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Wize University Calculus 1 Textbook > Applications of Differentiation for Science

Exponential Growth and Decay

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Exponential Growth and Decay

Any equation that involves an unknown function yy and its derivatives dy/dtdy/dt is called a differential equation.

Differential Equation for Exponential Growth


This equation states that the rate of change dy/dtdy/dt of a quantity is proportional to the current value of that quantity.
dydt=ky\boxed{\frac{dy}{dt}=ky}
where kk is a constant called the proportionality constant (k > 0 for growth, k < 0 for decay).\text{(k > 0 for growth, k < 0 for decay).} If y(0)y(0) is the value of the function yy at time t=0,t=0, the solution becomes
y(t)=y(0)ekt\boxed{y(t)=y(0)e^{kt}}

Wize Tip
The half-life of a radioactive substance is defined as the time t1/2t_{1/2} it takes the substance to decay to half of its original size.
t1/2=ln⁡2k\displaystyle t_{1/2}=\frac{\ln2}{k}

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Logistic Growth Model


A variant on the Exponential Growth Model is the Logistic Growth Model. This model's growth rate decreases as the population approaches the carrying capacity.
dPdt=kP(M−P)\displaystyle \boxed{\frac{dP}{dt}=kP\left(M-P\right)}
Where PPis the population at time tt, MMis the carrying capacity, and kk is the proportionality constant.
P(t)=M1+Ae(−Mk)t\boxed{P(t)=\frac{M}{1+Ae^{\left(-Mk\right)t}}}
Where AA is determined by the initial condition P(0)P(0).

Wize Tip
The population growth rate is fastest when the population is equal to M2\displaystyle \frac{M}{2}.

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Example: Radioactive Decay

A radioactive substance weighs initially 30 grams and decays at a half-life of 10 days. Find the amount of substance left after 20 days.



y(0)=30and y(10)=y(0)2=15y(0)=30 \quad \text{and} \quad\, y(10)=\dfrac{y(0)}{2}=15
dydt=ky\frac{dy}{dt}=ky

y(t)=y(0)ekt=30ekty(t)=y(0)e^{kt}=30e^{kt}
With 2nd condition, we have:

y(10)=15=30e10ky(10)=15=30e^{10k}

⇒10k=ln⁡(1/2)\Rightarrow10k=\ln(1/2)

⇒k=110ln⁡(1/2)\Rightarrow k=\frac{1}{10}\ln(1/2)

⇒y(t)=30e(1/10)ln⁡(1/2)t\Rightarrow y(t) = 30e^{(1/10)\ln(1/2)t}

y(20)=30e2ln⁡(1/2)=30eln⁡(12)2=30(12)2=7.5 grams\displaystyle y(20) = 30e^{2\ln(1/2)}=30e^{\ln{(\frac{1}{2})^2}}=30\Big(\frac{1}{2}\Big)^2=7.5 \text{ grams}


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Example: Logistic Growth Model

The growth of a population is given by the logistic differential equation

dPdt=0.02P(58−P)\displaystyle \frac{dP}{dt}=0.02P\left(58-P\right)

with time t measured in years.

a) What is the solution to this differential equation if P(0)=1P(0)=1?

b) At what population is the fastest growth rate?

a) P(t)=M1+Ae(−Mk)t=581+Ae−1.16tP(t)=\displaystyle\frac{M}{1+Ae^{\left(-Mk\right)t}}=\frac{58}{1+Ae^{-1.16t}}

P(0)=1P(0)=1

⇒1=581+Ae−1.16(0)\displaystyle \Rightarrow1=\frac{58}{1+Ae^{-1.16(0)}}

⇒1+A=58\Rightarrow 1+A=58

⇒A=57\Rightarrow A=57

P(t)=581+57e−1.16t\boxed{\displaystyle P(t)=\frac{58}{1+57e^{-1.16t}}}

b) Find the critical points of the differential equation.

dPdt=0.02P(58−P)=1.16P−0.02P2\displaystyle \frac{dP}{dt}=0.02P\left(58-P\right)=1.16P-0.02P^2

⇒1.16−0.04P=0\Rightarrow1.16-0.04P=0

⇒P=29 \Rightarrow \boxed{P=29 }
Alternatively, recall the fastest growth rate occurs at M2\displaystyle \frac{M}{2}.

