Question

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

• A. \(1\)
• B. \(5\)
• C. \(169\)
• D. \(49\)

Hint:

To find the minimum distance between two moving points, first write the distance formula in terms of the parameter, then minimize the resulting expression.

Answer 1

The Correct Option is A

Step 1: Write the coordinates of both particles.
The given points are \[ P(t,t^3-16t-3) \] and \[ Q(t+1,t^3-6t-6) \] Let the coordinates of \(P\) be \[ (x_1,y_1)=(t,t^3-16t-3) \] and the coordinates of \(Q\) be \[ (x_2,y_2)=(t+1,t^3-6t-6) \]

Step 2: Use the distance formula.
Distance between \(P\) and \(Q\) is \[ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \] Now, \[ x_2-x_1=(t+1)-t=1 \] Also, \[ y_2-y_1=(t^3-6t-6)-(t^3-16t-3) \] \[ =t^3-6t-6-t^3+16t+3 \] \[ =10t-3 \] Therefore, \[ d=\sqrt{1^2+(10t-3)^2} \] \[ d=\sqrt{1+(10t-3)^2} \]

Step 3: Minimize the distance.
Since \[ (10t-3)^2\geq 0 \] we have \[ 1+(10t-3)^2\geq 1 \] Thus, \[ d=\sqrt{1+(10t-3)^2}\geq 1 \] The minimum value occurs when \[ 10t-3=0 \] \[ t=\frac{3}{10} \] At this value of \(t\), \[ d_{\min}=\sqrt{1+0} \] \[ d_{\min}=1 \]

Step 4: Final conclusion.
Hence, the minimum distance between the two particles is \[ \boxed{1} \]