Standards with sections 6 4 initial value and exp growth and decay Sp2012 by 6286qGT


									Standards for Eleventh Calculus Unit: DIFFERENTIAL EQUATIONS

   solving separable differential equations and using them in modeling. In particular, studying the
    equation y’=ky and exponential growth
   finding specific antiderivatives using initial conditions, including applications to motion along a line

Textbook: Section 6.3 (pages 341 - 349)

An equation such as           is called a differential equation since the function is written in terms of a
derivative. The solution to the differential equation would be to find the function             whose
derivative is 4x. Using your antiderivative skills, you can probably tell that the answer is              .
This answer, which includes the +C, is the general solution to the differential equation.

Sometimes you will be told a particular point on the solution curve. That point is called an initial value.
For example, in our previous example, you might also be told                , which means (-3, 7) is a point on
the solution curve. If that happens, you can use that point to solve for C and find the specific solution to
the differential equation. If we substitute x=-3 and y=7 into our previous general solution, we find that C
= -11. That means our specific solution would be                   .

What are the differences between a general solution and a specific solution?

When we have more complicated differential equations, we will use a process called separation of
variables to solve the problem.

EXAMPLE: Find the specific solution to the differential equation if            and y(0)=5.

STEP #1: CROSS-MULTIPLY. All of your y’s and dy’s should go to the left-side of the equation and all
of your x’s and dx’s should go to the right-side. If you have any constants, leave them on the right-side
with the x’s and dx’s.

STEP #2: ANTIDERIVATIVES. Take the antiderivative of both sides of the equation using whatever
rules or methods necessary. We traditionally combine the +C on the left side with the +C on the right and
leave the +C on the right-side of the equation.

STEP #3: IN INTIAL VALUE PROBLEMS, SOLVE FOR C. Use your initial value to solve for C.

                               y 2  4 x 2  25
                                                              Since the point x = 0, y = 5 has to work, the
                               y   4 x 2  25
                                                              Final answer is y   4 x2  25
EXAMPLE: Find the specific solution to the differential equation if                                 and y(0)=2.

                                                                  e x ln 2
                                                          ln y           3
                                                       y  e x  eln 2   2e x
                                                                     3                    3

Since x = 0, y = 2 has to work, the final answer is y  2ex

EXAMPLE: If the acceleration of a particle is given by                           , and if the initial velocity is 64 ft
/ sec and the initial height is 32 feet, find (1) the equation of the particle’s velocity at time t, (2) the
equation of the particle’s height at time t, and (3) the maximum height of the particle.
                (1)                                              (2)
                 Solve for C: 64 = -32(0) + C, C = 64                            Solve for C: 32 = -16(0)+64(0)+C, C=32

(3) Find t by setting v(t) = 0 -> t = 2; x(2) = 96 feet

YOU TRY: Find the specific solution to each differential equation.

(A)                                          (B)                                                      (C)

(D) If a(t)= g, and an object’s initial velocity is       and its initial position is            , find the particle’s position,
velocity, and acceleration functions

                                  EXPONENTIAL GROWTH & DECAY
The specific solution to the differential equation     ky , when y  y0 at t  0 , is:
                                                      y  y0 e kt
This equation will be used to solve most exponential growth / decay problems. k is the rate constant.
k > 0 represents exponential growth and k < 0 represents exponential decay.
Anytime a question mentions the rate of change is proportional to the amount present, you are dealing with
either exponential growth or exponential decay and can automatically use the formula above.

 y0 represents the amount of material present at time t  0 (or whenever your time period begins). This
information is typically given to you in the problem so you can replace the y with a number; however, if no number
is given, leave y in your equation and your final amount, y, will be some multiple of y . For example, if your
                  0                                                                    0
population will double over time, you will leave y alone and use the value 2 y as your final amount.
                                                    0                          0

To solve for k, you generally need an initial amount at some time. Substitute in for y and for t and solve.

