Lesson 15: Exponential Growth and Decay

Section 3.4Exponential Growth and Decay

V63.0121.002.2010Su, Calculus I

New York University

June 2, 2010

Announcements

I Review in second half of class today

I Office Hours after class today

I Midterm tomorrow

Announcements

I Review in second half ofclass today

I Office Hours after classtoday

I Midterm tomorrow

V63.0121.002.2010Su, Calculus I (NYU) Section 3.4 Exponential Growth and Decay June 2, 2010 2 / 37

Objectives

Outline

Recall

The differential equation y ′ = ky

Modeling simple population growth

Modeling radioactive decayCarbon-14 Dating

Newton’s Law of Cooling

Continuously Compounded Interest

Derivatives of exponential and logarithmic functions

y y ′

ax (ln a)ax

loga x1

ln a· 1

Outline

Recall

What is a differential equation?

Definition

A differential equation is an equation for an unknown function whichincludes the function and its derivatives.

Example

I Newton’s Second Law F = ma is a differential equation, wherea(t) = x ′′(t).

I In a spring, F (x) = −kx , where x is displacement from equilibriumand k is a constant. So

−kx(t) = mx ′′(t) =⇒ x ′′(t) +k

mx(t) = 0.

I The most general solution is x(t) = A sinωt + B cosωt, whereω =

√k/m.

Definition

Example

−kx(t) = mx ′′(t) =⇒ x ′′(t) +k

mx(t) = 0.

√k/m.

Definition

Example

−kx(t) = mx ′′(t) =⇒ x ′′(t) +k

mx(t) = 0.

√k/m.

Definition

Example

−kx(t) = mx ′′(t) =⇒ x ′′(t) +k

mx(t) = 0.

√k/m.

The Equation y ′ = 2

Example

I Find a solution to y ′(t) = 2.

I Find the most general solution to y ′(t) = 2.

Solution

I A solution is y(t) = 2t.

I The general solution is y = 2t + C .

Remark

If a function has a constant rate of growth, it’s linear.

Example

Solution

Remark

Example

Solution

Remark

Example

Solution

Remark

The Equation y ′ = 2t

Example

I Find a solution to y ′(t) = 2t.

I Find the most general solution to y ′(t) = 2t.

Solution

I A solution is y(t) = t2.

I The general solution is y = t2 + C .

Example

Solution

Example

Solution

Example

I Find a solution to y ′(t) = y(t).

I Find the most general solution to y ′(t) = y(t).

Solution

I A solution is y(t) = et .

I The general solution is y = Cet , not y = et + C .

(check this)

Example

Solution

(check this)

Example

Solution

(check this)

Kick it up a notch

Example

I Find a solution to y ′ = 2y .

I Find the general solution to y ′ = 2y .

Solution

I y = e2t

I y = Ce2t

Kick it up a notch

Example

I Find a solution to y ′ = 2y .

I Find the general solution to y ′ = 2y .

Solution

I y = e2t

I y = Ce2t

In general

Example

I Find a solution to y ′ = ky .

I Find the general solution to y ′ = ky .

Solution

I y = ekt

I y = Cekt

Remark

What is C ? Plug in t = 0:

y(0) = Cek·0 = C · 1 = C ,

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

In general

Example

Solution

I y = ekt

I y = Cekt

Remark

y(0) = Cek·0 = C · 1 = C ,

In general

Example

Solution

I y = ekt

I y = Cekt

Remark

y(0) = Cek·0 = C · 1 = C ,

Constant Relative Growth =⇒ Exponential Growth

Theorem

A function with constant relative growth rate k is an exponential functionwith parameter k. Explicitly, the solution to the equation

y ′(t) = ky(t) y(0) = y0

isy(t) = y0ekt

Exponential Growth is everywhere

I Lots of situations have growth rates proportional to the current value

I This is the same as saying the relative growth rate is constant.

I Examples: Natural population growth, compounded interest, socialnetworks

Outline

Recall

Bacteria

I Since you need bacteria tomake bacteria, the amountof new bacteria at anymoment is proportional tothe total amount of bacteria.

I This means bacteriapopulations growexponentially.

Bacteria Example

Example

A colony of bacteria is grown under ideal conditions in a laboratory. At theend of 3 hours there are 10,000 bacteria. At the end of 5 hours there are40,000. How many bacteria were present initially?

Solution

Since y ′ = ky for bacteria, we have y = y0ekt . We have

10, 000 = y0ek·3 40, 000 = y0ek·5

Dividing the first into the second gives4 = e2k =⇒ 2k = ln 4 =⇒ k = ln 2. Now we have

10, 000 = y0e ln 2·3 = y0 · 8

So y0 =10, 000

8= 1250.

Bacteria Example

Example

Solution

10, 000 = y0ek·3 40, 000 = y0ek·5

10, 000 = y0e ln 2·3 = y0 · 8

So y0 =10, 000

8= 1250.

