Chapter 3: Elementary Applications MATH 374 Lecture 8 Chapter 3: Elementary Applications

3.1 Velocity of Escape From the Earth Find the minimum velocity needed for a particle to escape the earth, assuming motion in a radial direction. Also assume the only force on the particle is gravity. r R

Newton’s Law of Gravitation Recall Newton’s Law of Gravitation, where a is the acceleration of the particle, r is the distance from the particle to the center of the earth, and k is a proportionality constant.

Finding a value for k Now, when r = R, a = -g = 32 ft/sec2, so (1) tells us: Recalling that a = dv/dt and v = dr/dt, the chain rule, along with (2) and (1) give: Hence,

Solving Equation (3) Equation (3), which relates velocity v to distance r can be solved using separation of variables! One finds that Letting v = v0 when r = R, (4) tells us and thus, the particular solution to (3) with initial condition v = v0 when r = R is

Escape Velocity Notice that for the RHS of (5) to remain positive as r ! 1, vo2 – 2 g R  0 must hold., i.e. We call (6) the escape velocity of a planet with radius R and surface acceleration due to gravity, g.

3.2: Simple Chemical Conversion In certain chemical reactions, such as radioactive decay, the time rate of change of the amount x of substance A is proportional to x. Mathematically, with k a proportionality constant. If at time t = 0, x = x0, we find the particular solution to (1) is for any t. B A A x units of A at time t

Example 1 The half life of a radioactive element is the amount of time it takes for a given amount of the element to decay to half the given amount. Radioactive carbon 14 (14C) has a half life of 5730 years. Find the decay constant k for 14C.

Example 1 (continued) Solution: If at t = 0, there is x0 14C, then in 5730 years there will be ½ x0 14C. Thus, (2) implies Note that the units for k are yr-1.

Example 2 A piece of parchment is found that has 80% of the 14C found in living wood. Assuming that the amount of 14C in living matter is constant and starts to decay after death, estimate the age of the parchment.

Example 2 (continued) Solution: Using (2) and the result from Example 1, we have

3.3: Logistic Growth Suppose we wish to model how a population grows. Model 1: Assume the growth rate is proportional to the size of the population. Then where x = x(t) is the size of the population at time t and k is a proportionality constant. As in section 3.2, the solution to (1) is where x0 = x(0) is the initial population amount. In this model, the population grows exponentially.

A Better Model Model 2: Assume the average birth rate per individual is a positive constant and the average death rate per individual is proportional to the size of the population. Then the ODE with constants a > 0 and b > 0 is a model for the size of the population x(t) at time t. Equation (3) is called the logistic equation.

Solving the Logistic Equation Separation of variables leads to the general solution of (3): and if x(0) = x0, we find that

Solving the Logistic Equation Solving (4) for x, From (5) it is clear that as t ! 1, x ! b/a. Using (3) and the fact that solutions of (3) cannot cross (Theorem 1.1, Existence and Uniqueness), one can show that the behavior of solutions of (3) is determined by x(0) = x0.