Here are three other formulas you will need to work exercises in these sections (you saw some in Algebra 2):
                                         
Compounding interest:     A  P 1 r 
                                   n
A = Final amount in account                                P = Principal (original amount in account)
r = rate of growth (as a decimal – e.g. 7% growth means r = 0.07)
n = number of times interest is compounded each year       t = number of years of investment

Continuously Compounding interest:        A  Pert
A = Final amount in account                                     e = constant
r = rate of growth (as a decimal)                               t = number of years of investment

Newton’s Law of Cooling: T  TS  T0  TS  ekt
T = Final temperature           TS = surrounding temperature             T0 = Temperature of object at time t = 0
e = constant                    k = decay constant                       t = time

Example: The new theme restaurant in town (Rowdy Rita’s Eat and Hurl) is being tested by the
health department for cleanliness. Health inspectors find the men’s room floor to be a fertile ground
for growing bacteria. They have determined that the rate of bacterial growth is proportional to the
number of colonies. So, they plant 10 colonies and come back in 15 minutes; when they return, the
number of colonies has risen to 35. How many colonies will there be one full hour after they planted
the original 10?
Underlined section shows this is exponential growth: y  y0ekt
y0 =10 at t = 0        y  10ekt
                                                                 35 
y = 35 at t = 15 minutes means 35  10ek (15) . Solve for k: ln    15k and k  0.08352
                                                                 10 
                                                                      0.08352 60
To solve for the final answer, substitute t = 60 minutes: y  10e                     1,500.625

After 1 hour, there are almost 1501 colonies. (Check the amount after 24 hours, and you may want to avoid
this bathroom forever.)
EXAMPLE: The Easter Bunny has begun to express his more malevolent side. This year, instead of
hiding real Easter eggs, he’s hiding eggs made of a radioactive substance Nb-95, which has a half-life
of 35 days. If the danger eggs have a mass of 2 kg, and you don’t find the one hiding under your bed,
how long will it take that egg to decay to a “harmless” 50 g?

Half-life indicates exponential decay. Be careful to keep your weight units consistent!

The egg starts at 2,000 g       y  2000ekt
                                                         k  35
To find k, use the half-life information: 1000  2000e  , meaning k  0.01980

The formula is now y  2000e0.01980t . Solve: 50  2000e0.01980t , t 
                                                                         ln 1  
                                                                              40  186.287 days .

Good to go by Thanksgiving!

EXAMPLE: How long does it take for an investment to double at an annual interest rate of 8.5%
compounded continuously?

Continuously compounding interest:             . A will be 2 times P; r = .085; solve for t:

It will take 8 years and about 2 months for your investment to double.


   (E) The population of a country is growing at a rate proportional to its population. If the growth rate
       per year is 4% of the current population, how long will it take for the population to double? [Hint:
       use y0 , a variable, to represent the initial population. Also, k = 0.04]

   (F) The bacteria in a certain culture increase continuously at a rate proportional to the number present.
       (1) If the number triples in 6 hours, how many will there be in 12 hours? (2) In how many hours
       will the original number quadruple? [Use y0 to represent the initial population]

   (G) Radium-226 decays at a rate proportional to the quantity present. Its half-life is 1612 years. How
       long will it take for one quarter of a given quantity of radium-226 to decay?
      (H) At a yearly rate of 5% compounded continuously, how long does it take for an investment to triple?

      (I) In 1970 the world population was approximately 3.5 billion. Since then it has been growing at a
          rate proportional to the population, and the factor of proportionality has been 1.9% per year. At that
          rate, in how many years will there be one person per square foot of land? (The land area of Earth is
          approximately 200,000,000 mi2, or about 5.5 * 1015 ft2.)

      (J) The “Rule of 72” is a common investment formula used in the “real world.” It says if you want to
          know how long it will take for an investment to double in value, take the number 72 and divide it by
          the interest rate percentage (for example, use 6 and not 0.06 for 6%). For example, if you invest
          some money at 6% interest, you would expect your money to double in about               years. This
          rule gives a decent time estimate for interest compounded annually as long as the interest rate is in
          the 6 – 9% range (not likely in the beginning of 2011). Your question: What would be a better
          name for this rule if instead of the interest compounding annually, it was compounded
          continuously? The Rule of ______ (round to a 2-digit integer + 1 decimal place). [Use your
          compounding interest formula and solve for t. (Remember, at the end, to account for the interest
          rate being expressed as a percentage instead of a decimal.) NOTE: this rule would work for ANY
          interest rate!]


(A)                         (B)                  (C)                (D)

(E) about 17.33 years             (F) (1) 9 times the initial number; (2) about 7.6 hours

(G) about 669 years               (H) about 22 years            (I) about 750 years     (J) Rule of 69.3

CLASSWORK / HOMEWORK:                  Section 6.1 (49-50); Section 6.4 (1-5 odd, 7-10, 11-19 odd, 23, 25,
27, 30, 32, 35 {use half-life of 5700 years for C-14}, 41, 46)

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