IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

yogeendra

Because of viral infection, the shape of a certain cone-shaped cell is changing. The height is
increasing at the rate of 3 microns per min. For metabolic reasons, the volume remains constantly equal
to 20 cubic microns. At the moment that the radius is 5 microns, what is the rate of change of
the radius of the cell?

iitkgp_bipin

Volume is

r²h = 20

Differentiate both sides wrt time :

2hr.dr/dt + r².dh/dt = 0.............................(1)

given rate of increase of height, dh/dt = 3........(2)

When r=5 :

5²h = 20

h = 4/5

............................(3)

Putting (2) and (3) in (1) :

2(4/5

)(5)(dr/dt) + 5²(3) = 0

dr/dt = -75

/8 microns per second

(-ve sign is coming since radius is decreasing with time).

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