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

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

View Full Document Right Arrow Icon
) ( 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
Image of page 4

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

View Full Document Right Arrow Icon
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
Image of page 5
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
Image of page 6

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

View Full Document Right Arrow Icon
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
Image of page 7
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.
Image of page 8

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

View Full Document Right Arrow Icon
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern