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### MATH LECTURE OCT.27, 2011

Course: MATH 16A, Fall 2011
School: Berkeley
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LECTURE MATH OCT.27,2011 Today: Properties of the Natural Logarithm Chapter 4 Review Exponential Growth and Decay Ln(x) Previously we introduced the natural log, ln(x) as the inverse of the exponential function and established some of its properties By definition, ln(x)=y exactly when e^y=x The doman of ln(x) is (o, infinity) D[ln(x)]/dx=1/x By the Chain Rule, d[ln(x)]/dx=g(x)/g(x) We also drew the graph...

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LECTURE MATH OCT.27,2011 Today: Properties of the Natural Logarithm Chapter 4 Review Exponential Growth and Decay Ln(x) Previously we introduced the natural log, ln(x) as the inverse of the exponential function and established some of its properties By definition, ln(x)=y exactly when e^y=x The doman of ln(x) is (o, infinity) D[ln(x)]/dx=1/x By the Chain Rule, d[ln(x)]/dx=g(x)/g(x) We also drew the graph y=ln(x) by reflecting the graph y=e^x through the line y=x. Now we sketch the curve y=ln(x) directly Ln(x) has domain (0, infinity) and the following limits Lim x0=-infinity Lim xinfinity ln(x)=infinity Now, solve for ln(x)=0 Ln(x)=0 X=e^0 X=1 Further, F(x) is <0 if x<1 >0 if 1<x Ln(x) has domain (0, infininty), a vertical asymptote at x=0 and infinite limit as x inifinity Ln(x) has one x-intercept at x=1 Ln(x) is negative in (0,1) and positive in (1, infinity) Ln(x), (ln(x))=1/x Since (ln(x))=1/x and ln(x) has domain (0, infinity) and (ln(x)) is positive on the domain of ln(x) ln(x) is increasing on its domain (0, infinity) ln(x), (ln(x)) = 1/x, (ln(x)) = 1/x2 Since (ln(x)) = 1/x2 and ln(x) has domain (0,), (ln(x)) is negative on the domain of ln(x). ? ln(x) is concave down on its domain (0,). Summary of f(x)=ln(x) ? . ? ? ln(x ) has domain (0, ), a vertical asymptote at x = 0 and infinite limit as x ln(x ) has one x -intercept at x = 1. ln(x ) is negative in (0, 1) and positive in (1, ). ? f(x)=1/x ? ln(x) is increasing on its domain (0,). ? f(x)=1/x2 ? ln(x) is concave down on its domain (0,) Example: Differentiate f(x)=ln(x) Note, that x is positive for all lxl cannot equal 0, so the domain of ln(lxl) is the set of x such that x cannot equal 0. Since ln(lxl) is not defined for x=0, ln(lxl) is not differentiable at x=0. Otherwise, Lxl= -x, if x<0 X, if 0<x And we can apply the chain rule for each case 1. When x<0, d[ln(lxl)]/dx=d[ln(-x)]/dx=1/-x(-x)=-1/-x=1/x 2. When x>0, d[ln(lxl)]/dx=d[ln(x)]=1/x Laws of Logarithms: The Laws of Exponentiation that we have already isolated can be translated into the language of logarithms Ln(ab)=ln(a)+ln(b) Ln(1/a)=-ln(a) Ln(a/b)=ln(a)-ln(b) Ln(a^b)=blna Reason: 1. ln(ab)=ln(a)+ln(b) e^ln(ab) =ab =e^ln(a)e^ln(b) =e^ln(a)+ln(b) Since e^x is increasing, ln(ab)=ln(a)+ln(b) 2. ln(1/a)=-ln(a) e^ln(1/a) =1/a =1/e^ln(a) =e^-ln(a) So, ln(1/a)=-ln(a) 3. ln(a/b)=ln(a)-ln(b) Follows from the previous two 4. ln(a^b)=blna ln(a^b) =ln(e^ln(a))^b) =ln(e^bln(a)) =bln(a) Example: 1. Simplify the expression eln(x^2)+3ln(y) eln(x^2)+3ln(y) =(eln(x^2))(e3ln(y)) =x2(eln(y))3 =x2y3 2. Differentiate f(x)=ln(x(x+2)(x2+1)) One could differentiate immediately with the chain rule and iterated applications of the product rule OR With foresight, we can use the properties of the log and differentiate Ln(x(x+2)(x2+1))=ln(x)+ln(x+2)+ln(x2+1) (ln(x(x+2)(x2+1)) =(ln(x)=ln(x+2)+ln(x2+1)) =1/x+(1/x+2)(x+2)+(1/(x2+1))(x2+1) =1/x+1/(x+2)+(2x/(x2+1)) The Power General Rule When we first introduced the power rule, we checked it for exponents 2, 3, and indicated how to verify it for roots and powers. We have been using it for all powers ever since. Now, we can use the logarithm to verify that (xr)=rxr-1 (Xr) =(eln(x^r)) =(eln(x^r))(ln(xr)) =xr(rln(x)) =xr(r(1/x)) =rxr-1 Chapter 4 Recap: Topics of Chapter 4: Exponentiation o Laws of exponents o Exponential Functions The number e and the function ex Differentiation of ex o The differential equation y=ky The natural logarithm function ln(x) Differentiation of ln(x) Laws of logarithms CHAPTER 5 EXPONENTIAL GROWTH Many applications involve solving the differential equation Y=ky For k a constant We have shown that the solutions to this equation are the functions Y=Cekx For constant C Notation. ? When k > 0, y = Cekx describes exponential growth and k is called the growth constant. ? When k < 0, y = Cekx describes exponential decay and k is called the decay constant. Example The number of bacteria cells, N(t), increases at a rate proportional to N(t). Suppose that N(0) = N0 and solve for N(t). Solution. N(t) satisfies the equation N(t) = kN(t). So, N(t) = Cekt N (0) = Cek0=CN (0) = N0 given initial condition C = N0 N(t) = N0ekt Example: The amount of a radioactive isotope, A(t), decreases by radioactive decay at a rate proportional to A(t). That is A(t) = A(t)where is a positive constant. Solve for t such that A(t) = (1/2)A(0), the half-life of the isotope. Solution. Again, A(t) = Cet and C = A(0), the amount of material at t = 0. Now, we solve for the half-life A(0)e t = 1/2A(0) t e =1/2 t = ln(1/2)=-ln(2) ln(1) = ln(2) 2 t=ln(2)/ C C Example: Carbon Dating. 12 is the common stable isotope of carbon and 14 is a 14C radioactive isotope. The decay constant for is .00012 years1 . Thus, the half-life of 14C is ln(2)/.00012 5775 years. 14C 12C Example: The ratio of the natural occurrence of the isotopes and in the 12 atmosphere or in a living plant or animal is 14C/12C 10 . When an organism stops respirating, it no longer stays in equilibrium with the atmosphere. Let A(t) be the amount of equilibrium. 14C and B(t) be the amount of 12C in a specimen taken out of A(t)=A(0)e-0.00012t B(t) = B(0) The ratio of the two, R(t) satisfies R(t)=A(t )/B(t) =A(0)e .00012t /B(0) 00012t =R(0)e =10-12 e-0.00012t One can determine the time t at which a organism died by measuring the ratio of 12C in its remains, to the extent that is possible. 14C to Example: Determine the percentage (or the original amount) of 14C that remains in the wood of a 4500 year antique chest Solution: A(t)=A(0)e-0.00012t A(4,500) =A(0)e-.00012-4,500 This is about A(0)(0.58) So, approximately 58% of the original 14C remains
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