WARM-UP:1. Use Euler’s Method to find given that and Use .

(1,3) 0.5 3 1.5 (1.5,4.5)

(1.5, 4.5) 0.5 6.75 3.375 (2,7.875)

(2,7.875) 0.5 15.75 7.875 (2.5, 15.75)

(2.5, 15.75) 0.5 39.375 19.6875 (3,35.4375)

Euler’s Method approximates .

2. Use separation of variables to find given that and Compare your answer to #1.

Exponential Growth and

DecaySec. 6.4

The number of bighorn sheep in a population increases at a rate that is proportional to the number of sheep present (at least for awhile.)

So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account.

If the rate of change is proportional to the amount present, the change can be modeled by:

dyky

dt

dyky

dt

1 dy k dt

y

1 dy k dt

y

ln y kt C

Rate of change is proportional to the amount present.

Divide both sides by y.

Integrate both sides.

1 dy k dt

y

ln y kt C

Integrate both sides.

Exponentiate both sides.

When multiplying like bases, add exponents. So added exponents can be written as multiplication.

ln y kt Ce e

C kty e e

ln y kt Ce e

C kty e e

Exponentiate both sides.

When multiplying like bases, add exponents. So added exponents can be written as multiplication.

C kty e e

kty Ae Since is a constant, let .Ce Ce A

C kty e e

kty Ae Since is a constant, let .Ce Ce A

At , .0t 0y y00

ky Ae

0y A

1

0kty y e This is the solution to our original initial

value problem.

0kty y eExponential Change:

If the constant k is positive then the equation

represents growth. If k is negative then the equation represents decay.

Continuously Compounded Interest

If money is invested in a fixed-interest account where the interest is added to the account k times per year, the

amount present after t years is:

0 1kt

rA t A

k

If the money is added back more frequently, you will make a little more money.

The best you can do is if the interest is added continuously.

Of course, the bank does not employ some clerk to continuously calculate your interest with an adding machine.

We could calculate: 0lim 1kt

k

rA

k

but we won’t learn how to find this limit until chapter 8.

Since the interest is proportional to the amount present, the equation becomes:

Continuously Compounded Interest:

0rtA A e

You may also use:

rtA Pe

which is the same thing.

EXAMPLE: A man invests $10,000 in an account that pays 8.5% interest per year. What is the amount of money that he will have after 3 years if the interest is compounded…

a) annually?

b) quarterly?

c) weekly?

d) daily?

e) continuously?

$12,870.19

$12,772.89

$12,904.61

$12,901.92

$12,904.23

Radioactive Decay

The equation for the amount of a radioactive element left after time t is:

0kty y e

This allows the decay constant, k, to be positive.

The half-life is the time required for half the material to decay.

Half-life

0 0

1

2kty y e

1ln ln

2kte

ln1 ln 2 kt 0

ln 2 kt

ln 2t

k

Half-life:

ln 2half-life

k

EXAMPLE: Carbon-14 is a radioactive isotope of carbon that has a half life of 5600 years. It is used extensively in dating organic material that is tens of thousands of years old. What fraction of the original amount of Carbon-14 in a sample would be present after 10,000 years?

0kty y e

𝑦𝑦0

=𝑒−(1.2×10− 4)(10,000)≈ 0.301

So approximately 30% of the original Carbon-14 would be present in 10,000 years.

𝑦𝑦0

=𝑒−𝑘𝑡

EXAMPLE:

How long would it take for Cs137 to decay to 0.1% of its initial level, just after the explosion at Chernobyl?

0.001 𝑦 0=𝑦0𝑒− 0.023𝑡

0.001=𝑒−0.023 𝑡

t

𝑡=ln(0.001)−0.023

≈300.337

Therefore, it would take about 330 YEARS for the level of cesium-137 to decay to 0.1% of its initial level!!!!!!!!!!!!!!!

• An object’s temperature over time will approach the temperature of its surroundings (the medium)

• The greater the difference between the object’s temperature and the medium’s temperature, the greater the rate of change of the object’s temperature

• This change is a form of exponential decay

T0

Tm

Newton’s Law of Cooling

Newton’s Law of Cooling

Espresso left in a cup will cool to the temperature of the surrounding air. The rate of cooling is proportional to the difference in temperature between the liquid and the air.

(It is assumed that the air temperature is constant.)

If we solve the differential equation: sdT

k T Tdt

we get:Newton’s Law of Cooling

0kt

s sT T T T e

where is the temperature of the surrounding medium, which is a constant.

sT

p

Example: The great detective Sherlock Holmes and his assistant, Dr. Watson, are discussing the murder of actor Cornelius McHam. McHam was shot in the head, and his understudy, Barry Moore, was found standing over the body with the murder weapon in hand. Let’s listen in:

Watson:Open-and-shut case, Holmes. Moore is the murderer.

Holmes:Not so fast, Watson – you are forgetting Newton’s Law of Cooling!

Watson:Huh?

Holmes:Elementary, my dear Watson. Moore was found standing over McHam at 10:06 p.m., at which time the coroner recorded a body temperature of 77.9°F and noted that the room thermostat was set to 72°F. At 11:06 p.m. the coroner took another reading and recorded a body temperature of 75.6°F. Since McHam’s normal temperature was 98.6°F, and since Moore was on stage between 6:00 p.m. and 8:00 p.m., Moore is obviously innocent. Ask any calculus student to figure it out for you.

How did Holmes know that Moore was innocent?

Solution: NOTE: m is used in place of s in this example.

EXAMPLE: A detective is called to the scene of a crime where a dead body has just been found. She arrives on the scene at 10:23 pm and begins her investigation. Immediately, the temperature of the body is taken and is found to be 80o F. The detective checks the programmable thermostat and finds that the room has been kept at a constant 68o F for the past 3 days. After evidence from the crime scene is collected, the temperature of the body is taken once more and found to be 78.5o F. This last temperature reading was taken exactly one hour after the first one. The next day the detective is asked by another investigator, “What time did our victim die?” Assuming that the victim’s body temperature was normal (98.6o F) prior to death, what is her answer to this question?

To organize our thinking about this problem, let’s be explicit about what we are trying to solve for. We would like to know the time at which a person died.

In particular, we know the investigator arrived on the scene at 10:23 pm, which we will call τ hours after death.

At 10:23 (i.e. τ hours after death), the temperature of the body was found to be 80o F.

One hour later, τ + 1 hours after death, the body was found to be 78.5o F.

Our known constants for this problem are, Ts = 68o F and T0 = 98.6o F.

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Homework:

p. 357-359 #1-10, 15-18, 21, 30-32, 34, 45 +