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.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                   March 23, 2011


                                                         Notes
                    Sec on 3.4
    .
            Exponen al Growth and Decay
                            V63.0121.001: Calculus I
                          Professor Ma hew Leingang
                                   New York University


                                March 23, 2011


    .
                                                         .




                                                         Notes
        Announcements


           Quiz 3 next week in
           recita on on 2.6, 2.8, 3.1,
           3.2




    .
                                                         .




                                                         Notes
        Objectives

           Solve the ordinary
           differen al equa on
           y′ (t) = ky(t), y(0) = y0
           Solve problems involving
           exponen al growth and
           decay



    .
                                                         .

                                                                                  . 1
.
.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                   March 23, 2011


                                                          Notes
        Outline
         Recall
         The differen al equa on y′ = ky
         Modeling simple popula on growth
         Modeling radioac ve decay
           Carbon-14 Da ng
         Newton’s Law of Cooling
         Con nuously Compounded Interest

    .
                                                          .




        Derivatives of exponential and                    Notes
        logarithmic functions
                                     y          y′

                                    ex          ex

                                    ax      (ln a) · ax
                                                1
                                    ln x
                                                x
                                              1 1
                                   loga x        ·
                                             ln a x
    .
                                                          .




                                                          Notes
        Outline
         Recall
         The differen al equa on y′ = ky
         Modeling simple popula on growth
         Modeling radioac ve decay
           Carbon-14 Da ng
         Newton’s Law of Cooling
         Con nuously Compounded Interest

    .
                                                          .

                                                                                  . 2
.
.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                             March 23, 2011


                                                                          Notes
        What is a differential equation?
         Defini on
         A differen al equa on is an equa on for an unknown func on
         which includes the func on and its deriva ves.
         Example

              Newton’s Second Law F = ma is a differen al equa on, where
              a(t) = x′′ (t).
              In a spring, F(x) = −kx, where x is displacement from
              equilibrium and k is a constant. So
                                                          k
                         −kx(t) = mx′′ (t) =⇒ x′′ (t) +     x(t) = 0.
    .                                                     m
                                                                          .




                                                                          Notes
        Showing a function is a solution
         Example (Con nued)
         Show that x(t) = A sin ωt + B cos ωt sa sfies the differen al
                        k                   √
         equa on x′′ + x = 0, where ω = k/m.
                       m
         Solu on
         We have

                            x(t) = A sin ωt + B cos ωt
                           x′ (t) = Aω cos ωt − Bω sin ωt
                           x′′ (t) = −Aω 2 sin ωt − Bω 2 cos ωt
    .
                                                                          .




                                                                          Notes
        The Equation y′ = 2
         Example

              Find a solu on to y′ (t) = 2.
              Find the most general solu on to y′ (t) = 2.

         Solu on
              A solu on is y(t) = 2t.
              The general solu on is y = 2t + C.

         Remark
    .    If a func on has a constant rate of growth, it’s linear.

                                                                          .

                                                                                            . 3
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.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                    March 23, 2011


                                                                 Notes
        The Equation y′ = 2t
         Example

              Find a solu on to y′ (t) = 2t.
              Find the most general solu on to y′ (t) = 2t.

         Solu on
              A solu on is y(t) = t2 .
              The general solu on is y = t2 + C.


    .
                                                                 .




                                                                 Notes
        The Equation y′ = y
         Example

              Find a solu on to y′ (t) = y(t).
              Find the most general solu on to y′ (t) = y(t).

         Solu on
              A solu on is y(t) = et .
              The general solu on is y = Cet , not y = et + C.
         (check this)
    .
                                                                 .




                                                                 Notes
        Kick it up a notch: y′ = 2y
         Example

              Find a solu on to y′ = 2y.
              Find the general solu on to y′ = 2y.

         Solu on
              y = e2t
              y = Ce2t


    .
                                                                 .

                                                                                   . 4
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.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                                   March 23, 2011


                                                                                Notes
        In general: y′ = ky
         Example

              Find a solu on to y′ = ky.
              Find the general solu on to y′ = ky.

         Solu on                              Remark
              y=e  kt                         What is C? Plug in t = 0:
              y = Cekt                           y(0) = Cek·0 = C · 1 = C,

                                              so y(0) = y0 , the ini al value
                                              of y.
    .
                                                                                .




        Constant Relative Growth =⇒                                             Notes
        Exponential Growth
         Theorem
         A func on with constant rela ve growth rate k is an exponen al
         func on with parameter k. Explicitly, the solu on to the equa on

                             y′ (t) = ky(t)    y(0) = y0

         is
                                     y(t) = y0 ekt


    .
                                                                                .




