A spherical balloon is being inflated so that its volume is increasing at the rate of 10cm3/sec. Find the rate at which the diameter is increasing when the diameter is 20cm?

Derivative - related rates

1 Answer
Mar 17, 2018

#1/(20pi)# cm/s

Explanation:

The first thing to do is to write out what we do know about the problem.

We know the volume of the balloon is increasing at a rate of 10 #cm^3#/s

This is expressed as:

#(dV)/dt=10#

The volume of a sphere is given by:

#V=4/3pir^3#

The question asks to find the rate of which the diameter is increasing when the diameter is 20. The formula gives volume in terms of the radius, not the diameter, so we will halve the diameter to get the radius.

Radius = 10:

We need to find:

#(dr)/(dt)#

Since we don't have #bbt# in our equation, we will be differentiating implicitly. We can use the chain rule for this:

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

We already know #(dV)/(dt)=10#

#:.#

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

To find #(dV)/(dr)# we differentiate #V=4/3pir^3#

#(dV)/(dr)(4/3pir^3)=4pir^2#

#:.#

#10=4pir^2*(dr)/dt#

Dividing:

#10/(4pir^2)=(dr)/dt#

Radius is 10, so:

#10/(4pi(10)^2)=(dr)/dt#

#(dr)/(dt)=1/(40pi)#

We need the diameter, so:

#2*1/(40pi)=1/(20pi)#

The diameter is increasing at a rate of #1/(20pi)# cm/s