Math / Calculus

Question $\mathrm{t}=18$ days? Answer in appropriate units.

Studdy Solution

STEP 1

1. The formula given is for exponential decay, specifically for radioactive decay.
2. We are asked to find the rate of decay at $t = 18$ days, which involves finding the derivative of the function $A(t)$ with respect to $t$ .
3. The derivative will give us the rate of change of the amount of plutonium over time, which is the decay rate.

STEP 2

1. Differentiate the function $A(t)$ to find the rate of decay.
2. Evaluate the derivative at $t = 18$ days.
3. Interpret the result in appropriate units.

STEP 3

Differentiate the function $A(t) = 425 \left( \frac{1}{2} \right)^{\frac{t}{13.1}}$ with respect to $t$ .
To differentiate, we use the chain rule and the fact that the derivative of $a^u$ with respect to $u$ is $a^u \ln(a)$ .
Let $u = \frac{t}{13.1}$ , then:
$A(t) = 425 \left( \frac{1}{2} \right)^u$
The derivative of $A(t)$ with respect to $t$ is:
$A'(t) = 425 \cdot \left( \frac{1}{2} \right)^u \cdot \ln\left(\frac{1}{2}\right) \cdot \frac{1}{13.1}$

STEP 4

Evaluate the derivative $A'(t)$ at $t = 18$ days.
Substitute $t = 18$ into the expression for $u$ :
$u = \frac{18}{13.1}$
Then substitute $u$ into the derivative:
$A'(18) = 425 \cdot \left( \frac{1}{2} \right)^{\frac{18}{13.1}} \cdot \ln\left(\frac{1}{2}\right) \cdot \frac{1}{13.1}$
Calculate the numerical value of $A'(18)$ .

STEP 5

Interpret the result.
The value of $A'(18)$ represents the rate of decay of plutonium at $t = 18$ days. Since the derivative is negative, it indicates a decrease in the amount of plutonium.
The units of $A'(18)$ are grams per day, as it represents the rate of change of the amount of plutonium over time.
The rate at which the plutonium is decaying when $t = 18$ days is approximately $A'(18)$ grams per day.

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