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Question Number 102804 by bramlex last updated on 11/Jul/20

$$\mathrm{The}\mathrm{coordinates}\mathrm{of}\mathrm{two}\mathrm{points}\mathrm{A}$$ $$\mathrm{\&}\mathrm{B}\mathrm{are}(0,8)\mathrm{and}(9,4)$$ $$\mathrm{respectively}.\mathrm{The}\mathrm{point}\mathrm{P}\mathrm{with}$$ $$\mathrm{coordinate}(\mathrm{p},0)\mathrm{lies}\mathrm{on}\mathrm{the}$$ $$\mathrm{x}-\mathrm{axis}\mathrm{where}0<\mathrm{p}<9.\mathrm{Let}\mathrm{s}$$ $$\mathrm{denotes}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{length}\mathrm{of}$$ $$\mathrm{two}\mathrm{segments}\mathrm{PA}\mathrm{and}\mathrm{PB}.\mathrm{by}$$ $$\mathrm{expressing}\mathrm{s}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{p}$$ $$\mathrm{otherwise},\mathrm{show}\mathrm{that}\frac{\mathrm{ds}}{\mathrm{dp}}=$$ $$\frac{\mathrm{p}}{\sqrt{64+{\mathrm{p}}^2}}-\frac{9-\mathrm{p}}{\sqrt{16+{(9-\mathrm{p})}^2}}$$

Answered by bemath last updated on 11/Jul/20

$$s=PA+PB=\sqrt{p^2+64}$$ $$+\sqrt{{(9-p)}^2+16}$$ $$\frac{ds}{dp}=\frac{2p}{2\sqrt{64+p^2}}+$$ $$\frac{2(9-p)(-1)}{2\sqrt{16+{(9-p)}^2}}=$$ $$\frac{p}{\sqrt{64+p^2}}-\frac{(9-p)}{\sqrt{16+{(9-p)}^2}}◻$$

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