+77 Ive never seen this type of problem before but i got it for homework today.
A spherical ballon is being inflated in such a way that its radius is increasing at the constant rate of 3cm/min. If the volume of the balloon is 0 at time 0, at what rate is the Volume increasing after 5 minutes?
I think the equation is sorta like V(r) = 4r3/3 but that is for feet and I need cm.
+583 A spherical ballon is being inflated in such a way that its radius is increasing at the constant rate of 3cm/min.\(\Rightarrow \frac{dr}{dt}=3cm/min\)
The relation between the volume and radius of the spherical ballon is \(v(r)=4*{r}^{3}/3\)
The dervative of volume v respect to variable v is \(\frac{dv}{dr}=4{r}^{2}\)
At time 5 minutes \(\frac{dv}{dr}=4*5^2=100\)cm^2
Applying chain rule,\(\frac{dv}{dt}=\frac{dv}{dr}*\frac{dr}{dt}\)
\(\frac{dv}{dt}=100*cm^2*3cm/min=300cm^3/min\)
This is a kind of problem relative rate of change,application of differvative.
