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The growth of a quantity $N(t)$ at any instant $t$ is given by
$\frac{d N(t)}{d t}=\alpha N(t)$. Given that $N(t)=c e^{k \pi}, c$ is a constant. What
is the value of $\alpha ?$
Options:
$\frac{d N(t)}{d t}=\alpha N(t)$. Given that $N(t)=c e^{k \pi}, c$ is a constant. What
is the value of $\alpha ?$
Solution:
1607 Upvotes
Verified Answer
The correct answer is:
$k$
Given $\mathrm{N}(\mathrm{t})=c e^{k t}$
Diff. both side w.r.t. ' $t^{\prime}$
$\begin{aligned} \therefore \quad \frac{d N(t)}{d t} &=\frac{d}{d t} c e^{k t}=k\left(c e^{k t}\right) & & \\=k[N(t)] & &(\text { by Defn. of } \mathrm{N}(t)) \end{aligned}$
But $\frac{d N(t)}{d t}=\alpha N(t) \quad$ (given)
$\Rightarrow \alpha=k$
Diff. both side w.r.t. ' $t^{\prime}$
$\begin{aligned} \therefore \quad \frac{d N(t)}{d t} &=\frac{d}{d t} c e^{k t}=k\left(c e^{k t}\right) & & \\=k[N(t)] & &(\text { by Defn. of } \mathrm{N}(t)) \end{aligned}$
But $\frac{d N(t)}{d t}=\alpha N(t) \quad$ (given)
$\Rightarrow \alpha=k$
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