Bacteria Example

Example

Solution

10, 000 = y0ek·3 40, 000 = y0ek·5

10, 000 = y0e ln 2·3 = y0 · 8

So y0 =10, 000

8= 1250.

Could you do that again please?

We have

10, 000 = y0ek·3

40, 000 = y0ek·5

Dividing the first into the second gives

40, 000

10, 000=

=⇒ 4 = e2k

=⇒ ln 4 = ln(e2k) = 2k

=⇒ k =ln 4

2 ln 2

2= ln 2

Outline

Recall

Modeling radioactive decay

Radioactive decay occurs because many large atoms spontaneously give offparticles.

This means that in a sample of abunch of atoms, we can assume acertain percentage of them will“go off” at any point. (Forinstance, if all atom of a certainradioactive element have a 20%chance of decaying at any point,then we can expect in a sampleof 100 that 20 of them will bedecaying.)

Modeling radioactive decay

Radioactive decay occurs because many large atoms spontaneously give offparticles.

This means that in a sample of abunch of atoms, we can assume acertain percentage of them will“go off” at any point. (Forinstance, if all atom of a certainradioactive element have a 20%chance of decaying at any point,then we can expect in a sampleof 100 that 20 of them will bedecaying.)

Radioactive decay as a differential equation

The relative rate of decay is constant:

where k is negative.

y ′ = ky =⇒ y = y0ekt

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

where k is negative. So

y ′ = ky =⇒ y = y0ekt

again!

It’s customary to express the relative rate of decay in the units of half-life:the amount of time it takes a pure sample to decay to one which is onlyhalf pure.

where k is negative. So

y ′ = ky =⇒ y = y0ekt

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

Computing the amount remaining of a decayingsample

Example

The half-life of polonium-210 is about 138 days. How much of a 100 gsample remains after t years?

Solution

We have y = y0ekt , where y0 = y(0) = 100 grams. Then

50 = 100ek·138/365 =⇒ k = −365 · ln 2

Thereforey(t) = 100e−

365·ln 2138

t = 100 · 2−365t/138.

Notice y(t) = y0 · 2−t/t1/2 , where t1/2 is the half-life.

Example

Solution

50 = 100ek·138/365 =⇒ k = −365 · ln 2

365·ln 2138

t = 100 · 2−365t/138.

Example

Solution

50 = 100ek·138/365 =⇒ k = −365 · ln 2

365·ln 2138

t = 100 · 2−365t/138.

Carbon-14 Dating

The ratio of carbon-14 tocarbon-12 in an organism decaysexponentially:

p(t) = p0e−kt .

The half-life of carbon-14 isabout 5700 years. So theequation for p(t) is

p(t) = p0e−ln25700

Another way to write this wouldbe

p(t) = p02−t/5700

Computing age with Carbon-14 content

Example

Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is 10%of that in a living organism. How old is the fossil?

Solution

We are looking for the value of t for which

p(0)= 0.1

From the equation we have

2−t/5700 = 0.1 =⇒ − t

5700ln 2 = ln 0.1 =⇒ t =

ln 0.1

ln 2· 5700 ≈ 18, 940

So the fossil is almost 19,000 years old.

Example

Solution

p(0)= 0.1

2−t/5700 = 0.1 =⇒ − t

5700ln 2 = ln 0.1 =⇒ t =

ln 0.1

ln 2· 5700 ≈ 18, 940

Example

Solution

p(0)= 0.1

2−t/5700 = 0.1 =⇒ − t

5700ln 2 = ln 0.1 =⇒ t =

ln 0.1

ln 2· 5700 ≈ 18, 940

Example

Solution

p(0)= 0.1

2−t/5700 = 0.1 =⇒ − t

5700ln 2 = ln 0.1 =⇒ t =

ln 0.1

ln 2· 5700 ≈ 18, 940

Outline

Recall

I Newton’s Law of Coolingstates that the rate ofcooling of an object isproportional to thetemperature differencebetween the object and itssurroundings.

I This gives us a differentialequation of the form

dt= k(T − Ts)

(where k < 0 again).

I Newton’s Law of Coolingstates that the rate ofcooling of an object isproportional to thetemperature differencebetween the object and itssurroundings.

I This gives us a differentialequation of the form

dt= k(T − Ts)

(where k < 0 again).

General Solution to NLC problems

To solve this, change the variable y(t) = T (t)− Ts . Then y ′ = T ′ andk(T − Ts) = ky . The equation now looks like

dt= k(T − Ts) ⇐⇒ dy

dt= ky

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 solution to the equation T ′(t) = k(T (t)− Ts), T (0) = T0 is

T (t) = (T0 − Ts)ekt + Ts

dt= ky

Now we can solve!

y ′ = ky

=⇒ y = Cekt =⇒ T − Ts = Cekt =⇒ T = Cekt + Ts

Theorem

dt= ky

Now we can solve!

y ′ = ky =⇒ y = Cekt

=⇒ T − Ts = Cekt =⇒ T = Cekt + Ts

Theorem

dt= ky

Now we can solve!

y ′ = ky =⇒ y = Cekt =⇒ T − Ts = Cekt

=⇒ T = Cekt + Ts

Theorem

dt= ky

Now we can solve!

