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Two particles $P$ and $Q$ located at the points with coordinates $P\left(t, t^3-16 t-3\right)$, $Q\left(t+1, t^3-6 t-6\right)$ are moving in a plane. The minimum distance between them in their motion is
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Verified Answer
The correct answer is:
1
Here,
$$
\begin{aligned}
& P Q=\sqrt{(t+1-t)^2+\left(t^3-6 t-6-t^3+16 t+3\right)^2} \\
& =\sqrt{1+(10 t-3)^2} \\
\therefore & P Q^2=1+(10 t-3)^2 \geq 1
\end{aligned}
$$
Hence, minimum value of $P Q$ is 1 .
$$
\begin{aligned}
& P Q=\sqrt{(t+1-t)^2+\left(t^3-6 t-6-t^3+16 t+3\right)^2} \\
& =\sqrt{1+(10 t-3)^2} \\
\therefore & P Q^2=1+(10 t-3)^2 \geq 1
\end{aligned}
$$
Hence, minimum value of $P Q$ is 1 .
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