Two particles P and Q located at the points P | vedclass

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(A) The coordinates of the particles are $P(t, t^3 - 16t - 3)$ and $Q(t + 1, t^3 - 6t - 6)$.The distance $PQ$ between the two particles is given by th

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$1$

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(A) The coordinates of the particles are $P(t, t^3 - 16t - 3)$ and $Q(t + 1, t^3 - 6t - 6)$.The distance $PQ$ between the two particles is given by the distance formula:$PQ = \sqrt{((t + 1) - t)^2 + ((t^3 - 6t - 6) - (t^3 - 16t - 3))^2}$$PQ = \sqrt{(1)^2 + (t^3 - 6t - 6 - t^3 + 16t + 3)^2}$$PQ = \sqrt{1 + (10t - 3)^2}$To find the minimum distance,we minimize the square of the distance,$y = PQ^2 = 1 + (10t - 3)^2$.For the minimum value,we set the derivative with respect to $t$ to zero:$\frac{dy}{dt} = 2(10t - 3) 	imes 10 = 20(10t - 3) = 0$This gives $10t - 3 = 0$,or $t = \frac{3}{10}$.Substituting $t = \frac{3}{10}$ into the expression for $PQ$:$PQ_{min} = \sqrt{1 + (10(\frac{3}{10}) - 3)^2} = \sqrt{1 + (3 - 3)^2} = \sqrt{1 + 0} = 1$.Thus,the minimum distance is $1$.

Two particles $P$ and $Q$ located at the points $P(t, t^3 - 16t - 3)$ and $Q(t + 1, t^3 - 6t - 6)$ are moving in a plane. The minimum distance between the points during their motion is:

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