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As a balloon in the shape of a sphere is being blown up, the radius is increasing #1/pi# inches per second. at what rate is the volume increasing when the radius is 1 inch

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Answer 1 · 2017-03-05T00:21:09

when # r=1 => (dV)/dt = 4 #

Let us set up the following variables:

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

We are told that:

# (dr)/dt = 1/pi # (constant)

And we aim to find:

# (dV)/dt # when #r=1#

As the balloon is a sphere, then:

# V=4/3pir^3 #

Differentiating wrt #r# we get:

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

And applying the chain rule we have:

# (dV)/dt = (dV)/(dr) * (dr)/dt #
# " " = (4pir^2) * (1/pi) #
# " " = 4r^2 #

So when # r=1 => (dV)/dt = 4 #