P=582=29P=\dfrac{58}{2}=29
The number of a certain type of bacteria grows exponentially from 100 to 200 in 3 hours. In a fermentation process, 50 of these bacteria live. After 9 hours, 150 bacteria die due to feed rate shortage. Approximately how long will it take them to grow from that time to 1410 bacteria?



Extra Practice
Population Growth with 2 Values
The size of a certain chicken population at time t (in years) can be modeled by y(t)=y0 ekty\left(t\right)=y_0\ e^{kt} for some constant kk and y(0)=y0y\left(0\right)=y_0 . At time t=2t=2 , the population size is 300 and at time t=4t=4 , the population size increases to 1200. Find the values of y0y_0 and kk . When will the population exceed 300(2)7300\left(2\right)^7 ?
Exponential Growth and Decay: Radioactive Decay with Given Value
A radioactive substance has a half-life of 50 years. At the current time t=75 years, 200g of the substance remains. How long will it take from now for the substance to be reduced down to 25 g?
Enter your answer in years, do not include units.
Exponential Growth and Decay
A certain bacteria culture starts with 2000 bacteria, and the number of bacteria doubles every 10 minutes. If this continues indefinitely, how long (in hours) will it take for the population to reach 8×10158 \times 10^{15} ? What is the differential equation governing the behaviour of the population of the bacteria?
Exponential Growth and Decay
A certain bacteria culture starts with 2000 bacteria, and the number of bacteria doubles every 10 minutes. If this continues indefinitely, how long (in hours) will it take for the population to reach 8×10158 \times 10^{15} ? What is the differential equation governing the behaviour of the population of the bacteria?
Exponential Growth and Decay
Use the fact that the amount of a drug in a patient at time tt is given by D(t)=D0e−ln⁡(12)tD\left(t\right)=D_0e^{-\ln\left(12\right)t} mg with the initial dose of the drug being 60 mg. The next dosage of the drug cannot be administered until there is only 5 mg of the drug left in the patient's body. After how long, in hours, should the patient wait to take the next dose?
Exponential Growth and Decay
Use the fact that the amount of a drug in a patient at time tt is given by D(t)=D0e−ln⁡(12)tD\left(t\right)=D_0e^{-\ln\left(12\right)t} mg with the initial dose of the drug being 60 mg. The next dosage of the drug cannot be administered until there is only 5 mg of the drug left in the patient's body. After how long, in hours, should the patient wait to take the next dose?
A bacteria population follows the equation y(t)=y0 ekty\left(t\right)=y_0\ e^{kt}. At time t=3t=3 months, the population size is 550550. At time t=6t=6 , the population rises to 13751375. What will the population be after one year? Round to the nearest whole number.
(a) 8210 (b) 2750 (c) 8001 (d) 8594 (e) 9172
Suppose a rabbit population is growing exponentially and quadruples every three months. If there are 4 rabbits to begin with, what will the population be after one year? (Round to the nearest whole number, if necessary)
(a) 512 (b) 2048 (c) 1024 (d) 256 (e) ∞\infin
A certain bacteria culture starts with 2000 bacteria, and the number of bacteria doubles every 10 minutes. If this continues indefinitely, how long (in hours) will it take for the population to reach 8×10158 \times 10^{15} ? What is the differential equation governing the behaviour of the population of the bacteria?
Practice
Use the fact that the amount of a drug in a patient at time tt in hours is given by D(t)=D0e−ln⁡(12)tD\left(t\right)=D_0e^{-\ln\left(12\right)t} mg with the initial dose of the drug being 200 mg. The next dosage of the drug cannot be administered until there is only 10 mg of the drug left in the patient's body. After how long, in hours, should the patient wait to take the next dose?
A bacteria population follows the equation y(t)=y0 ekty\left(t\right)=y_0\ e^{kt}. At time t=3t=3 months, the population size is 550550. At time t=6t=6 , the population rises to 13751375. What will the population be after one year?
Suppose a rabbit population is growing exponentially and quadruples every three months. If there are 4 rabbits to begin with, what will the population be after two years?
A certain bacteria culture starts with 2000 bacteria, and the number of bacteria doubles every 10 minutes. If this continues indefinitely, how long (in hours) will it take for the population to reach 8×10158 \times 10^{15} ? What is the differential equation governing the behaviour of the population of the bacteria?
Practice
