The amount of oil pumped from one of the wells decreases at the continuous rate of 10% per year. When will the

Exercise 5 · Q4

Q. The amount of oil pumped from one of the wells decreases at the continuous rate of 10% per year. When will the wells output fall to one-fourth of its present value?

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Concept understanding — Exponential Decay Half Life

Exponential Decay Half-Life: From Intuition to Precision

Imagine you have a freshly poured cup of coffee. It's hot — say 80°C. The room is at 25°C. You know the coffee cools down, but it doesn't cool at a constant rate. In the first minute, it drops a lot. In the second minute, it drops less. By the tenth minute, it's barely changing. That slowing-down pattern — where the rate of change is proportional to the amount left — is exponential decay.

Now, the half-life is a single number that captures the speed of that decay. It answers the question: How long does it take for half of whatever we're measuring to disappear?

The Intuition: A Simple Story

Suppose you have 100 radioactive atoms of a certain element. After one half-life, 50 atoms remain. After another half-life, 25 remain. After another, 12.5 (on average). Notice the pattern: each half-life, the quantity is cut in half. It doesn't matter whether you start with 1000 atoms or 10 atoms — the time to go from any number to half that number is always the same.

This is the key insight: the half-life is constant. It doesn't depend on how much you have. That's what makes it so useful. If a drug has a half-life of 4 hours in your body, then every 4 hours, half of what's left gets eliminated. After 8 hours, only a quarter remains. After 12 hours, an eighth.

Note

Exponential decay is not linear. In linear decay, you'd lose the same amount each hour (e.g., 10 mg per hour). In exponential decay, you lose the same fraction each hour (e.g., 50% per hour). That fraction is what the half-life encodes.

The Precise Statement

Let $N_0$ be the initial quantity (atoms, drug concentration, temperature difference, etc.). Let $N(t)$ be the quantity after time $t$ . Let $T_{1/2}$ be the half-life.

The defining property of half-life is:

N(T_{1/2}) = \frac{1}{2} N_0

After two half-lives:

N(2T_{1/2}) = \frac{1}{4} N_0

After $n$ half-lives:

N(nT_{1/2}) = \left(\frac{1}{2}\right)^n N_0

This works even for fractional half-lives. After 0.5 half-lives, you'd have about 70.7% remaining (since $\sqrt{1/2} \approx 0.707$ ).

The Mathematical Formula

Exponential decay follows the equation:

N(t) = N_0 e^{-\lambda t}

where $\lambda$ is the decay constant — the probability per unit time that a single atom decays. The half-life and decay constant are related by a simple formula:

T_{1/2} = \frac{\log 2}{\lambda}

Why $\log 2$ ? Because we want $N(T_{1/2}) = \frac{1}{2} N_0$ . Plugging into the decay equation:

N_0 e^{-\lambda T_{1/2}} = \frac{1}{2} N_0

Cancel $N_0$ :

e^{-\lambda T_{1/2}} = \frac{1}{2}

Take natural log of both sides:

-\lambda T_{1/2} = \log\left(\frac{1}{2}\right) = -\log 2

So:

\lambda T_{1/2} = \log 2

And therefore:

$T_{1/2} = \frac{\log 2}{\lambda}$ …

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With continuous decay at 10% per year the output is $y=y_0e^{-0.1t}$ ; setting $y=\tfrac14 y_0$ gives $t=\frac{\log 4}{0.1}\approx 13.86$ years.

y(t)=y_0e^{-kt}

Model. The output decreases at a continuous rate of 10% per year, so $k=0.10$ and $y=y_0e^{-0.1t}$ .
Condition. Output falls to one-fourth: $\dfrac14 y_0=y_0e^{-0.1t}\Rightarrow e^{-0.1t}=\dfrac14$ . …

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Q.The amount of oil pumped from one of the wells decreases at the continuous rate of 10% per year. When will the wells output fall to one-fourth of its present value?

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