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The radius of a cylinder is increasing at the rate of `3ms^(-1)` and its altitude is decreasing at the rate of `4ms^(-1)`. The rate of change of volume when radius is `4m` and altitude is 6m isA. `80 pi cu m//s`B. `144 pi cu m//s`C. `80 cu m//s`D. `64 cu m//s` |
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Answer» Correct Answer - A Let h and r be the height and radius of cylinder respectively. Given that `(dr)/(dt) = 3m//s` and `(dh)/(dt) = - 4m//s` Also, `" "V = pi r^(2)h` On differentiating w.r.t , we get `(dV)/(dt) = pi [r^(2) (dh)/(dt )+ h2r (dr)/(dt)]` `(dV)/(dt) = pi [r^(2) (dh)/(dt )+ h2r (dr)/(dt)]` At r = 4 m and h = 6, m , them `(dV)/(dt) = pi [ - 64 + 144]` `80 pi cu m//s` |
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