Use the fact that the amount of a drug in a patient at time tt in hours is given by D(t)=D0e−ln⁡(12)tD\left(t\right)=D_0e^{-\ln\left(12\right)t} mg with the initial dose of the drug being 60 mg. The next dosage of the drug cannot be administered until there is only 5 mg of the drug left in the patient's body. After how long, in hours, should the patient wait to take the next dose?
A particular item sells for $4,000. However, according to market statistics, the value of the item will decline continuously in value at a rate of 4% every year. How many years are required for the value of the item to depreciate to $1,500?
A drug in the form of a 200 mg pill is given to a patient at time t=0t=0. The drug diminish at a rate proportional to the amount of drug remaining in the body. Suppose that after 5 hours only 100 mg of the drug remains. (t is in hours)
Q:\bf{Q:} A certain bacteria has an exponential growth rate. At point (8,5)(8,5), the slope of the growth curve is 44. Find the constant CC and kk such that y(t)=Cekty(t)=Ce^{kt}.
A bacteria population follows the equation y(t)=y0 ekty\left(t\right)=y_0\ e^{kt}. At time t=3t=3 months, the population size is 550550. At time t=6t=6 , the population rises to 13751375. What will the population be after one year?
Suppose a rabbit population is growing exponentially and quadruples every three months. If there are 4 rabbits to begin with, what will the population be after one year? (Round to the nearest whole number, if necessary)
Exponential Growth and Decay: Half Life
A scientist isolates 48 grams of a radioactive substance in the lab at 1PM. At 5PM it weights 6 grams. What is the half-life of the substance?
Practice: Radioactive Decay with Given Value
Q:\textbf{Q:} A radioactive substance has a half-life of 50 years. At the current time t=75 years, 200g of the substance remains. How long will it take from now for the substance to be reduced down to 25 g?
Enter your answer in years, do not include units.
Exponential Growth and Decay
Use the fact that the amount of a drug in a patient at time tt is given by D(t)=D0e−ln⁡(12)tD\left(t\right)=D_0e^{-\ln\left(12\right)t} mg with the initial dose of the drug being 60 mg. The next dosage of the drug cannot be administered until there is only 5 mg of the drug left in the patient's body. After how long, in hours, should the patient wait to take the next dose?
Exponential Growth and Decay
Suppose a rabbit population is growing exponentially and quadruples every three months. If there are 4 rabbits to begin with, what will the population be after one year? (Round to the nearest whole number, if necessary)
Half-Life: Exponential Growth and Decay
The radioactive decay of carbon 14 is given by the equation:
M(t)=M0e−ln⁡(2)5370tM\left(t\right)=M_0e^{-\frac{\ln\left(2\right)}{5370}t}
Where M0 is the initial mass, t is the time measured in years, and M(t) is the remaining mass at time t.
Exponential growth and decay
A radioactive material has a half-life of 400 years. 75%75\%of the
original material is remaining. Determine how many years have passed since it started decaying.
Exponential Growth and Decay
A drug in the form of a 200 mg pill is given to a patient at time t=0t=0. The drug diminish at a rate proportional to the amount of drug remaining in the body. Suppose that after 5 hours only 100 mg of the drug remains. (t is in hours)
Application to Growth and Decay
The initial dose of antibiotic ampicillin given to a patient is 250 mg. Assuming that the drug decays at a rate of 0.3 hr−10.3 \space hr^{-1}, the number of hours it takes for the drug to decay to 14\frac{1}{4} of the initial dose is given by
Use the fact that the amount of a drug in a patient at time tt is given by D(t)=D0e−ln⁡(12)tD\left(t\right)=D_0e^{-\ln\left(12\right)t} mg with the initial dose of the drug being 60 mg. The next dosage of the drug cannot be administered until there is only 5 mg of the drug left in the patient's body. After how long, in hours, should the patient wait to take the next dose?
Half-Life
The radioactive decay of carbon 14 is given by the equation:
M(t)=M0e−ln⁡(2)5370tM\left(t\right)=M_0e^{-\frac{\ln\left(2\right)}{5370}t}
Where M0 is the initial mass, t is the time measured in years, and M(t) is the remaining mass at time t.
A radioactive substance has a half-life of 110 years. How much of the 500g substance will remain after 200 years? (enter as exact value in base 1/2, do not include units)
A bacteria population follows the equation y(t)=y0 ekty\left(t\right)=y_0\ e^{kt}. At time t=3t=3 months, the population size is 550550. At time t=6t=6 , the population rises to 13751375. What will the population be after one year? (enter as simplified fraction, do not include any units)
