Question-123212

Question Number 123212 by 676597498 last updated on 24/Nov/20

Commented by 676597498 last updated on 24/Nov/20

$${pls}\:{i}\:{need}\:{help} \\ $$

Answered by MJS_new last updated on 24/Nov/20

mouse ((x),(y) ) = ((0),(1) ) + (((6p)),((−2p)) ) cat ((x),(y) ) = ((0),((−1)) ) + (((8q)),(q) ) the paths intersect at 6p=8q∧1−2p=−1+q ⇒ p=(8/(11))∧q=(6/(11)) ⇒ ⇒ ((x),(y) ) = (((48/11)),((−5/11)) ) but they don′t meet because p≠q. p>q means the mouse is late for the time when they are the nearest let p=q=t the position of the mouse is (((6t)),((1−2t)) ) the position of the cat is (((8t)),((−1+t)) ) their distance is (√((6t−8t)^2 +(1−2t−(−1+t)^2 ))= =(√(13t^2 −12t+4)) the minimum distance is at t=(6/(13)) and is ((4(√(13)))/(13))≈1.11m

$$\mathrm{mouse} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{6}{p}}\\{−\mathrm{2}{p}}\end{pmatrix} \\ $$$$\mathrm{cat} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}}\\{−\mathrm{1}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{8}{q}}\\{{q}}\end{pmatrix} \\ $$$$\mathrm{the}\:\mathrm{paths}\:\mathrm{intersect}\:\mathrm{at} \\ $$$$\mathrm{6}{p}=\mathrm{8}{q}\wedge\mathrm{1}−\mathrm{2}{p}=−\mathrm{1}+{q}\:\Rightarrow\:{p}=\frac{\mathrm{8}}{\mathrm{11}}\wedge{q}=\frac{\mathrm{6}}{\mathrm{11}}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{48}/\mathrm{11}}\\{−\mathrm{5}/\mathrm{11}}\end{pmatrix} \\ $$$$\mathrm{but}\:\mathrm{they}\:\mathrm{don}'\mathrm{t}\:\mathrm{meet}\:\mathrm{because}\:{p}\neq{q}.\:{p}>{q}\:\mathrm{means} \\ $$$$\mathrm{the}\:\mathrm{mouse}\:\mathrm{is}\:\mathrm{late} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{time}\:\mathrm{when}\:\mathrm{they}\:\mathrm{are}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{let} \\ $$$${p}={q}={t} \\ $$$$\mathrm{the}\:\mathrm{position}\:\mathrm{of}\:\mathrm{the}\:\mathrm{mouse}\:\mathrm{is}\:\begin{pmatrix}{\mathrm{6}{t}}\\{\mathrm{1}−\mathrm{2}{t}}\end{pmatrix} \\ $$$$\mathrm{the}\:\mathrm{position}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cat}\:\mathrm{is}\:\begin{pmatrix}{\mathrm{8}{t}}\\{−\mathrm{1}+{t}}\end{pmatrix} \\ $$$$\mathrm{their}\:\mathrm{distance}\:\mathrm{is} \\ $$$$\sqrt{\left(\mathrm{6}{t}−\mathrm{8}{t}\right)^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{2}{t}−\left(−\mathrm{1}+{t}\right)^{\mathrm{2}} \right.}= \\ $$$$=\sqrt{\mathrm{13}{t}^{\mathrm{2}} −\mathrm{12}{t}+\mathrm{4}} \\ $$$$\mathrm{the}\:\mathrm{minimum}\:\mathrm{distance}\:\mathrm{is}\:\mathrm{at}\:{t}=\frac{\mathrm{6}}{\mathrm{13}}\:\mathrm{and}\:\mathrm{is} \\ $$$$\frac{\mathrm{4}\sqrt{\mathrm{13}}}{\mathrm{13}}\approx\mathrm{1}.\mathrm{11m} \\ $$

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