                                                                                Notes
        Exponential Growth is everywhere
              Lots of situa ons have growth rates propor onal to the current
              value
              This is the same as saying the rela ve growth rate is constant.
              Examples: Natural popula on growth, compounded interest,
              social networks




    .
                                                                                .

                                                                                                  . 5
.
.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                                   March 23, 2011


                                                                                Notes
        Outline
         Recall
         The differen al equa on y′ = ky
         Modeling simple popula on growth
         Modeling radioac ve decay
           Carbon-14 Da ng
         Newton’s Law of Cooling
         Con nuously Compounded Interest

    .
                                                                                .




                                                                                Notes
        Bacteria
              Since you need bacteria
              to make bacteria, the
              amount of new bacteria
              at any moment is
              propor onal to the total
              amount of bacteria.
              This means bacteria
              popula ons grow
              exponen ally.

    .
                                                                                .




                                                                                Notes
        Bacteria Example
         Example
         A colony of bacteria is grown under ideal condi ons in a laboratory.
         At the end of 3 hours there are 10,000 bacteria. At the end of 5
         hours there are 40,000. How many bacteria were present ini ally?

         Solu on




    .
                                                                                .

                                                                                                  . 6
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.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                             March 23, 2011


                                                                          Notes
        Bacteria Example Solution
         Solu on (Con nued)
         We have




    .
                                                                          .




                                                                          Notes
        Outline
         Recall
         The differen al equa on y′ = ky
         Modeling simple popula on growth
         Modeling radioac ve decay
           Carbon-14 Da ng
         Newton’s Law of Cooling
         Con nuously Compounded Interest

    .
                                                                          .




                                                                          Notes
        Modeling radioactive decay
         Radioac ve decay occurs because many large atoms spontaneously
         give off par cles.
         This means that in a sample of a
         bunch of atoms, we can assume a
         certain percentage of them will “go
         off” at any point. (For instance, if all
         atom of a certain radioac ve element
         have a 20% chance of decaying at any
         point, then we can expect in a
         sample of 100 that 20 of them will be
         decaying.)
    .
                                                                          .

                                                                                            . 7
.
.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                                     March 23, 2011


                                                                                  Notes
        Radioactive decay as a differential equation
          The rela ve rate of decay is constant:
                                         y′
                                            =k
                                         y
          where k is nega ve. So

                                y′ = ky =⇒ y = y0 ekt

          again!
          It’s customary to express the rela ve rate of decay in the units of
          half-life: the amount of me it takes a pure sample to decay to one
          which is only half pure.
    .
                                                                                  .




                                                                                  Notes
        Computing the amount remaining
          Example
          The half-life of polonium-210 is about 138 days. How much of a
          100 g sample remains a er t years?




    .
                                                                                  .




                                                                                  Notes
        Carbon-14 Dating
                                       The ra o of carbon-14 to carbon-12 in
                                       an organism decays exponen ally:

                                                     p(t) = p0 e−kt .

                                       The half-life of carbon-14 is about 5700
                                       years. So the equa on for p(t) is

                                             p(t) = p0 e− 5700 t = p0 2−t/5700
                                                           ln2




    .
                                                                                  .

                                                                                                    . 8
.
.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                                 March 23, 2011


                                                                              Notes
        Computing age with Carbon-14
         Example
         Suppose a fossil is found where the ra o of carbon-14 to carbon-12
         is 10% of that in a living organism. How old is the fossil?

         Solu on




    .
                                                                              .




                                                                              Notes
        Outline
         Recall
         The differen al equa on y′ = ky
         Modeling simple popula on growth
         Modeling radioac ve decay
           Carbon-14 Da ng
         Newton’s Law of Cooling
         Con nuously Compounded Interest

    .
                                                                              .




                                                                              Notes
        Newton’s Law of Cooling
              Newton’s Law of Cooling states
              that the rate of cooling of an
              object is propor onal to the
              temperature difference between
              the object and its surroundings.
              This gives us a differen al
              equa on of the form
                       dT
                          = k(T − Ts )
                       dt
              (where k < 0 again).
    .
                                                                              .

