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: Write the given current function.
\[ i(t) = K t e^{-\alpha t} \]
Step 2: Differentiate $i(t)$ with respect to time.
\[ \frac{di(t)}{dt} = K \left( e^{-\alpha t} - \alpha t e^{-\alpha t} \right) \]
Step 3: Set the derivative equal to zero for maximum value.
\[ e^{-\alpha t} (1 - \alpha t) = 0 \]
Since $e^{-\alpha t} \neq 0$,
\[ 1 - \alpha t = 0 \]
Step 4: Solve for $t$.
\[ t = \frac{1}{\alpha} \]
Step 5: Conclusion.
The current $i(t)$ attains its maximum value at
\[ \boxed{t = \frac{1}{\alpha}} \]

Answer 2

Instead of differentiating \(i(t)\) directly and setting the result to zero, the maximum can be located using logarithmic differentiation, which turns the product rule into a simpler sum, and then each candidate value of \(t\) can be checked against the resulting condition.

Taking the natural log of \(i(t) = Kte^{-\alpha t}\) (for \(t>0\), where \(i(t)>0\)): \[ \ln i(t) = \ln K + \ln t - \alpha t \] Differentiating both sides with respect to \(t\): \[ \frac{1}{i}\frac{di}{dt} = \frac{1}{t} - \alpha \] At a maximum, \(di/dt = 0\), and since \(i \ne 0\) at the point of interest, this requires \(\dfrac{1}{t}-\alpha = 0\).

1. \(t=\alpha\): Substituting gives \(\dfrac{1}{\alpha}-\alpha\), which is zero only if \(\alpha=1\), not for general \(\alpha\).
2. \(t=2\alpha\): Substituting gives \(\dfrac{1}{2\alpha}-\alpha\), which is nonzero for general \(\alpha\).
3. \(t=1/\alpha\): Substituting gives \(\dfrac{1}{1/\alpha}-\alpha = \alpha-\alpha=0\), satisfying the maximum condition for any value of \(\alpha\).
4. \(t=\alpha^2\): Substituting gives \(\dfrac{1}{\alpha^2}-\alpha\), which is nonzero in general.

Only \(t=1/\alpha\) makes the logarithmic derivative vanish for every value of \(\alpha\), confirming it as the location of the maximum.

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