Question

Radioactive Decay The amount \(A\) (in grams) of radioactive plutonium remaining in a 20 -gram sample after \(t\) days is given by the formula $$ A=20 \cdot(1 / 2)^{t / 140} $$ At what rate is the plutonium decaying when \(t=2\) days? Answer in appropriate units. rate \(\approx 0.098\) grams/day

Short answer

The rate at which the plutonium is decaying when \(t=2\) days is approximately \(0.098\) grams/day.

Step-by-step solution

Identify the rate function

First, identify the quantity that is changing. Here, the amount of plutonium \(A\) is decreasing over time \(t\). The function describing this action is \(A=20 \cdot(1 / 2)^{t / 140}\).

Differentiate the function

The next step is to evaluate the rate at which the amount \(A\) is changing. This requires differentiating the function with respect to \(t\). Apply chain rule to find its derivative. Hence, \( dA/dt = 20 \cdot (1/2)^{t/140} \cdot ln(1/2) \cdot (1/140) \) .

Substitute for \(t\)

Substitute the value of \(t = 2\) days into the derivative to find the rate at \(t = 2\) days. Therefore, rate at \(t = 2\) is \( dA/dt = 20 \cdot (1/2)^{2/140} \cdot ln(1/2) \cdot (1/140) \). Calculate this expression to find the rate.

Interpretation of the Result

The value calculated from the previous step represents the rate at which the amount of plutonium is decaying at \(t = 2\) days. Since \( dA/dt\) represents rate of change of \(A\) with \(t\), the negative number indicates that \(A\) is decreasing. The rate should be expressed in appropriate units, in this case, grams/day.

Key concepts

Differentiation

Picture a smooth curve on a graph, representing some quantity changing over time—this could be the height of a rocket after launch, your speed during a morning run, or as in our textbook example, the amount of radioactive plutonium left over time. Differentiation is the mathematical tool that helps us understand rates of change at any given point on such curves. It's like having a speedometer for any process that can be graphed. In calculus, the derivative of a function at a point is the rate at which the function's value is changing at that point. For example, if the function describes the amount of plutonium remaining over time, the derivative of that function with respect to time would tell us how quickly the plutonium is decaying at any given moment.

To find the rate of change for the function we used the differentiation process which, simply put, involves taking the original amount function of the radioactive substance, given by the equation \( A = 20 \cdot (1/2)^{t/140} \), and applying rules of calculus to find its derivative with respect to time \(t\). The derivative, denoted as \(\frac{dA}{dt}\), represents the instantaneous rate of decay of plutonium. In our exercise, differentiation was key to understand how fast the remaining amount of plutonium is changing at a specific time—precisely what we needed to find out.

Exponential Decay

Exponential decay models describe processes that decrease at a rate proportional to their current value. This kind of decay is everywhere—radioactive substances lose mass, populations shrink, even the brightness of a star diminishes over time, all following exponential patterns. Why? Because the greater the amount (or mass, or brightness), the faster the decline, creating a curve that swoops downwards on a graph.

Particularly with radioactive substances, such as plutonium in our exercise, the concept of half-life is fundamental to exponential decay. The half-life is the time it takes for half of the substance to decay. In the provided formula \( A = 20 \cdot (1/2)^{t/140} \), 140 is the half-life of our plutonium sample in days, which is essentially the 'speed' of the exponential decay. Understanding exponential decay is crucial when solving for the rate at which a quantity is decreasing exponentially, as it affects how we interpret changes over time and helps us predict future values.

Chain Rule

Imagine unpicking a knotted chain—one link leads invariably to another. That's somewhat how the chain rule in differentiation works; it helps us handle composite functions where one function is nested inside another. Think of it like opening up a Russian nesting doll, where each doll is a function within a function, and you apply differentiation step by step.

In the context of radioactive decay and the exercise at hand, the chain rule allowed us to differentiate a composite function—the exponent \(t/140\) within the decay function \( A=20 \cdot (1/2)^{t/140} \). Here's why we needed the chain rule: when differentiating, the outer function (the exponential decay) changes according to its own rule, and then, multiplied by the derivative of the inner function (the linear change of time in days), it unveils the full rate of change. This way, we provided a complete picture of how the plutonium's rate of decay depends on time. Understanding the chain rule was essential to calculate the exact decay rate when \(t = 2\) days.