A cone whose height always equal to its diameter is increasing in volume at the rate `40 cm^3`/`sec` .At what rate is the radius increasing when its circular base area is `1m^2` ?

Options

A. `1`
B. `0.001`
C. `2`
D. `0.002`

Answer

`0.003`

Explanation

Given that, `h=2r`

`(dV)/(dt)=40 cm^3`/`sec`

`V=(1)/(3)pir^2 h`

`V=(2)/(3)pir^3`

`(dV)/(dt)=2 times 10^4 (dr)/(dt)`

`40=2 times 10^4 (dr)/(dt)`

`(dr)/(dt)=(40)/(2 times 10^4)`

`(dr)/(dt)=0.002cm`/`sec`

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