Math Problem Statement

Suppose that the amount, in grams, of plutonium present in a given sample is determined by the function A(t) = 3.00e^(-0.052t), where t is measured in years. What was the initial amount present?

Solution

The given function for the plutonium decay is:

A(t)=3.00e0.052tA(t) = 3.00e^{-0.052t}

The problem asks for the initial amount of plutonium present, which is the value of A(t)A(t) when t=0t = 0.

Substitute t=0t = 0 into the function:

A(0)=3.00e0.052(0)A(0) = 3.00e^{-0.052(0)}

Since e0=1e^0 = 1, the expression simplifies to:

A(0)=3.001=3.00gramsA(0) = 3.00 \cdot 1 = 3.00 \, \text{grams}

Thus, the initial amount of plutonium present is:

3.00grams\boxed{3.00 \, \text{grams}}

Would you like me to explain the decay function further or clarify any steps? Here are some related questions for practice:

  1. How much plutonium is left after 10 years?
  2. What does the decay constant (-0.052) represent in this function?
  3. How long will it take for the plutonium to decay to half of its initial value (half-life)?
  4. What is the value of A(t)A(t) when t=5t = 5 years?
  5. How would the function change if the initial amount were 5 grams instead of 3 grams?

Tip: Exponential decay functions are common in radioactive decay, population studies, and financial depreciation problems.

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Math Problem Analysis

Mathematical Concepts

Exponential Functions
Radioactive Decay
Initial Value

Formulas

A(t) = A_0 * e^(-kt)

Theorems

Properties of Exponential Functions
Exponential Decay Models

Suitable Grade Level

Grades 9-12