The charge flowing in a conductor changes with time as `Q(t)=alpha t-beta t^(2)+yt^(3)`.Where `alpha,beta` and `gamma` are constants.Minimum value of current is:

Options

A. `alpha-(gamma^(2))/(3 beta)`
B. `beta-(alpha^(2))/(3 gamma)`
C. `alpha-(beta^(2))/(3 gamma)`
D. `alpha-(3 beta^(2))/(gamma)`

Answer

To find the minimum value of the current flowing through a conductor where the charge \( Q(t) \) is given by the equation:

\[ Q(t) = \alpha t - \beta t^2 + \gamma t^3 \]

we first need to determine the expression for current \( I(t) \). The current is defined as the rate of change of charge with respect to time, which can be expressed mathematically as:

\[ I(t) = \frac{dQ}{dt} \]

### Step 1: Differentiate \( Q(t) \)

We differentiate \( Q(t) \) with respect to \( t \):

\[ I(t) = \frac{d}{dt}(\alpha t - \beta t^2 + \gamma t^3) \]

Using the power rule of differentiation:

\[ I(t) = \alpha - 2\beta t + 3\gamma t^2 \]

### Step 2: Find the critical points

To find the minimum current, we need to find the critical points by setting the derivative of the current \( I(t) \) with respect to \( t \) equal to zero:

\[ \frac{dI}{dt} = -2\beta + 6\gamma t = 0 \]

Solving for \( t \):

\[ 6\gamma t = 2\beta \implies t = \frac{\beta}{3\gamma} \]

### Step 3: Determine the second derivative

To confirm that this critical point corresponds to a minimum, we compute the second derivative of the current:

\[ \frac{d^2I}{dt^2} = 6\gamma \]

The sign of \( 6\gamma \) will determine the concavity. If \( \gamma > 0 \), then \( \frac{d^2I}{dt^2} > 0 \), indicating a local minimum.

### Step 4: Substitute \( t \) back into \( I(t) \)

Now we substitute \( t = \frac{\beta}{3\gamma} \) back into the expression for \( I(t) \):

\[ I\left(\frac{\beta}{3\gamma}\right) = \alpha - 2\beta\left(\frac{\beta}{3\gamma}\right) + 3\gamma\left(\frac{\beta}{3\gamma}\right)^2 \]

Calculating each term:

1. The first term is simply \( \alpha \).
2. The second term becomes:

\[ -2\beta\left(\frac{\beta}{3\gamma}\right) = -\frac{2\beta^2}{3\gamma} \]

3. The third term becomes:

\[ 3\gamma\left(\frac{\beta^2}{9\gamma^2}\right) = \frac{\beta^2}{3\gamma} \]

### Step 5: Combine the terms

Now, combining these results:

\[ I\left(\frac{\beta}{3\gamma}\right) = \alpha - \frac{2\beta^2}{3\gamma} + \frac{\beta^2}{3\gamma} \]

This simplifies to:

\[ I\left(\frac{\beta}{3\gamma}\right) = \alpha - \frac{\beta^2}{3\gamma} \]

### Conclusion

Thus, the minimum value of the current is:

\[ \boxed{\alpha - \frac{\beta^2}{3\gamma}} \]