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Question Number 132407 by bramlexs22 last updated on 14/Feb/21

$$\mathrm{on}\mathrm{circle}{\mathrm{x}}^2+{\mathrm{y}}^2=25,\mathrm{find}\mathrm{the}\mathrm{point}$$$$\mathrm{closest}\mathrm{to}(1,1).$$

Answered by EDWIN88 last updated on 14/Feb/21

$$\mathrm{let}\mathrm{P}(\mathrm{x},\mathrm{y})\mathrm{be}\mathrm{a}\mathrm{point}\mathrm{on}\mathrm{a}\mathrm{circle}$$$$\mathrm{let}\mathrm{x}\mathrm{be}\mathrm{a}\mathrm{distance}\mathrm{of}\mathrm{plint}(1,1)\mathrm{to}(\mathrm{x},\mathrm{y})$$$${\mathrm{d}}^2={(\mathrm{x}-1)}^2+{(\mathrm{y}-1)}^2$$$${\mathrm{d}}^2={(\mathrm{x}-1)}^2+25-{\mathrm{x}}^2-2\sqrt{25-{\mathrm{x}}^2}+1$$$${\mathrm{d}}^2={\mathrm{x}}^2-2\mathrm{x}+1+25-{\mathrm{x}}^2-2\sqrt{25-{\mathrm{x}}^2}+1$$$${\mathrm{d}}^2=27-2\mathrm{x}-2\sqrt{25-{\mathrm{x}}^2}$$$$\mathrm{differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}\mathrm{x}$$$${\left({\mathrm{d}}^2\right)}^{\prime }=-2-2\left(\frac{-\mathrm{x}}{\sqrt{25-{\mathrm{x}}^2}}\right)=0$$$$\frac{\mathrm{x}}{\sqrt{25-{\mathrm{x}}^2}}=1\iff {2\mathrm{x}}^2=25\Rightarrow \mathrm{x}=\pm \frac{5}{\sqrt{2}}$$$$\mathrm{and}\mathrm{y}=\pm \sqrt{25-\frac{25}{2}}=\pm \frac{5}{\sqrt{2}}\mathrm{the}\mathrm{point}$$$$\mathrm{on}\mathrm{circle}\mathrm{such}\mathrm{that}\mathrm{closest}\mathrm{to}(1,1)\mathrm{is}(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}})$$

Commented by EDWIN88 last updated on 14/Feb/21

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