- #1
Asphyxiated
- 263
- 0
Homework Statement
A spherical balloon is being inflated at the rate of 10 cu in/sec. Find the rate of change of the area when the balloon has a radius of 6 in.
Homework Equations
V = \frac {4}{3} \pi r^{3} and A = 4 \pi r^{2}
The Attempt at a Solution
\frac {dV}{dt} = \frac{4}{3} \pi 3r^{2}\frac{dr}{dt}
\frac {dA}{dt} = 4 \pi 2r\frac{dr}{dt}
the value of dV/dt is given in the question so
\frac {dV}{dt} = 10 in^{3}/sec
If we substitute the value into the volume equation we can find dr/dt like so
10 = \frac {4}{3} \pi 3r^{2}\frac{dr}{dt}
\frac {10}{\frac {4}{3} \pi 3r^{2}} = \frac {dr}{dt}
then set r = 6 we get
\frac {dr}{dt} = \frac {10}{452.39} = .0221
Then move on to solve this equation for dA/dt
\frac {dA}{dt} = 4 \pi 2r\frac{dr}{dt}
substituting dr/dt value from other equation and setting r = 6 again
\frac {dA}{dt} = 10/3 ~= 3.33 in^{2}/sec
Word, problem solved
Last edited: