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If surface area of a cube is changing at a rate of `5 m^(2)//s`, find the rate of change of body diagonal at the moment when side length is `1 m`.A. `5 m//s`B. `5sqrt(3) m//s`C. `5/2sqrt(3) m//s`D. `5/(4sqrt(3)) m//s` |
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Answer» Correct Answer - C::D Surface area of cube `S=6a^(2)` (where a=side of cube) Body diagonal `1=sqrt(3)a`. Therefore `S=2l^(2)` Differentiating it w.r.t. time `(dS)/(dt)=2(2l)(dl)/(dt)rArr (dl)/(dt)=(1)/(4(sqrt(3)a))(dS)/(dt)=5/(4sqrt(3)) m//s` |
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