6.9
Applications
Equation for Radio-active Decay
Radio-active substances are used extensively to trace various chemical and biological reactions in living organisms. Accidents such as those at Cher- nobyl and Windscale (Sellafield) also provide ample opportunity to examine the effects of radio-active pollution. Here we derive the fundamental equa- tion for exponential decay.
Let R(t) Bq be the radiation at time t and R0 Bq the initial radiation.1
The rate of increase (dR
dt) of radio-active material will be negative, and
proportional to the amount of radio-active substance present :-
dR
dt ∝ −R or dR
dt = kR
where k is a negative constant, the decay constant, specific to the material. From our experience, or reference to the earlier example (equation (6.11)), we can remember that
d dte
kt= kekt
where k is a constant, so that if R = ekt and k is negative dR
dt = kR
so that the radiation at time t may be described by the function :
R = R0ekt (6.12)
where k is the decay constant and R0 is the original level of radio-activity.
Half-life
The half-life of a radio-active substance is defined to be the time taken for the level of radiation to fall by one half. Thus if thalf is the half-life we can
use equation (6.12) to give :-
R0ekthalf R0 = 1 2 ∴ ekthalf = 1/2 kthalf = ln(1/2) thalf = ln(1/2) k = −0.693/k
The half-life of polonium 210 is 138 days. What is its decay constant?
k = thalf/ − 0.693 = 138/ − 0.693 = −0.005
84 CHAPTER 6. DIFFERENTIAL CALCULUS
Linear Regression - The Method of Least Squares
Here, for completeness we derive the equations used in section (4.9.2) for fitting the best straight line through a set of data points.
Suppose we have a set of experimental values
(x1, y1), (x2, y2), . . . (xi, yi) . . . (xn, yn)
and we suspect that the two variables are linearly related (y = a + bx). Draw the line y= a + bx as shown (at present we don’t know the values of a and b - we just guess them!).
- 6 (((((((( (((((((( (((((((( (((((( ? 6 • • • • • • y x di= yi− (a + bxi) (xi, yi) di y = a + bx
We can define the quantities dias the distance of the experimental points
from the line (in the y direction) i.e. the difference between the experimental value of y (yi) and the value of y predicted by the straight line (a + bxi).
di = yi− (a + bxi)
A measure of the goodness of fit can be obtained by adding the squares of all such di. ∴ G = n X i=1 di2
We use the squares of the differences in order that the contribution from each difference is positive and therefore G will always be positive. If G is
6.9. APPLICATIONS 85 large the fit is bad, if G is small the fit is good.
G = n X i=1 [yi− (a + bxi)]2 = n X i=1 [yi− a − bxi]2 = n X i=1 [yi2+ a2+ b2xi2− 2yia − 2yibxi+ 2abxi] = n X i=1 yi2+ na2+ b2Xx2i − 2aXyi− 2b X xiyi+ 2ab X xi
Note the term na2 which is the result of adding n copies of a2
We can minimise G in order to obtain the best fit:
i) by changing a, keeping b constant (moving the line up or down) ii) by changing b, keeping a constant (rotating the line)
∂G
∂a = 2na − 2Σy + 2bΣx
at minimum na − Σy + bΣx = 0 (6.13)
∂G
∂b = 2bΣx
2− 2Σxy + 2aΣx
at minimum bΣx2− Σxy + aΣx = 0 (6.14) from (6.13) a = Σy − bΣx
n (6.15)
substitute a from (6.15) into (6.14)
bΣx2− Σxy +(Σy − bΣx)Σx n = 0 nbΣx2− nΣxy + ΣyΣx − bΣxΣx = 0 b(nΣx2− ΣxΣx) = nΣxy − ΣxΣy b = nΣxy − ΣxΣy nΣx2− (Σx)2 (6.16)
Thus we calculate b from equation (6.16) and then use its value in equa-
86 CHAPTER 6. DIFFERENTIAL CALCULUS
Cylinder of Minimum Surface Area
A hot water tank consists of a closed cylinder of height H and radius R. In order to minimise heat loss, and hence save the planet and our heating bill, we would like the tank to have minimum surface area. If the volume is fixed, what ratio of H to R will give the mimimum total surface area? The same principles apply to microbial cells or cylindrical cows!
