(a) Find the function that describes the fraction of light bulbs failing by time t e (0.1)x dx = [ e (0.1)x ] t 0 = 1 e (0.1)t.

Transcription

1 1 M 13-Lecture March 8, 216 Contents: 1) Differential Equations 2) Unlimited Population Growth 3) Terminal velocity and stea states Voluntary Quiz: The probability density function of a liht bulb failin at time t, t <, is iven by p(t) = 1 1 e (.1)t, where t is in days. (a) Find the function that describes the fraction of liht bulbs failin by time t. (b) After how many days will 8% percent of the LSK s liht bulbs stop workin? Sol: For part a), this is iven by the cummulative function, F (t). Hence our function is F (t) = p(x)dx = 1 1 e (.1)x dx = [ e (.1)x ] t = 1 e (.1)t. For b), usin the probability density we must find t so that p(x)dx =.8. In terms of the cummulative functions this equally means F (t) =.8 or 1 e (.1)t =.8 e (.1)t =.2 t 1 = ln(2) t = 1 ln(2). So after approximately 16 days 8% percent of the LSK s liht bulbs stop workin. Differential Equations A differential equation is a relationship between some (unknown) function and one of its derivatives. In this course, since we are concerned only with functions that depend on a sinle variable, we discuss March 8, 216

2 2 ordinary differential equations (ODEs), whereas later, after a multivariate calculus course where partial derivatives are introduced, a wider class, of partial differential equations (PDEs) can be studied. Such equations are encountered in many areas of science, and in any quantitative analysis of systems where rates of chane are linked to the state of the system. Most laws of physics are of this form; for example, applyin the familiar Newtons law, F = ma, links the position of a pendulums mass to its acceleration (second derivative of position). Many bioloical processes are also described by differential equations. The rate of rowth of a population dn/ depends on the size of that population at the iven time N(t). Unlimited Population Growth We consider a population with per capita birth and mortality rates that are constant, irrespective of ae, disease, environmental chanes, or other effects. Let y(t) represent the size of a population at time t. We will assume that at time t =, the population level is specified, i.e. y() = y is some iven constant. We want to find the population at later times, iven information about birth and mortality rates, (both of which are here assumed to be constant over time). The population chanes throuh births and mortality. Suppose that b > is the per capita averae birth rate, and m > the per capita averae mortality rate. The assumption that b, m are both constants is a simplification that nelects many bioloical effects, but will be used for simplicity in this first example. The statement that the population increases throuh births and decreases due to mortality, can be restated as: rate of chane of y = rate of chane of birth - rate of chane of mortality. The rate of births is iven by the product of the per capita averae birth rate b and the population size y. Similarly, the rate of mortality is iven by my. Translatin the rate of chane into the correspondin derivative of y leads to = by my = (b m)y. Since b m is a constant, we see that these simple models of unlimited rowth are of the form = ky, y() = y for some constant k. The condition y() = y is called the initial condition and the problem to be solved is referred to as the intial value problem. March 8, 216

3 3 Challene Question: If the cases k =,k > and k <? = ky models the rate of chane of a population size, how can you interpret Separation of Variables Notice that indeed the expression = ky involves both our function of interest and its derivative. Our oal is to solve for y usin our interaiton techniques. To this end, we consider and as differentials (as we discussed in lecture notes from February 9) so that we can describe the relation ship equally as = ky. y = k. Notice that we have put all terms involvin the dependent variable y on the left hand side and all terms involvin the independent variable t on the riht hand side. For this reason, this approach is known as separation of variables. We may now interate both sides of the equality. Notice that if we interate t from to some final time T, then y(t) will be interated from y() = y to so that Interatin the left hand side we obtain y y y = [ln y ]y(t ) y Interatin the riht hand side we obtain so that ln y T y = k. = ln ln y = ln k = [kt] T = kt, y = kt y = e kt = y e kt. This computation work for any time T, so we may use our independent variable aain and have obtained y(t) = y e kt.. March 8, 216

4 4 Challene Question: Now that we have the solution, can you justify the name Unlimited Population Growth? Challene Question: How do the cases k =,k > and k < chane the behaviour of the population? Are there any unrealistic situations? So in this situation, we must be iven the initial condition y() = y. Notice that indeed our obtained y(t) satisfies our differential equation since = d (y e kt ) = ky e kt = ky(t), and furthermore the initial condition is also satisfied Terminal velocity and stea states y() = y e = y. We now consider a slihtly more complicated differential equation and its application to computin the velocity of a fallin object (with initial velocity ) due to the influence of ravity and friction. Recall that ealier in the course we considered similar computations. In this case, fallin objects accelerate accordin to the force of ravity so that if a(t) ives our acceleration at time t and is the raviatational force then a(t) =, and from before we know that such constant acceleration cases ive velocity function, v(t), as v(t) = t + v() = t. However, if we consider this model notice that as time increases indefinitely so will v(t). This is unrealistic because in reality frictional forces do not allow fallin objects to accelerate indefinitely. So we now consider a more realistic model. To account for frictional forces, we may consider the followin ma(t) = m γv(t) m = m γv(t) where γ is the frictional coefficient and m is the mass of the object. If the mass is constant, then we equally write = γ m v(t). March 8, 216

5 5 To simplify further, notice that γ m is a constant, so that these initial value problems are of the form Usin separation of variables we obtain = kv(t), v() =, v(t ) v() = kv(t) kv = kv = Usin substitution u = kv we obtain du = k so that the equality becomes kv(t ) kv(t ) 1 k ln 1 du k u = ln u kv(t ) kv(t ). = T = kt. = e kt kv(t ) = e kt v(t ) = k (1 e kt ). This is the solution to our posed initial value problem. It predicts the velocity of the fallin object throuh time. Note that as t increases, the term e kt decreases rapidly, so that the velocity approaches a constant whose value is which is known as the terminal velocity. v(t) k, Challene Question: Check that our obtained solution does satisfy the initial value problem. March 8, 216