The volume of the cone at h metres is

`V = int_0^u r^2 = 1/3r^3|_0^u = 1/3u^3` cubic metres

where `u = R h/H = 45h/6`

So `V = 1/3(45/6)^3h^3 = 140.6h^3`

`(dV)/(dt) = 50 m^3` per minute

Now, using the chain rule

`(dt)/(dh) = ((dt)/(dV))((dV)/(dh))`

` `Differentiating V(h) we have `(dV)/(dh) = 3(140.6)h^2 = 421.9h^2`

and we also have `(dt)/(dV) = 1/50` by simply inverting `(dV)/(dt)`

So `(dt)/(dh) = (1/50)(421.9h^2)`

`implies (dh)/(dt) = 50/(421.9h^2)`

So, when h = 5m,

`(dh)/(dt) = 50/((421.9)(25)) = 0.00474` metres per minute

`= 0.474` centimetres per minute

The equation for the radius at h metres is

`u = R h/H = 45h/6 = 7.5h`

Again using the chain rule

`(du)/(dt) = ((du)/(dh))((dh)/(dt))`

` `Now, `(du)/(dh) = 7.5`

So when h = 5

` ` `(du)/(dt) = 7.5/0.474 = 15.82` centimetres per minute

**1) 0.474 cm/min 2) 15.82 cm/min**

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