A closed circular cylinder has a base radius r cm and a height 5r cm. If r is increasing in mm/s at a rate of `(2)/(pi)`

Find expressions in terms of r for

a) the rate of change of the volume of the cylinder

b) the rate of change of the surface aras of the cylinder

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Volume of the cylinder is given by:

`V=pir^2h` where h=5r cm

`rArr V=pir^2*5r=5pir^3`

Differentiating,

`(dV)/(dt)=5pi*3r^2(dr)/(dt)`

`=15pi*r^2(dr)/(dt)`

`=15pi*r^2*2/(10pi)` (`(dr)/(dt)=2/pi (mm)/s = 2/(10pi) (cm)/s` )

`=3r^2`

a) the rate of change of the volume of the cylinder is: `(dV)/(dt)=3r^2`

Surface area of the closed cylinder is given by:

`A=2pir^2+2pirh`

`=2pir^2+2pir*5r` (h=5r)

`=12pir^2`

Differentiating,

`(dA)/(dt)=12pi*2r(dr)/(dt)`

`=24pir(dr)/(dt)`

`=24pir*2/(10pi)`

`=4.8r`

b) the rate of change of the surface area of the cylinder is:

`(dA)/(dt)=4.8r`

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