chapter4

# T kn dt dn solution is ntae kt where an0 ae 6k n6 2n0

This preview shows pages 4–9. Sign up to view the full content.

) ( t kN dt dN = Solution is N(t)=Ae kt , where A=N(0) Ae 6k =N(6) =2N(0) = 2A or e 6k =2 or k = 6 1 ln 2 Find t when N(t)=3A=3N(0) or N(0) e kt =3N(0) or t e ) 2 (ln 6 1 3 = or ln 3= 6 ) 2 (ln t 81

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
or t= 2 ln 3 ln 6 9.6 years (approximately 9 years 6 months) Example 4.3 Let population of country be decreasing at the rate proportional to its population. If the population has decreased to 25% in 10 years, how long will it take to be half? Solution: This phenomenon can be modeled by ) t ( kN dt dN = Its solution is N(t)=N(0) e kt , where N(0) in the initial population For t=10, N(10)= 4 1 N(0) 4 1 N(0) = N(0) e 10k or e 10k = 4 1 or k= 10 1 ln 4 1 Set N(t)= 4 1 N(0) ) 0 ( N 2 1 e ) 0 ( N t 4 1 ln 10 1 = or t= 4 1 ln 10 1 2 1 ln 8.3 years approximately. Example 4.4 Let N(t) be the population at time t and Let N 0 denote the initial population, that is, N(0)=N 0 . 82
Find the solution of the model 2 ) ( ) ( t bN t aN dt dN - = with initial condition N(0)=N o Solution: This is a separable differential equation, and its solution is t ds ds ) s ( bN ) s ( aN ) s ( dN t 0 2 t 0 = = - bN a B N A ) bN a ( N 1 bN aN 1 2 - + = - = - To find A and B, observe that ) ( ) ( ) ( ) ( bN a N N bA B Aa bN a N BN bN a A bN a B N A - - + = - + - = - + Therefore, Aa+(B-bA)N=1. Since this equation is true for all values of N, we see that Aa=1 and B-bA=0. Consequently, A= a 1 , B=b/a, and ) bs a ( s ds N N 0 - = ds ) bs a b s 1 ( a 1 N N o - + - - + = bN a o bN a o N N a ln ln 1 | bN a bN a | N N ln a 1 o o - - Thus 83

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
at = ln bN a bN a N N 0 0 - - It can be verified that ) t ( bN a bN a o - - is always positive for 0<t<∞. Hence at = ln bN a bNo a N N o - - Taking exponentials of both sides of this equation gives e at = bN a o bN a o N N - - N 0 (a-bN)e at = (a-bN 0 )N Bringing all terms involving N to the left-hand side of this equation, we see that [a-bN o + bN 0 e at ] N(t) = aN 0 e at or N(t)= at e o bN o bN a at e o aN + - 4.3 Radio-active Decay and Carbon Dating As discussed in Section 1.4.2. a radioactive substance decomposes at a rate proportional to its mass. This rate is called the decay rate. If m(t) represents the mass of a substance at any time, then the decay rate dt dm is proportional to m(t). Let us recall that the half-life of a substance is the amount of time for it to decay to one-half of its initial mass. 84
Example 4.5. A radioactive isotope has an initial mass 200mg, which two years later is 50mg. Find the expression for the amount of the isotope remaining at any time. What is its half-life? Solution: Let m be the mass of the isotope remaining after t years, and let -k be the constant of proportionality. Then the rate of decomposition is modeled by dt dm = - km, where minus sign indicates that the mass is decreasing. It is a separable equation. Separating the variables, integrating, and adding a constant in the form lnc, we get lnm+lnc = - kt Simplifying, lnmc = - kt (4.3) or mc = e -kt or m = c 1 e -kt , where c 1 = c 1 To find c 1 , recall that m =200 when t=0. Putting these values of m and t in (4.3) we get 200 = c 1 e -ko = c 1 .1 or c 1 =200 and m = 200e -kt (4.4) The value of k may now be determined from (4.4) by substituting t=2, m=150.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern