Exponential Word Problems: Growth & Decay
Growth Formula: y = a (1 + r)t
Decay Formula: y = a (1 β r)t
where a = original number
r = rate (% in decimal form)
t = time periods
What is the half life ?
Half-life is the time required for a quantity to reduce to half its initial value.
The term is used generally to characterize any type of exponential decay. When solving half-life problems, use the following formula:
A = Amount remaining
A0 = Initial amount
t = time elapsed
H = half life
Doubling time :
An initial amount doubles every d.
For each problem, create a function to model the scenario.
Problem 1 :
A population π of 500 people doubles every 35 years π‘.
Solution :
a = 500
Doubles every 35 years. So, d = 35
f(t) = a(2)t/d
f(t) = 500 (2)t/35
Problem 2 :
Mr. Kelly bought a new tractor for his farm in New York. It cost him $150,000. Unfortunately, itβs value π£ depreciates in value by 5.4% per year π‘.
Solution :
Exponential decay :
v(t) = p(1 - r%)t
p = 150000, r = 5.4%
v(t) = 150000(1 - 5.4%)t
v(t) = 150000(1 - 0.054)t
= 150000 (0.946)t
Problem 3 :
A baseball card is worth $50 and its value π£ increases at a rate of 23.5% per year π‘
Solution :
Exponential growth :
v(t) = p(1 + r%)t
p = 50, r = 23.5%
v(t) = 50(1 + 23.5%)t
= 50(1 + 0.235)t
= 50(1.235)t
Problem 4 :
There is 500 grams π of radioactive material. Its halflife is 5,700 years, π‘
Solution :
A = A0 (1/2) t/h
Here A0 = 500, h = 5700
A = 500 (1/2) t/5700
Problem 5 :
700 grams of radioactive material π decays at a rate of 2.4% per year π‘. How much material will be there after 100 years ?
Solution :
Exponential decay :
m(t) = p(1 - r%)t
p = 700, r = 2.4%
m(t) = 700(1 - 2.4%)t
= 700(1 - 0.024)t
m(t) = 700(0.976)t
The quantity of material after 100 years :
m(100) = 700(0.976)100
= 61.67 grams.
So, after 100 years 61.67 grams of material will be there.
Problem 6 :
The new tires on a truck have a tread depth of 0.5 inches and decays at the rate of 1.6% per week. How deep will the tread be after 52 weeks ?
Solution :
Exponential decay :
m(t) = p(1 - r%)t
p = 0.5, r = 1.6%
m(t) = 0.5(1 - 1.6%)t
= 0.5(1 - 0.016)t
m(t) = 0.5(0.984)t
The deep of tread after 52 weeks :
t = 52
m(52) = 0.5(0.984)52
= 0.216
Approximately 0.22 inches
Problem 7 :
A car that is worth $25000, decreases in value by 15% per year. How much will the car be after 5 years ?
Solution :
Exponential decay :
v(t) = p(1 - r%)t
p = 25000, r = 15%
v(t) = 25000(1 - 15%)t
= 25000(1 - 0.15)t
= 25000(0.85)t
Value after 5 years :
When t = 5
v(5) = 25000(0.85)5
= 25000(0.4437)
= 11092.63
So, the worth of car after 5 years is $11092.63.
Problem 8 :
Mr. Brust IQ currently 173, but it is decaying at a rate of 4.5% every year. What will Mr. Brust's IQ be in 20 years ?
Solution :
Exponential decay :
v(t) = p(1 - r%)t
p = 173, r = 4.5 %
v(t) = 173(1 - 4.5%)t
= 173(1 - 0.045)t
v(t) = 173(0.955)t
After 20 years :
v(t) = 173(0.955)t
v(20) = 173(0.955)20
= 68.88
Problem 9 :
A plague of mice has hit Australia! Starting with only 30 mice, their population π increases by 650% every month, π.
Solution :
Exponential growth :
m(t) = p(1 + r%)m
p = 30, r = 650 %
m(t) = 30(1 + 650%)m
= 30(1 + 6.5)m
= 30(7.5)m
Problem 10 :
The rodent population π in a large city is being controlled by a new poison that kills half the population every 6 months π. There are currently 2,000,000 rodents.
Solution :
A = A0 (1/2) t/h
Here A0 = 2,000,000, h = 6
A = 2,000,000 (1/2) t/6
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