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 two positions.
Particle $P$ is at $\left(t,\;t^3-16t-3\right)$ and particle $Q$ is at $\left(t+1,\;t^3-6t-6\right)$, both depending on the common parameter $t$.
Step 2: Compute the coordinate gaps.
The horizontal gap is $(t+1)-t=1$. The vertical gap is $(t^3-6t-6)-(t^3-16t-3)=10t-3$.
Step 3: Form the squared distance.
Working with $D=d^2$ avoids square roots: $D(t)=1^2+(10t-3)^2=1+(10t-3)^2$.
Step 4: Minimise the squared distance.
Differentiating, $\frac{dD}{dt}=2(10t-3)\cdot 10$. Setting this to zero gives $10t-3=0$, so $t=\frac{3}{10}$.
Step 5: Confirm the minimum.
Since $(10t-3)^2\ge 0$ and equals $0$ at $t=\frac{3}{10}$, this clearly gives the least value of $D$.
Step 6: Evaluate the minimum distance.
At $t=\frac{3}{10}$, $D=1+0=1$, so the minimum distance is $d=\sqrt{1}=1$. \[ \boxed{1} \]