The volume of a spherical balloon is increasing at a constant rate of #0.25m^3s^-1#. Find the rate at which the radius is increasing at the instant when the volume is #10m^3#?

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Answer 1 · 2018-01-30T13:38:07

#0.0149 \ ms^-1# (3sf )

Let us set up the following variables:

# { (t,"time elapsed", s), (r, "radius of the balloon at time "t, m), (V, "Volume of the balloon at time "t, m^3s^(-1)) :} #

Using the standard formula for the volume of a sphere, we have:

# V = 4/3pir^3 #

If we differentiate wrt #r# then we get:

# (dV)/(dr) = 4pir^2 #

And applying the chain rule, we have:

# (dV)/(dt) = (dV)/(dr) * (dr)/(dt) #

And so we have:

# (dV)/(dt) = 4pir^2 (dr)/(dt) #

Now, the question tells us that the "volume is increasing at a constant rate of #0.25 \ m^3s^-1#", which mathematically means that:

# (dV)/(dt) = 0.25 #

Therefore:

# 4pir^2 (dr)/(dt) = 0.25 => (dr)/(dt)= 1/(16pir^2)#

And we are asked to find "rate at which the radius is increasing at the instant when the volume is #10 \ m^3#. If #V=10#, then:

# 4/3pir^3 = 10 #
# :. r^3 = 15/(2pi) #
# :. r = root(3)(15/(2pi)) #

And with this value of #r# we can calculate #(dr)/dt:#

# (dr)/(dt) = 1/(16pi \ (root(3)(15/(2pi)))^2) #
# \ \ \ \ \ = 0.0149 \ ms^-1# (3sf )