Growth Decay Calculator: Physics Formula & Uses

This calculator works out exponential growth or decay from a starting value, a rate, and a time period, using the standard physics and mathematics formula. You enter the initial amount, the growth or decay rate (as a percentage per time unit), and the elapsed time, and you get the final amount after that time. It also shows the same result expressed as a multiplier so you can check the math yourself.

Formula

The core equation is N(t) = N₀ · e^kt, where:

• N(t) = the amount after time t (same unit as N₀)
• N₀ = the initial amount at time zero (any unit: grams, cells, dollars, etc.)
• k = the growth rate constant (per unit time). If you enter a percentage rate r, then k = r/100. For decay, k is negative.
• t = elapsed time (seconds, hours, years, etc., as long as it matches the rate’s time unit)
• e ≈ 2.71828, the base of natural logarithms

This is the continuous exponential model, used in physics for radioactive decay, in biology for unrestricted population growth, and in finance for continuously compounded interest. It assumes the rate is constant and proportional to the current amount.

Worked Example

Using the default inputs: N₀ = 100, r = 5% per time unit, t = 10 time units.

1. Convert the percentage rate to a decimal: k = 5/100 = 0.05 per time unit.
2. Multiply the rate by time: k·t = 0.05 × 10 = 0.5.
3. Raise e to that power: e^0.5 ≈ 1.64872.
4. Multiply by the initial amount: 100 × 1.64872 = 164.872.
5. Rounded to 4 decimal places, the final amount is 164.872, and the multiplier is 1.6487.

For decay, you would enter a negative rate (e.g., -5%), and the multiplier would be less than 1, giving a final amount smaller than the start.

When to Use This Calculator

Use this model when a quantity changes at a rate proportional to its current value. Common real cases:

• Radioactive decay: The number of unstable nuclei decreases exponentially. The half-life is related to k by t_1/2 = ln(2)/|k|.
• Population growth (unrestricted): Bacteria, cells, or small populations with abundant resources grow exponentially until limits kick in.
• Cooling or heating: Newton’s law of cooling uses a similar exponential approach to the ambient temperature, but with an offset; the simple form here works only if the final temperature is zero.
• Discharging capacitor: Voltage across a capacitor discharging through a resistor decays exponentially with time constant RC = 1/|k|.
• Drug elimination: Many drugs leave the bloodstream at a rate proportional to concentration, giving exponential decay.

If the rate is given as a doubling time or half-life, convert it first: k = ln(2)/t_double for growth, or k = -ln(2)/t_half for decay.

How to Read the Result

The final amount is the value after exactly the time you entered. The multiplier tells you how many times the initial amount remains. For growth, the multiplier is greater than 1; for decay, it is between 0 and 1. For example, a multiplier of 0.5 means half remains. The calculator uses continuous compounding, which is slightly different from discrete compounding (e.g., yearly or monthly). If your problem states a rate “per year” but compounds monthly, you need to adjust the rate to a continuous equivalent, or use a discrete formula instead.

Common Mistakes and Limits

• Unit mismatch: The rate and time must use the same unit. If the rate is per year and you enter time in months, you must convert months to years first (divide by 12).
• Sign of the rate: A positive rate gives growth; a negative rate gives decay. Forgetting the negative sign turns decay into growth.
• Percentage vs. decimal: The calculator expects the rate as a percentage (e.g., 5 for 5%). If you enter 0.05, you get a result that is far too small.
• Continuous vs. discrete: This formula assumes continuous change. For discrete compounding (like interest added once per year), use N(t) = N₀(1 + r/100)^t instead.
• Model limits: Exponential growth never stops, so it cannot model populations with limited resources or reactions that slow down. It also assumes the rate constant is truly constant over time, which is an idealization.

Relation to Half-Life and Doubling Time

The half-life (t_1/2) is the time for a decaying quantity to fall to half its starting value. It is calculated as t_1/2 = ln(2)/|k|. For growth, the doubling time is t_double = ln(2)/k. These are constant for exponential processes, so you can use them to find k if you know the half-life or doubling time. For instance, if a radioactive isotope has a half-life of 5 years, k = -ln(2)/5 ≈ -0.1386 per year. Enter that as the rate (as -13.86%) to find the remaining amount after any time.

Frequently Asked Questions

What is the difference between exponential growth and decay?

Growth means the quantity increases over time, so the rate constant k is positive. Decay means the quantity decreases, so k is negative. In both cases, the change per unit time is proportional to the current amount, which produces the characteristic exponential curve.

Can I use this for compound interest?

Yes, but only for continuously compounded interest. If interest is compounded monthly or annually, use the discrete formula A = P(1 + r/n)^nt. The continuous model gives a slightly higher final amount because interest is added constantly rather than in steps.

What does the rate constant k actually mean?

It is the fractional change per unit time. A k of 0.05 per year means the quantity grows by 5% of its current value each year (in continuous terms). Because growth is proportional, the actual increase gets larger as the quantity grows.

Why is e used instead of just a growth factor?

Using e makes the math continuous and elegant. It arises naturally when you compound growth infinitely often. The multiplier e^kt is equivalent to (1 + r/n)^nt as n approaches infinity. For most practical purposes, the difference between continuous and daily compounding is tiny.

How do I find the time to reach a certain amount?

Rearrange the formula to solve for t: t = ln(N/N₀) / k. For example, to find when a population doubles, set N/N₀ = 2, giving t = ln(2)/k. The calculator here does not solve for t directly, but you can try different time values until the final amount matches your target.

Does this work for negative initial amounts?

Mathematically, the formula works for any initial value, but in physics and biology, negative amounts usually don’t make sense. If you are modeling something like a temperature difference that can be negative, you can use it, but be careful interpreting the result.

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in a0 | Initial amount | - | 100 | 0
in r | Rate (% per time unit) | % | 5 | -100
in t | Time elapsed | time units | 10 | 0
out final | Final amount | - | a0 * exp((r/100) * t) | 4
out multiplier | Growth multiplier | - | exp((r/100) * t) | 4