Exercise · ZW-EXPONENTIAL-GROWTH-AND-DECAY-010 · ★★★ Advanced
A quantity doubles every 3 hours. Which equation does it satisfy?
Choose one option.
Show answer
- C. \(y'=\tfrac{\ln2}{3}\,y\)
Explanation
We need \(y_0e^{3k}=2y_0\), i.e. \(k=\ln2/3\). A rate of \(2/3\) does not mean doubling every 3 hours.
Review the concept: Exponential growth and decay
A quantity \(y\) with initial value \(y_0\) grows or decays exponentially when its rate of change is proportional to itself, with constant \(k\): it grows if \(k>0\), decays if \(k<0\).
\[y'=ky\;\Longrightarrow\; y(t)=y_0\,e^{kt}\]
More practice problems: exponential growth and decay
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- ZW-EXPONENTIAL-GROWTH-AND-DECAY-006 In y'=ky with k<0 and y(0)>0, what happens to y(t) as t→∞?
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