Question

The response of a network is $i(t) = K t e^{-\alpha t}$ for $t \ge 0$, where $\alpha$ is real and positive. The value of $t$ at which $i(t)$ will become maximum is,

• A. $\alpha$
• B. $2\alpha$
• C. $1/\alpha$
• D. $\alpha^2$

Hint:

For functions of the form $t e^{-\alpha t}$, the maximum always occurs at $t = 1/\alpha$.

Answer 1

The Correct Option is C

Step 1: Take the logarithmic derivative: \(\ln i = \ln K+\ln t - \alpha t\), so \(\dfrac{i'}{i} = \dfrac{1}{t}-\alpha\).
Step 2: Set the derivative to zero (since \(i \ne 0\) at the maximum): \(\dfrac{1}{t}=\alpha\).
\[ \boxed{t = \frac{1}{\alpha}} \]

Answer 2

The function \(i(t)=Kte^{-\alpha t}\) is a product of a term \(t\) that grows steadily and a term \(e^{-\alpha t}\) that decays exponentially. The maximum occurs at the instant where the relative rate of growth of the \(t\) factor exactly equals the relative rate of decay of the exponential factor, since before that point growth dominates and after it decay dominates.

The relative growth rate of \(t\) is \(\dfrac{d}{dt}(\ln t) = \dfrac{1}{t}\), and the relative decay rate of \(e^{-\alpha t}\) is \(\alpha\) (constant). Setting these equal: \[ \frac{1}{t} = \alpha \]

1. \(t=\alpha\): This would only balance the two rates if \(\alpha=1\), not in general.
2. \(t=2\alpha\): Gives a relative growth rate of \(1/(2\alpha)\), which is smaller than \(\alpha\) except in special cases, so the rates are not balanced.
3. \(t=1/\alpha\): Gives a relative growth rate of \(\alpha\), exactly matching the exponential's decay rate, so this is where growth and decay exactly trade off.
4. \(t=\alpha^2\): Gives a relative growth rate of \(1/\alpha^2\), which generally does not equal \(\alpha\).

Therefore, the correct answer is \(1/\alpha\).