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Last updated on October 6, 2025
In mathematics and physics, the decay formula is used to model the decrease of a quantity over time. This can pertain to various contexts, such as radioactive decay, depreciation of assets, or diminishing returns in investments. In this topic, we will learn about the decay formula and how it is applied in different scenarios.
Decay is a process where a quantity decreases over time. Let's learn the formula to calculate decay in various contexts.
Exponential decay describes a process where a quantity decreases at a rate proportional to its current value.
It is calculated using the formula:\( [ N(t) = N_0 \cdot e^{-kt} ] \)where \(( N(t))\) is the quantity at time \(( t ), ( N_0 ) \)is the initial quantity, k is the decay constant, and e is the base of the natural logarithm.
Radioactive decay is a specific type of exponential decay.
The formula for radioactive decay is: \([ N(t) = N_0 \cdot e^{-\lambda t} ]\) where\( ( \lambda )\) is the decay constant specific to the substance.
The half-life\( ( t_{1/2} )\) is related to \(( \lambda ) \) by the formula\( ( t_{1/2} = \frac{\ln(2)}{\lambda} ).\)
Depreciation models the decrease in value of an asset over time. One common method is straight-line depreciation, calculated as:\( [ \text{Depreciation} = \frac{\text{Cost} - \text{Salvage Value}}{\text{Useful Life}} ]\)
In exponential decay terms, depreciation can also be modeled with: \([ V(t) = V_0 \cdot (1 - r)^t ] \)where \(( V(t) )\) is the value at time\( ( t ), ( V_0 )\) is the initial value, and \(( r ) \)is the depreciation rate.
Decay formulas are crucial in many scientific and financial fields. Here are some important uses of decay formulas:
Students often find decay formulas tricky. Here are some tips to master them:
Students often make errors when using decay formulas. Here are some mistakes and ways to avoid them:
A radioactive substance has a half-life of 10 years. If the initial quantity is 100 grams, how much is left after 20 years?
The remaining amount is 25 grams.
Using the half-life formula, we know after one half-life (10 years), 50 grams remain. After another half-life (20 years total), 25 grams remain.
A car worth $20,000 depreciates at an annual rate of 10%. What is its value after 3 years?
The car's value is approximately $14,580.
Using the depreciation formula: \([ V(t) = 20,000 \cdot (1 - 0.10)^3 = 20,000 \cdot 0.9^3 = 14,580 ]\)
A radioactive isotope decays with a decay constant of 0.693 per year. What is its half-life?
The half-life is 1 year.
Using the formula \(( t_{1/2} = \frac{\ln(2)}{\lambda} ): [ t_{1/2} = \frac{\ln(2)}{0.693} = 1 \text{ year} ]\)
A machine costs $50,000 and has a salvage value of $10,000 after 10 years. What is the straight-line depreciation per year?
The straight-line depreciation is $4,000 per year.
Using the straight-line depreciation formula:\( [ \text{Depreciation} = \frac{50,000 - 10,000}{10} = 4,000 ]\)
If a population decreases from 500 to 250 in 5 years, what is the decay constant assuming exponential decay?
The decay constant is approximately 0.1386.
Using the formula\( ( N(t) = N_0 \cdot e^{-kt} ): [ 250 = 500 \cdot e^{-5k} ] [ e^{-5k} = 0.5 ] [ -5k = \ln(0.5) ] [ k = -\frac{\ln(0.5)}{5} \approx 0.1386 ]\)
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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