How quickly is the volume increasing when the radius is 6 cm?

Suppose the radius of a spherical balloon is growing at the rate of 50 cm/s. How quickly is the volume increasing when the radius is 6 cm

1 Answer
Steve M
Feb 20, 2018

# [(dV)/(dt)]_(r=6) = 7200pi ~~ 22619 \ cm^3s^(-1) #

Explanation:

Let us set up the following variables:

# { (t,"time elapsed", s), (r, "radius of the balloon at time "t, cm), (V, "Volume of the balloon at time "t, cm^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 "radius is increasing at a constant rate of #50 \ cms^-1#", which mathematically means that:

# (dr)/(dt) = 50 #

Therefore:

# (dV)/(dt) = 4pir^2 * 50 = 200pir^2 #

And we are asked to find "rate at which the volume is increasing at the instant when the radius is #6 \ cm#. If #r=6#, then:

# [(dV)/(dt)]_(r=6) = 200pi * 6^2 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 200pi * 36 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 7200pi #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ~~ 22619 \ cm^3s^(-1) #

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