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Question

(A) Let $O$ be the origin $(0,0)$. Let the position of $X$ at time $t$ be $x(t) = 4t$ along $OA$ and the position of $Y$ at time $t$ be $y(t) = 3t$ al

Vedclass · Accepted Answer

$\sqrt{37} 	ext{ km/h}$

Vedclass · Answer

(A) Let $O$ be the origin $(0,0)$. Let the position of $X$ at time $t$ be $x(t) = 4t$ along $OA$ and the position of $Y$ at time $t$ be $y(t) = 3t$ along $OB$.Let $A$ be the shortest distance between $X$ and $Y$ at time $t$.Using the Law of Cosines in $	riangle OXY$:$A^2 = (4t)^2 + (3t)^2 - 2(4t)(3t) \cos(120^{\circ})$Since $\cos(120^{\circ}) = -\frac{1}{2}$,we have:$A^2 = 16t^2 + 9t^2 - 24t^2 \left(-\frac{1}{2}ight)$$A^2 = 25t^2 + 12t^2 = 37t^2$$A = \sqrt{37}t$Differentiating with respect to $t$:$2A \frac{dA}{dt} = 37(2t)$$\frac{dA}{dt} = \frac{37t}{A}$Substituting $A = \sqrt{37}t$:$\frac{dA}{dt} = \frac{37t}{\sqrt{37}t} = \sqrt{37} 	ext{ km/h}$.Thus,the rate at which the distance is increasing is $\sqrt{37} 	ext{ km/h}$.

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