Differential Calculus - Word Problems

[Asphyxiated]

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

 

[Dick]

Second line. dV/dt=(4/3)*pi*(3*r^2*dr/dt). The r is squared.

 

[Asphyxiated]

oh yeah it totally is!

 

[Asphyxiated]

Yeah that was all, thanks man, totally solved. if you could take a look at the other one, I think that I have solved the cone problem as much as I could.. either way though, thanks