## Question

###### The radius of a sphere is increasing at the rate of 2 m/sec. find the rate of change of the volume with respect to time when radius is 4m?

The radius of a sphere is increasing at the rate of 2 m/sec. find the rate of change of the volume with respect to time when radius is 4m?

## Answers

Sphere Volume is given by

#V=4/3pir^3# where r is given in meters for example (whichever linear unit can be used)Radius varies in function of time

#r=2t# If radius increase at rate of

#2m/sec# in t seconds it will be#2m/(sec) t sec=2t# meters in t secondsIn our volume equation

#V=4/3pi(2t)^3=32/3pit^3# The volume varies each t seconds acording to this formula

#sf(128picolor(white)(x)m^3"/s")# ## Explanation:

The volume of a sphere is given by:

#sf(V=4/(3)pir^3)# We are told that:

#sf((dr)/(dt)=2color(white)(x)"m/s")# Differentiating implicitely with respect to

t::

#sf((dV)/(dt)=4/(3)pi.3r^2.(dr)/dt)# When

#sf(r=4color(white)(x)m)# this becomes:

#sf((dV)/(dt)=4/(cancel(3))picancel(3)xx16xx2=128picolor(white)(x)m^3"/s")#

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