Two particles P and Q located at the points Pt,t^3-16t-3,Qt+1,t^3-6t-6 are moving in a plane, the minimum distance between the points in their motion…

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Two particles P and Q located at the points Pt,t^3-16t-3,Qt+1,t^3-6t-6 are moving in a plane, the minimum distance between the points in their motion is

MARKS Content Team · Accepted Answer

Correct answer: (A) 1. Pt,t^3-16t-3,Qt+1,t^3-6t-6 ⇒ PQ=√((t+1-t)^2+t3-6t-6-t3+16t+3^2) = √(1+(10t-3)^2) Let, y=PQ^2=1+10t-3^2 Minimum value of y will be at dy/dt=0 i.e. 210t-3=0 or t=3/10 ∴ minimum distance = √(1+10×310-3^2)=1

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

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Relevant Questions

Two particles P and Q located at the points Pt,t3-16t-3,Qt+1,t3-6t-6 are moving in a plane, the minimum distance between the points in their motion is

Solution

Relevant Questions

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