A spherical balloon is inflated so that its volume is increasing at the rate of 3.9 ft^3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.7 feet?

Answer 1

Hi Sasha,

This is a classic related rates style problem. The typical process for a related rates question is as follows:

1) Find an expression that relates the relevant changing variables. In this case, the changing variables are volume and diameter (or radius) of the balloon. So, a useful equation here would be one for volume of a sphere:

V = 4/3*π*r^3

2) Differentiate this expression with respect to whatever is changing in the problem. Usually this is time, but beware that sometimes the question will ask for things like the rate of change of the area or volume with respect to change in radius, etc. In this case, the volume and diameter are changing with respect to time, as can be seen by the per minute units. Differentiating the volume equation with respect to time gives us the following:

d/dt(V) = d/dt(4/3*π*r^3)

dv/dt = 4/3*π*r^2*3*dr/dt = 4*π*r^2*dr/dt

3) Apply knowns from the problem statement or other relationships:

dV/dt = 3.9 ft^3/min

d = 1.7 ft = 2*r

r = 0.85 ft

4) Solve for the unknown:

3.9 = 4*π*(0.85^2)*dr/dt

dr/dt = 0.43 ft/min

That is our change in radius with respect to time, but the problem asked for change in diameter with respect to time. So, to relate those, we have the following equation:

d = 2*r

Differentiating this with respect to time:

d/dt(d) = d/dt(2*r)

dd/dt = 2*dr/dt = 2*0.43 ft/min

Final answer: dd/dt = 0.86 ft/min

Hope this helps! Let me know if you have any further questions.