y₁ = 2 (∆x) , y₂ = y₁ + 2 (∆x) =12 +2(.1)=12.2 ,

that is y(3.1) = y₂ = 12.2

Problem 6. Given V= g(u) and u =f(x) , by the chain rule find

dV/dx . dV/dx= (dg/du) (du/dx) or alternatively

=(dg/du) df/dx

Problem 7. Let f(x)= x⁴+2 , g(u) =u² and apply it to prob 6.

df/dx= 4 x³ , dg/du = 2 u = 2f(x)= 2x⁴+4 . Hence

dV/dx= (2x⁴+4)( 4 x³ ) = 8x⁷ + 16x³

Problem 8 & 9 are examples of exponential growth.

d C(t)/dt = k C(t) is the same kind equation as dy/dt = k y.

So the answer is C(t) = A exp (kt) The constant A is found from an intial value.

First case : given C(t=0) ≡ C₀ =2moles/liter= A exp( 0) =A

So the solution is C(t) = (2 moles/liter) * exp( kt) .

Second case: the concentration is specified at t=1 sec.

C(t=1) = A exp(k (1) ) = 2moles/liter

A = 2 exp(-k) ~ moles/liter

So C(t) = 2 exp(-k) * exp( kt) . If we call the intial instant t₁ , where t₁ is arbitrary then a general expression would be, C(t) = 2 exp( k (t- t₁ ) ).

Prob 13. Given y = A exp(-kt) and y(4) = (1/2) y(2) , find k.

This tells us that the half life is 2 seconds. From what is explained in the previous problems or examples ( see above)

k ≡ λ = .693 /t_1/2 = .693 /(2) = 0.346 /sec

Prob 11. In the following expressions what must be the dimensions of k:

a) exp( -kt² ) , t ~ seconds ; implies k ~ 1/sec²

b) C=( F/k) t , k ~F t/C ~ (liters/sec)* (sec)*(liters/mol)~liters²/mol

c)R=(F+K) exp(-F *t /V ) given F~liter/sec , K must have the same dimensions ~ liters/sec and of course R ~ liters/sec

Prob 16. Given the second order differential equation (DE)

d²C/dt ²+ B dC/dt +D C = E and

C ~ concentration say ~ moles/liter . What are the dimensions

of each term in the DE.

A derivative d/dt ~ 1/time , d² /dt² ~ 1/time²

d²C/dt ² ~ moles/(liter-time²)

B ~ { moles/(liter-time²)}/ dC/dt ~{ moles/(liter-time²)}*(liter-time/moles)

~1/time

D ~ 1/time² , E ~ moles/(liter-time²)

In a problem with a given data we can associate particular time scales related to the coefficients B and D ; T_B ≡ (1/B) ; T_D = (1/D)^1/2 .

Prob 17. Find the derivatives of the following.

a) y = x⁴ + 3 x³ + x +2

Analytically, and by hand, this is quite a straight forward. The rule to apply is d xⁿ /dx = n x^n-1.

dy/dx = 4 x³ + 9 x² + 1 + 0 .

Using MATLAB is also quite easy.

syms x y;

y=x^4 +3*x^3 +x + 2 ;

diff(y,x)

ans =

4*x^3+9*x^2+1

These are analytical answers . A specific value of x _p can be inserted in the equations to find the numerical values of the derivative at that point.

For example at x= 1 , dy/dx= 14 . No units or dimensions have been specified in this purely mathematical example.

Find the second derivative of y = x⁴ + 3 x³ + x +2. We have already taken the first derivative dy/dx= 4 x³ + 9 x² + 1 . The second derivative is

d²y/dx² = 12 x² + 18x . Or using Matlab

syms x y;

y=x^4 +3*x^3 +x +2;

dydx=diff(y,x),

d2ydx=diff(y,x,2)

d3ydx=diff(y,x,3)

Answers:

dydx = 4*x^3+9*x^2+1

d2ydx =12*x^2+18*x

d3ydx =24*x+18

17. (b) find the derivative of y =(x³ –x) /(x³ +x) ≡ u(x) v(x)^-1

where u(x) = (x³ –x) and v(x) = (x³ +x)

dy/dx = du/dx * v^-1 + u d u^-1 /x

=(3x² -1) v^-1 + u (-1) v^-2 dv/dx

=( 3x² -1) v^-1 – ( u/v²) (3x² +1) .

Or using MATLAB

syms x y;

y=(x^3-x)/(x^3+x);

dydx=diff(y,x)

dydx =

(3*x^2-1)/(x^3+x)-(x^3-x)/(x^3+x)^2*(3*x^2+1)

17. (c) y= A exp(x² )

dy/dx = A d exp[exp(x² )] /dx = A exp (x²) d (x²) /dx

= 2x A exp(x²)

18 (d) y = A x exp(x²)

Using MATLAB

syms x y A;

y=A*x*exp(x^2) ;

dydx=diff(y,x)

dydx =

A*exp(x^2)+2*A*x^2*exp(x^2)

Let in 18 (d) A =2 .By numerical means calculate an approximation to dy/dx in the range 0 ≤ x ≤ 2 and compare with the exact values

obtained from dydx= A*exp(x² )+2Ax² exp(x² ) .

Let ∆x =1.e-3

and approximate dy/dx by {y(x+∆x) – y(x)} /∆x.

The last column is the exact value of the derivative.

x,deltay/deltax , dy/dx= 0.0000E+00 0.2000E+01 0.2000E+01

x,deltay/deltax , dy/dx= 0.2000E+00 0.2249E+01 0.2248E+01

x,deltay/deltax , dy/dx= 0.4000E+00 0.3101E+01 0.3098E+01

x,deltay/deltax , dy/dx= 0.6000E+00 0.4937E+01 0.4931E+01

x,deltay/deltax , dy/dx= 0.8000E+00 0.8661E+01 0.8648E+01

x,deltay/deltax , dy/dx= 0.1000E+01 0.1634E+02 0.1631E+02

x,deltay/deltax , dy/dx= 0.1200E+01 0.3281E+02 0.3275E+02

x,deltay/deltax , dy/dx= 0.1400E+01 0.7000E+02 0.6986E+02

x,deltay/deltax , dy/dx= 0.1600E+01 0.1587E+03 0.1583E+03

x,deltay/deltax , dy/dx= 0.1800E+01 0.3829E+03 0.3820E+03

x,deltay/deltax , dy/dx= 0.2000E+01 0.9851E+03 0.9828E+03

Fortran code for the derivative

data A ,xi ,xf ,nstep/2.,0.,2., 10/

data deltax/1.e-3/

y(x) =a*x*exp(x**2)

dydx(x)=A*exp(x**2 )+2.*A*x**2*exp(x**2)

dx=(xf-xi)/float(nstep)

do 10 i=0,nstep

x=xi+dx*float(i)

print 100 ,x, (y(x+deltax)-y(x))/deltax ,dydx(x)

10 continue

100 format(1x,'x,deltay/deltax , dy/dx=', 3(3x,e11.4))

stop

end

END OF CHAPTER 1.