Question-159748

Question Number 159748 by 07814040830 last updated on 20/Nov/21

Answered by 07814040830 last updated on 20/Nov/21

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Answered by TheSupreme last updated on 21/Nov/21

(4t−0)^2 +(2t)^2 +(2t−12)^2 (∂d/∂t)=0 → 2×4×(4t)+2×2(2t)+2×2(2t−12)=48t−48 t=1 4^2 +2^2 +100=(√(120)) (5+3t)^2 +(5+4t)^2 +(−3−5t)^2 2(5+3t)3+2(5+4t)4+2(3+5t)5=0 30+18t+40+32t+30+50t=0 100+81t=0 t=1 8^2 +9^2 +8^2 =128+81=(√(209))

$$\left(\mathrm{4}{t}−\mathrm{0}\right)^{\mathrm{2}} +\left(\mathrm{2}{t}\right)^{\mathrm{2}} +\left(\mathrm{2}{t}−\mathrm{12}\right)^{\mathrm{2}} \\ $$$$\frac{\partial{d}}{\partial{t}}=\mathrm{0}\:\rightarrow\:\mathrm{2}×\mathrm{4}×\left(\mathrm{4}{t}\right)+\mathrm{2}×\mathrm{2}\left(\mathrm{2}{t}\right)+\mathrm{2}×\mathrm{2}\left(\mathrm{2}{t}−\mathrm{12}\right)=\mathrm{48}{t}−\mathrm{48} \\ $$$${t}=\mathrm{1} \\ $$$$\mathrm{4}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{100}=\sqrt{\mathrm{120}} \\ $$$$ \\ $$$$\left(\mathrm{5}+\mathrm{3}{t}\right)^{\mathrm{2}} +\left(\mathrm{5}+\mathrm{4}{t}\right)^{\mathrm{2}} +\left(−\mathrm{3}−\mathrm{5}{t}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{5}+\mathrm{3}{t}\right)\mathrm{3}+\mathrm{2}\left(\mathrm{5}+\mathrm{4}{t}\right)\mathrm{4}+\mathrm{2}\left(\mathrm{3}+\mathrm{5}{t}\right)\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{30}+\mathrm{18}{t}+\mathrm{40}+\mathrm{32}{t}+\mathrm{30}+\mathrm{50}{t}=\mathrm{0} \\ $$$$\mathrm{100}+\mathrm{81}{t}=\mathrm{0} \\ $$$${t}=\mathrm{1} \\ $$$$\mathrm{8}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} =\mathrm{128}+\mathrm{81}=\sqrt{\mathrm{209}} \\ $$

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Question-159748

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