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Question Number 56914 by rahul 19 last updated on 26/Mar/19

$$Theshortestdistancebetweenthepoint$$ $$(\frac{3}{2},0)andthecurvey=\sqrt{x},(x>0)is?$$

Answered by Smail last updated on 26/Mar/19

$$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$$ $$=\sqrt{{(x-\frac{3}{2})}^2+{(\sqrt{x}-0)}^2}=\sqrt{x^2-2x+\frac{9}{4}}$$ $$d^{\prime }=\frac{2x-2}{2\sqrt{x^2-2x+\frac{9}{2}}}$$ $$CVisx=1$$ $$d_{min}=\sqrt{1^2-2\times 1+\frac{9}{4}}=\frac{\sqrt{5}}{2}$$

Commented byrahul 19 last updated on 26/Mar/19

thank you sir!

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Mar/19

$$let(x,\sqrt{x})beapointonthecurve.$$ $$distancel=\sqrt{{(x-\frac{3}{2})}^2+{(\sqrt{x}-0)}^2}$$ $$l^2=x^2-2\times x\times \frac{3}{2}+\frac{9}{4}+x$$ $$l^2=x^2-2x+\frac{9}{4}$$ $$l=\sqrt{x^2-2x+\frac{9}{4}}$$ $$\frac{dl}{dx}=\frac{1}{2\sqrt{x^2-2x+\frac{9}{4}}}\times (2x-2)$$ $$\frac{dl}{dx}=\frac{x-1}{\sqrt{x^2-2x+\frac{9}{4}}}$$ $$formax/min\frac{dl}{dx}=0sox=1$$ $$$$ $$\frac{d^{2}l}{{dx}^2}=\frac{\sqrt{x^2-2x+\frac{9}{4}}\times 1-(x-1)\times \frac{1}{2\sqrt{x^2-2x+\frac{9}{4}}}\times (2x-2)}{(x^2-2x+\frac{9}{4})}$$ $${\left(\frac{d^{2}l}{{dx}^2}\right)}_{x=1}=\frac{\sqrt{\frac{5}{4}}\times 1}{\left(\frac{5}{4}\right)}=\sqrt{\frac{4}{5}}>0$$ $$sominimumdistanceisbetweenpoint(x,\sqrt{x})$$ $$thatmeans(1,1)and(\frac{3}{2},0)is$$ $$l=\sqrt{{(\frac{3}{2}-1)}^2+{(1-0)}^2}$$ $$=\sqrt{\frac{1}{4}+1}$$ $$=\sqrt{\frac{5}{4}}$$ $$=\frac{\sqrt{5}}{2}\leftarrow thisisthesnswer$$ $$$$ $$$$ $$$$

Commented byrahul 19 last updated on 26/Mar/19

thank you sir!

Answered by mr W last updated on 26/Mar/19

$$d=\sqrt{{(x-\frac{3}{2})}^2+{(\sqrt{x}-0)}^2}=\sqrt{x^2-2x+\frac{9}{4}}$$ $$=\sqrt{{(x-1)}^2+\frac{5}{4}}⩾\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$$ $$\Rightarrow d_{min}=\frac{\sqrt{5}}{2}atx=1$$

Commented byrahul 19 last updated on 26/Mar/19

thank you sir!

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