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InterviewSolution
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The radius of the base of a cone is increasing at the rate of 3 cm/minand the altitude is decreasing at the rate of 4 cm/min. The rate of change oflateral surface when the radius is 7 cm and altitude is 24cm is`108pic m^2//min`(b) `7pic m^2//min``27pic m^2//min`(d) none of theseA. `54pi cm^(2)//min`B. `7 pi cm^(2)//min`C. `27 cm^(2)//min`D. none of these |
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Answer» Correct Answer - A Let r, l and h denote respectively the radius, slant height and height of the cone at any time l. Then, `l^(2)=r^(2)+h^(2)` `implies 2l(dl)/(dt)=2r(dr)/(dt)+2h(dh)/(dt)` `implies l(dl)/(dt)=r(dr)/(dt)+h(dh)/(dt)` `l(dl)/(dt)=7xx3+24xx-4 " "[because(dh)/(dt)=-4and (dr)/(dt)=3]` `implies l(dl)/(dt)=-75` When r = 7 and h=24, we have `l^(2)=7^(2)+24^(2)" "[because l^(2)=r^(2)+h^(2)]` implies l = 25 `l(dl)/(dt)=-75implies(dl)/(dt)=-3` Let S denote the laternal surface area. Then, `S=pi r l` `implies (dS)/(dt)=pi{(dr)/(dt)l+r(dl)/(dt)}=pi{3xx25+7xx-3}=54pi` |
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