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Prove that the semi-vertical angle of the right circular cone of givenvolume and least curved surface is `cot^(-1)(sqrt(2))dot` |
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Answer» `V=1/3pir^2h` `h=(3V)/(pir^2)` `CSA=pirl` `=pirsqrt(h^2+r^2)` `=pirsqrt((9V^2+pi^2r^6)/(pir^2)` `CSA=sqrt((9V^2+pi^2r^6)/r` Differentiate with respect to r `d/(dr)(CSA)=(3pi^2r^6-9V^2-pi^2r^6)/(r^2sqrt(9V^2+pi^2r^6)=0` `2pi^2r^6-9V^2=0` `r^6=(9V^2)/(2pi^2)` `r=(9/2)^(1/6)*(V/pi)^(1/3)` `cottheta=B/P=r/h=r/((3V)/(pir^2))` `=pi/(3V)*[(n/2)^(1/6)*(V/pi)^(1/3)]^3` `tantheta=1/sqrt2` `cottheta=sqrt2` ``theta=cot^(-1)sqrt2`. |
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