                                                                                                . 9
.
.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                                   March 23, 2011


                                                                                Notes
        General Solution to NLC problems
                                   ′ ′
         To solve this, change the variable y(t) = T(t) − Ts . Then y = T and
         k(T − Ts ) = ky. The equa on now looks like
                             dT                     dy
                                = k(T − Ts ) ⇐⇒        = ky
                             dt                     dt
         Now we can solve!
            y′ = ky =⇒ y = Cekt =⇒ T − Ts = Cekt =⇒ T = Cekt + Ts
         Plugging in t = 0, we see C = y0 = T0 − Ts . So
         Theorem
         The solu on to the equa on T′ (t) = k(T(t) − Ts ), T(0) = T0 is

    .                           T(t) = (T0 − Ts )ekt + Ts
                                                                                .




                                                                                Notes
        Computing cooling time with NLC
         Example
         A hard-boiled egg at 98 ◦ C is put in a sink of 18 ◦ C water. A er 5
         minutes, the egg’s temperature is 38 ◦ C. Assuming the water has
         not warmed appreciably, how much longer will it take the egg to
         reach 20 ◦ C?
         Solu on
         We know that the temperature func on takes the form

                         T(t) = (T0 − Ts )ekt + Ts = 80ekt + 18

         To find k, plug in t = 5 and solve for k.
    .
                                                                                .




                                                                                Notes
        Finding k
         Solu on (Con nued)




    .
                                                                                .

                                                                                                  . 10
.
.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                   March 23, 2011


                                                    Notes
        Finding t
          Solu on (Con nued)




    .
                                                    .




                                                    Notes
        Computing time of death with NLC
        Example
        A murder vic m is discovered at
        midnight and the temperature of the
        body is recorded as 31 ◦ C. One hour
        later, the temperature of the body is
        29 ◦ C. Assume that the surrounding
        air temperature remains constant at
        21 ◦ C. Calculate the vic m’s me of
        death. (The “normal” temperature of
        a living human being is approximately
        37 ◦ C.)
    .
                                                    .




          Solu on                                   Notes




    .
                                                    .

                                                                                  . 11
.
.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                                            March 23, 2011


                                                                                         Notes
        Outline
          Recall
          The differen al equa on y′ = ky
          Modeling simple popula on growth
          Modeling radioac ve decay
            Carbon-14 Da ng
          Newton’s Law of Cooling
          Con nuously Compounded Interest

    .
                                                                                         .




                                                                                         Notes
        Interest
               If an account has an compound interest rate of r per year
               compounded n mes, then an ini al deposit of A0 dollars
               becomes                    (    r )nt
                                        A0 1 +
                                               n
               a er t years.
               For different amounts of compounding, this will change. As
               n → ∞, we get con nously compounded interest
                                            (    r )nt
                              A(t) = lim A0 1 +        = A0 ert .
                                     n→∞         n

               Thus dollars are like bacteria.
    .
                                                                                         .




                                                                                         Notes
        Continuous vs. Discrete Compounding of interest
          Example
          Consider two bank accounts: one with 10% annual interested compounded
          quarterly and one with annual interest rate r compunded con nuously. If they
          produce the same balance a er every year, what is r?

          Solu on
          The balance for the 10% compounded quarterly account a er t years
          is
                           A1 (t) = A0 (1.025)4t = P((1.025)4 )t
          The balance for the interest rate r compounded con nuously
          account a er t years is
                                      A2 (t) = A0 ert
    .
                                                                                         .

                                                                                                           . 12
.
.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                                  March 23, 2011


                                                                               Notes
        Solving
         Solu on (Con nued)


                                 A1 (t) = A0 ((1.025)4 )t
                                 A2 (t) = A0 (er )t

         For those to be the same, er = (1.025)4 , so

                       r = ln((1.025)4 ) = 4 ln 1.025 ≈ 0.0988

         So 10% annual interest compounded quarterly is basically equivalent
         to 9.88% compounded con nuously.
    .
                                                                               .




        Computing doubling time with                                           Notes
        exponential growth
         Example
         How long does it take an ini al deposit of $100, compounded
         con nuously, to double?

         Solu on




    .
                                                                               .




                                                                               Notes
        I-banking interview tip of the day
                           ln 2
             The frac on        can also
                             r
             be approximated as
             either 70 or 72 divided by
             the percentage rate (as a
             number between 0 and
             100, not a frac on
             between 0 and 1.)
             This is some mes called
             the rule of 70 or rule of
             72.
             72 has lots of factors so
    .        it’s used more o en.
                                                                               .

                                                                                                 . 13
.
.   V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay
    .                                                                          March 23, 2011


                                                                       Notes
        Summary

          When something grows or decays at a constant rela ve rate,
          the growth or decay is exponen al.
          Equa ons with unknowns in an exponent can be solved with
          logarithms.




    .
                                                                       .




                                                                       Notes




    .
                                                                       .




