A certain radioactive material decays at a rate proportional to the amount present. At time t = 0, there are 50 milligrams. We use the following equation to model this decay: dP = KP. dt P(t) is the amount of this material at time t. After two years, 20% has decayed. (a) (b) How much is left after 5 years? (Round to four decimal places.) What is the half-life of this material? That is, find 71/2 so that P(71/2) = } P(0). (Round to four decimal places )
A certain radioactive material decays at a rate proportional to the amount present. At time t = 0, there are 50 milligrams. We use the following equation to model this decay: dP = KP. dt P(t) is the amount of this material at time t. After two years, 20% has decayed. (a) (b) How much is left after 5 years? (Round to four decimal places.) What is the half-life of this material? That is, find 71/2 so that P(71/2) = } P(0). (Round to four decimal places )
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem 1**
A certain radioactive material decays at a rate proportional to the amount present. At time \( t = 0 \), there are 50 milligrams. We use the following equation to model this decay:
\[
\frac{dP}{dt} = KP
\]
\( P(t) \) is the amount of this material at time \( t \). After two years, 20% has decayed.
(a) How much is left after 5 years? (Round to four decimal places.)
(b) What is the half-life of this material? That is, find \( t_{1/2} \) so that \( P(t_{1/2}) = \frac{1}{2}P(0) \). (Round to four decimal places.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11d9adce-31d3-450e-a738-16d52e01b365%2F972c41e7-bb9d-456f-a22e-23f034d988d8%2Fgq6ghis_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 1**
A certain radioactive material decays at a rate proportional to the amount present. At time \( t = 0 \), there are 50 milligrams. We use the following equation to model this decay:
\[
\frac{dP}{dt} = KP
\]
\( P(t) \) is the amount of this material at time \( t \). After two years, 20% has decayed.
(a) How much is left after 5 years? (Round to four decimal places.)
(b) What is the half-life of this material? That is, find \( t_{1/2} \) so that \( P(t_{1/2}) = \frac{1}{2}P(0) \). (Round to four decimal places.)
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