Use the fact that the amount of a drug in a patient at time tt is given by D(t)=D0e−ln⁡(12)tD\left(t\right)=D_0e^{-\ln\left(12\right)t} mg with the initial dose of the drug being 60 mg. The next dosage of the drug cannot be administered until there is only 5 mg of the drug left in the patient's body. After how long, in hours, should the patient wait to take the next dose?
Q:\bf{Q:} A certain bacteria has an exponential growth rate. At point (8,5)(8,5), the slope of the growth curve is 44. Find the constant CC and kk such that y(t)=Cekty(t)=Ce^{kt}.
Q:\bf{Q:} The half-life of a radioactive substance is 108 years. What percentage of the present amount is remaining after 270 years?
Q:\bf{Q:} A type of bacteria doubles every 1010 minutes. How long does it take to have 3276832768 thousand bacteria if initially there were 11 thousand bacteria?
Exponential Growth and Decay: Radioactive Decay with Given Value
A radioactive substance has a half-life of 50 years. At the current time t=75 years, 200g of the substance remains. How long will it take from now for the substance to be reduced down to 25 g?
Enter your answer in years, do not include units.
Population Growth with 2 Values
The size of a certain chicken population at time t (in years) can be modeled by y(t)=y0 ekty\left(t\right)=y_0\ e^{kt} for some constant kk and y(0)=y0y\left(0\right)=y_0 . At time t=2t=2 , the population size is 300 and at time t=4t=4 , the population size increases to 1200. Find the values of y0y_0 and kk . When will the population exceed 300(2)7300\left(2\right)^7 ?
Practice: Radioactive Decay
Q:\textbf{Q:} An unknown radioactive particle has a half-life of 200 years. An initial 50 gram sample was placed in a dish. How much of the substance will remain after 230 years?
Enter your answer as an exact value, do not include units.
A type of bacteria doubles every 10 minutes. How long doe it take to have 3276832768 thousand bacteria if initially there were 1 thousand bacteria?
The latest phone model sells for $4,000. However, according to market statistics, the value of the phone will decline continuously in value at a rate of 4% every year. How many years are required for the value of the phone to depreciate to $1,500?
The number of a certain type of bacteria grows exponentially from 100 to 200 in 3 hours. In a fermentation process, 50 of these bacteria live. After 9 hours, 150 bacteria die due to feed rate shortage. Approximately how long will it take them to grow from that time to 1410 bacteria?
The half-life of a radioactive substance is 108 years. What percentage of the present amount is remaining after 270 years?
Exponential Growth and Decay
A certain bacteria culture starts with 2000 bacteria, and the number of bacteria doubles every 10 minutes. If this continues indefinitely, how long (in hours) will it take for the population to reach 8×10158 \times 10^{15} ? What is the differential equation governing the behaviour of the population of the bacteria?
Exponential growth and decay
A sample of radioactive material decayed to 15% of its original amount after 10 years. What percentage is remaining after another 5 years?
Exponential growth and decay
A bacteria culture grows exponentially. After one day, there are 40×10440\times10^4 bacteria. After 5 days, there are 40×10640\times10^6 bacteria.
Find the population of bacteria after ttdays.
Exponential growth and decay
A bacteria culture grows exponentially. After one day, there are 40×10440\times10^4 bacteria. After 5 days, there are 40×10640\times10^6 bacteria.
Find the initial amount of bacteria.
Exponential growth and decay
A sample of radioactive material decayed to 15% of its original amount after 10 years. What is the half-life of the material?
Exponential growth and decay
Two cities, Growth and Decay, have populations that are respectively increasing and decreasing at different rates that are proportional to their current population. Growth’s population is now 4 millions and was 2 millions 10 years ago. Decay’s population is now 6 millions and was 8 millions 10 years ago.
In how many years from now will Growth and Decay’s populations be equal?
Exponential growth and decay
Two cities, Growth and Decay, have populations that are respectively increasing and decreasing at different rates that are proportional to their current population. Growth’s population is now 4 millions and was 2 millions 10 years ago. Decay’s population is now 6 millions and was 8 millions 10 years ago.
In how many years from now will Growth’s population equal 5 millions?
Exponential Growth and Decay
After the Fukushima nuclear meltdown, 40 grams of radioactive Iodine 131 (half-life of 8 days) was released into the atmosphere. If it takes 30 days for this material to travel to Vancouver, how much radioactive material is left when it arrives to Vancouver?