                                                                       Notes




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                                                                       .

                                                                                         . 14
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Lesson 15: Exponential Growth and Decay (handout)

  • 1. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Sec on 3.4 . Exponen al Growth and Decay V63.0121.001: Calculus I Professor Ma hew Leingang New York University March 23, 2011 . . Notes Announcements Quiz 3 next week in recita on on 2.6, 2.8, 3.1, 3.2 . . Notes Objectives Solve the ordinary differen al equa on y′ (t) = ky(t), y(0) = y0 Solve problems involving exponen al growth and decay . . . 1 .
  • 2. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . Derivatives of exponential and Notes logarithmic functions y y′ ex ex ax (ln a) · ax 1 ln x x 1 1 loga x · ln a x . . Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . . 2 .
  • 3. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes What is a differential equation? Defini on A differen al equa on is an equa on for an unknown func on which includes the func on and its deriva ves. Example Newton’s Second Law F = ma is a differen al equa on, where a(t) = x′′ (t). In a spring, F(x) = −kx, where x is displacement from equilibrium and k is a constant. So k −kx(t) = mx′′ (t) =⇒ x′′ (t) + x(t) = 0. . m . Notes Showing a function is a solution Example (Con nued) Show that x(t) = A sin ωt + B cos ωt sa sfies the differen al k √ equa on x′′ + x = 0, where ω = k/m. m Solu on We have x(t) = A sin ωt + B cos ωt x′ (t) = Aω cos ωt − Bω sin ωt x′′ (t) = −Aω 2 sin ωt − Bω 2 cos ωt . . Notes The Equation y′ = 2 Example Find a solu on to y′ (t) = 2. Find the most general solu on to y′ (t) = 2. Solu on A solu on is y(t) = 2t. The general solu on is y = 2t + C. Remark . If a func on has a constant rate of growth, it’s linear. . . 3 .
  • 4. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes The Equation y′ = 2t Example Find a solu on to y′ (t) = 2t. Find the most general solu on to y′ (t) = 2t. Solu on A solu on is y(t) = t2 . The general solu on is y = t2 + C. . . Notes The Equation y′ = y Example Find a solu on to y′ (t) = y(t). Find the most general solu on to y′ (t) = y(t). Solu on A solu on is y(t) = et . The general solu on is y = Cet , not y = et + C. (check this) . . Notes Kick it up a notch: y′ = 2y Example Find a solu on to y′ = 2y. Find the general solu on to y′ = 2y. Solu on y = e2t y = Ce2t . . . 4 .
  • 5. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes In general: y′ = ky Example Find a solu on to y′ = ky. Find the general solu on to y′ = ky. Solu on Remark y=e kt What is C? Plug in t = 0: y = Cekt y(0) = Cek·0 = C · 1 = C, so y(0) = y0 , the ini al value of y. . . Constant Relative Growth =⇒ Notes Exponential Growth Theorem A func on with constant rela ve growth rate k is an exponen al func on with parameter k. Explicitly, the solu on to the equa on y′ (t) = ky(t) y(0) = y0 is y(t) = y0 ekt . . Notes Exponential Growth is everywhere Lots of situa ons have growth rates propor onal to the current value This is the same as saying the rela ve growth rate is constant. Examples: Natural popula on growth, compounded interest, social networks . . . 5 .
  • 6. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . Notes Bacteria Since you need bacteria to make bacteria, the amount of new bacteria at any moment is propor onal to the total amount of bacteria. This means bacteria popula ons grow exponen ally. . . Notes Bacteria Example Example A colony of bacteria is grown under ideal condi ons in a laboratory. At the end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are 40,000. How many bacteria were present ini ally? Solu on . . . 6 .
  • 7. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Bacteria Example Solution Solu on (Con nued) We have . . Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . Notes Modeling radioactive decay Radioac ve decay occurs because many large atoms spontaneously give off par cles. This means that in a sample of a bunch of atoms, we can assume a certain percentage of them will “go off” at any point. (For instance, if all atom of a certain radioac ve element have a 20% chance of decaying at any point, then we can expect in a sample of 100 that 20 of them will be decaying.) . . . 7 .
