The growth of a quantity N (t) at any constant t is given by `(d N (t))/(dt) = alpha N (t)` . Given that N (t) = ` ce^(kt) .c` is a constant . What is the value of `alpha` ?

Options

A. c
B. k
C. `c + k`
D. ` c- k`

Answer

To solve the problem, we need to find the value of \( \alpha \) given the equation for the growth of a quantity \( N(t) \) and its derivative. Let's break it down step by step.

### Step 1: Write down the given equation
We start with the differential equation that describes the growth of the quantity:
\[ \frac{dN(t)}{dt} = \alpha N(t) \]
We are also given that:
\[ N(t) = c e^{kt} \]
where \( c \) is a constant.

### Step 2: Differentiate \( N(t) \)
Next, we differentiate \( N(t) \) with respect to \( t \):
\[ \frac{dN(t)}{dt} = \frac{d}{dt}(c e^{kt}) \]
Using the chain rule, we differentiate:
\[ \frac{dN(t)}{dt} = c \cdot k e^{kt} \]
This simplifies to:
\[ \frac{dN(t)}{dt} = k c e^{kt} \]

### Step 3: Substitute \( N(t) \) into the derivative
Since we know that \( N(t) = c e^{kt} \), we can substitute this into our derivative:
\[ \frac{dN(t)}{dt} = k N(t) \]

### Step 4: Compare with the original equation
Now we have:
\[ \frac{dN(t)}{dt} = k N(t) \]
We can compare this with the original equation:
\[ \frac{dN(t)}{dt} = \alpha N(t) \]
From this comparison, we see that:
\[ \alpha = k \]

### Conclusion
Thus, the value of \( \alpha \) is:
\[ \alpha = k \]