V = πR2H = constant (6.17) A = 2πRH + 2πR2 (6.18) from equation (6.17) H = V πR2 Substitute in equation (6.18) A = 2πR V πR2 + 2πR2 = 2V R + 2πR 2 dA dR = − 2V R2 + 4πR At minimum dA dR = 0, hence 4πR = 2V R2
Replace V eq. (6.17) to give 4πR = 2πR2H
R2
and hence for minimum area H
R = 2
Exercises
1. Find the following derivatives. (a) dtd sin t (b) dtd cos2t (c) dtd sin(t2) (d) d dz3zez (e) dxd ln(3x2)
2. The milk yield of the cows in a herd has been found to follow a curve of the type
y = Ate−Bt
where y = yield in litres/day and t is the time in days from the start of lactation. A specific animal was recorded as giving 15.9 litres on day 10 and 24.7 litres on day 20.
What will be the milk yield for this animal at day 100?
6.9. APPLICATIONS 87 3. (a) From first principles prove that dxd x2 = 2x
(b) A crop is observed to be infected with a form of rust. A survey shows that the number of plants infected at present is 3 per sq. metre. The disease is known to spread according to the equation
dN dt = 2
√ N
where N is the number of infected plants per sq. metre and t is the time in days. When the infected rate reaches 100/m2 it is felt
that the crop cannot be saved. How many days do we have in which to effect a cure? [Hint:- the two parts of the question are related].
4. Find the derivative of each of the following (a) x3/2
(b) x2e3x
(c) sin 3x (d) ln 2x
(e) tan x(= sin x cos x)
(f) ex sin x
5. Find the following derivatives (a) dxd (a + bx + cx2) (b) dtd(sin wt) (c) dxd (sin xx ) (d) d d sin x(sin2x) (e) dxd (3x2+ x) sin(2x) (f) dxd ecos(ln(x))
6. The specific weight of water at toC is given by
w = 1 + (5.3 × 10−5)t − (6.53 × 10−6)t2+ (1.4 × 10−8)t3 Find the temperature at which water has maximum specific weight. 7. An animal is fed on a diet in which the concentration (F) of an added
factor is varied. It is found that the daily consumption (D) of the diet is related to the concentration of added factor as follows:-
D = A − aF
where A is the consumption when no factor is added and a is a con- stant. What value of F should be chosen so that the animal consumes the maximum amount of the factor?
88 CHAPTER 6. DIFFERENTIAL CALCULUS Answers 1. (a) cos t (b) −2 sin t cos t (c) 2t cos t2 (d) 3zez+ 3ez= 3(z + 1)ez (e) 2/x
2. A = 2.04704,B = 0.02527,16.36 litres,29.8 litres on day 39 3. (a) (b) 8.27 days 4. (a) 32x1/2 (b) e3x(2x + 3x2) (c) 3 cos 3x (d) x1 (e) cos12x
(f) (sinx + x cos x)ex sin x
5. (a) b + 2cx (b) w cos wt (c) cos xx −sin xx2
(d) 2 sin x
(e) 2(3x2+ x) cos(2x) + (6x + 1) sin(2x)
(f) −1xsin(ln(x))ecos(ln(x)) 6. 4.113oC
Chapter 7
Integral Calculus
7.1
Introduction
Integration (anti-differentiation) is the inverse of differentiation. We begin with an expression defining the rate of change (growth rate, heat flux, rate of infection, glucose in and glucose out . . . ) and use it to define the state of the system (size, temperature, magnitude of the epidemic, amount of glucose in pool . . . ).
If we have two functions y(x) and g(x) so that
d
dxy(x) = g(x)
i.e. the derivative of y(x) is g(x), then y(x) is the integral of g(x), which we write as follows:
if d
dxy(x) = g(x) then
Z
g dx = y + c (7.1)
Here c is an arbitrary constant, which must be included because the deriva- tive of a constant is zero. Equation (7.1) is the basis of integral calculus. An integral, such as this, in which there is an unknown constant is known as an indefinite integral. The value of c can usually be found because there will be a boundary condition at which the values of both x and y will be known so that substituting these values will allow the calculation of c. Here are some examples which you have seen in the previous chapter.
d dxx 2 = 2x ∴ Z 2x dx = x2+ c and d dxe x= ex ∴ Z exdx = ex+ c and d dx sin x = cos x ∴ Z cos x dx = sin x + c 89
90 CHAPTER 7. INTEGRAL CALCULUS
where c is an unknown constant.
If, at this point you say to yourself “So what?”, I for one have sympathy with you - all we have gained so far is another definition, but we don’t understand what it is all about. Accepting things because somebody says so is hardly science, so let us try to see why and how integration works.
First of all let us see if we can make sense of the notation: Z
y dx
which is read as “the integral (R) of the function (y) with respect to x (dx)”. Here x is an independent variable upon which the variable y depends. The above expression represents the sum of all the values of y dx over the range of x. For example if y = 1, R1 dx = R dx = x which is the sum of all
the infinitely small bits of x, which of course will be equal to x. Similarly R
dy = y orR d anything = anything.