Theorem

dt= ky

Now we can solve!

Theorem

Computing cooling time with NLC

Example

A hard-boiled egg at 98 ◦C is put in a sink of 18 ◦C water. After 5minutes, the egg’s temperature is 38 ◦C. Assuming the water has notwarmed appreciably, how much longer will it take the egg to reach 20 ◦C?

Solution

We know that the temperature function takes the form

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

To find k, plug in t = 5:

38 = T (5) = 80e5k + 18

and solve for k.

Computing cooling time with NLC

Example

A hard-boiled egg at 98 ◦C is put in a sink of 18 ◦C water. After 5minutes, the egg’s temperature is 38 ◦C. Assuming the water has notwarmed appreciably, how much longer will it take the egg to reach 20 ◦C?

Solution

We know that the temperature function takes the form

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

To find k, plug in t = 5:

38 = T (5) = 80e5k + 18

and solve for k.

Finding k

38 = T (5) = 80e5k + 18

20 = 80e5k

4= e5k

=⇒ k = −1

5ln 4.

Now we need to solve

20 = T (t) = 80e−t5ln 4 + 18

for t.

Finding k

38 = T (5) = 80e5k + 18

20 = 80e5k

4= e5k

=⇒ k = −1

5ln 4.

Now we need to solve

20 = T (t) = 80e−t5ln 4 + 18

for t.

Finding t

20 = 80e−t5ln 4 + 18

2 = 80e−t5ln 4

40= e−

t5ln 4

− ln 40 = − t

=⇒ t =ln 4015 ln 4

=5 ln 40

ln 4≈ 13min

Computing time of death with NLC

Example

A murder victim is discovered atmidnight and the temperature ofthe body is recorded as 31 ◦C.One hour later, the temperatureof the body is 29 ◦C. Assumethat the surrounding airtemperature remains constant at21 ◦C. Calculate the victim’stime of death. (The “normal”temperature of a living humanbeing is approximately 37 ◦C.)

Solution

I Let time 0 be midnight. We know T0 = 31, Ts = 21, and T (1) = 29.We want to know the t for which T (t) = 37.

I To find k:

29 = 10ek·1 + 21 =⇒ k = ln 0.8

I To find t:

37 = 10et·ln(0.8) + 21

1.6 = et·ln(0.8)

t =ln(1.6)

ln(0.8)≈ −2.10hr

So the time of death was just before 10:00pm.

Solution

I To find k:

29 = 10ek·1 + 21 =⇒ k = ln 0.8

I To find t:

37 = 10et·ln(0.8) + 21

1.6 = et·ln(0.8)

t =ln(1.6)

ln(0.8)≈ −2.10hr

Solution

I To find k:

29 = 10ek·1 + 21 =⇒ k = ln 0.8

I To find t:

37 = 10et·ln(0.8) + 21

1.6 = et·ln(0.8)

t =ln(1.6)

ln(0.8)≈ −2.10hr

Outline

Recall

Interest

I If an account has an compound interest rate of r per yearcompounded n times, then an initial deposit of A0 dollars becomes

)ntafter t years.

I For different amounts of compounding, this will change. As n→∞,we get continously compounded interest

A(t) = limn→∞

= A0ert .

I Thus dollars are like bacteria.

Interest

)ntafter t years.

A(t) = limn→∞

)nt= A0ert .

Interest

)ntafter t years.

A(t) = limn→∞

)nt= A0ert .

Computing doubling time with exponential growth

Example

How long does it take an initial deposit of $100, compoundedcontinuously, to double?

Solution

We need t such that A(t) = 200. In other words

200 = 100ert =⇒ 2 = ert =⇒ ln 2 = rt =⇒ t =ln 2

For instance, if r = 6% = 0.06, we have

t =ln 2

0.06≈ 0.69

6= 11.5 years.

Computing doubling time with exponential growth

Example

How long does it take an initial deposit of $100, compoundedcontinuously, to double?

Solution

We need t such that A(t) = 200. In other words

200 = 100ert =⇒ 2 = ert =⇒ ln 2 = rt =⇒ t =ln 2

For instance, if r = 6% = 0.06, we have

t =ln 2

0.06≈ 0.69

6= 11.5 years.

I-banking interview tip of the day

I The fractionln 2

rcan also be

approximated as either 70 or72 divided by the percentagerate (as a number between 0and 100, not a fractionbetween 0 and 1.)

I This is sometimes called therule of 70 or rule of 72.

I 72 has lots of factors so it’sused more often.

Summary

I When something grows or decays at a constant relative rate, thegrowth or decay is exponential.

I Equations with unknowns in an exponent can be solved withlogarithms.

I Your friend list is like culture of bacteria (no offense).

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