  • 8. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Radioactive decay as a differential equation The rela ve rate of decay is constant: y′ =k y where k is nega ve. So y′ = ky =⇒ y = y0 ekt again! It’s customary to express the rela ve rate of decay in the units of half-life: the amount of me it takes a pure sample to decay to one which is only half pure. . . Notes Computing the amount remaining Example The half-life of polonium-210 is about 138 days. How much of a 100 g sample remains a er t years? . . Notes Carbon-14 Dating The ra o of carbon-14 to carbon-12 in an organism decays exponen ally: p(t) = p0 e−kt . The half-life of carbon-14 is about 5700 years. So the equa on for p(t) is p(t) = p0 e− 5700 t = p0 2−t/5700 ln2 . . . 8 .
  • 9. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Computing age with Carbon-14 Example Suppose a fossil is found where the ra o of carbon-14 to carbon-12 is 10% of that in a living organism. How old is the fossil? Solu on . . Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . Notes Newton’s Law of Cooling Newton’s Law of Cooling states that the rate of cooling of an object is propor onal to the temperature difference between the object and its surroundings. This gives us a differen al equa on of the form dT = k(T − Ts ) dt (where k < 0 again). . . . 9 .
  • 10. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes General Solution to NLC problems ′ ′ To solve this, change the variable y(t) = T(t) − Ts . Then y = T and k(T − Ts ) = ky. The equa on now looks like dT dy = k(T − Ts ) ⇐⇒ = ky dt dt Now we can solve! y′ = ky =⇒ y = Cekt =⇒ T − Ts = Cekt =⇒ T = Cekt + Ts Plugging in t = 0, we see C = y0 = T0 − Ts . So Theorem The solu on to the equa on T′ (t) = k(T(t) − Ts ), T(0) = T0 is . T(t) = (T0 − Ts )ekt + Ts . Notes Computing cooling time with NLC Example A hard-boiled egg at 98 ◦ C is put in a sink of 18 ◦ C water. A er 5 minutes, the egg’s temperature is 38 ◦ C. Assuming the water has not warmed appreciably, how much longer will it take the egg to reach 20 ◦ C? Solu on We know that the temperature func on takes the form T(t) = (T0 − Ts )ekt + Ts = 80ekt + 18 To find k, plug in t = 5 and solve for k. . . Notes Finding k Solu on (Con nued) . . . 10 .
  • 11. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Finding t Solu on (Con nued) . . Notes Computing time of death with NLC Example A murder vic m is discovered at midnight and the temperature of the body is recorded as 31 ◦ C. One hour later, the temperature of the body is 29 ◦ C. Assume that the surrounding air temperature remains constant at 21 ◦ C. Calculate the vic m’s me of death. (The “normal” temperature of a living human being is approximately 37 ◦ C.) . . Solu on Notes . . . 11 .
  • 12. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Outline Recall The differen al equa on y′ = ky Modeling simple popula on growth Modeling radioac ve decay Carbon-14 Da ng Newton’s Law of Cooling Con nuously Compounded Interest . . Notes Interest If an account has an compound interest rate of r per year compounded n mes, then an ini al deposit of A0 dollars becomes ( r )nt A0 1 + n a er t years. For different amounts of compounding, this will change. As n → ∞, we get con nously compounded interest ( r )nt A(t) = lim A0 1 + = A0 ert . n→∞ n Thus dollars are like bacteria. . . Notes Continuous vs. Discrete Compounding of interest Example Consider two bank accounts: one with 10% annual interested compounded quarterly and one with annual interest rate r compunded con nuously. If they produce the same balance a er every year, what is r? Solu on The balance for the 10% compounded quarterly account a er t years is A1 (t) = A0 (1.025)4t = P((1.025)4 )t The balance for the interest rate r compounded con nuously account a er t years is A2 (t) = A0 ert . . . 12 .
  • 13. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Solving Solu on (Con nued) A1 (t) = A0 ((1.025)4 )t A2 (t) = A0 (er )t For those to be the same, er = (1.025)4 , so r = ln((1.025)4 ) = 4 ln 1.025 ≈ 0.0988 So 10% annual interest compounded quarterly is basically equivalent to 9.88% compounded con nuously. . . Computing doubling time with Notes exponential growth Example How long does it take an ini al deposit of $100, compounded con nuously, to double? Solu on . . Notes I-banking interview tip of the day ln 2 The frac on can also r be approximated as either 70 or 72 divided by the percentage rate (as a number between 0 and 100, not a frac on between 0 and 1.) This is some mes called the rule of 70 or rule of 72. 72 has lots of factors so . it’s used more o en. . . 13 .
  • 14. . V63.0121.001: Calculus I Sec on 3.4: Exponen al Growth and. Decay . March 23, 2011 Notes Summary When something grows or decays at a constant rela ve rate, the growth or decay is exponen al. Equa ons with unknowns in an exponent can be solved with logarithms. . . Notes . . Notes . . . 14 .