Answer:
[tex]\displaystyle \frac{dS}{dt}=\frac{3}{50r}[/tex]
Step-by-step explanation:
Water is being pumped into an inflated rubber sphere at a constant rate of 0.03 cubic meters per second.
So, dV/dt = 0.03.
We want to show that dS/dt is directly proportional to 1/r.
In other words, we want to establish the relationship that dS/dt = k(1/r), where k is some constant.
First, the volume of a sphere V is given by:
[tex]\displaystyle V=\frac{4}{3}\pi r^3[/tex]
Therefore:
[tex]\displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}[/tex]
Next, the surface area of a sphere S is given by:
[tex]\displaystyle S=4\pi r^2[/tex]
Therefore:
[tex]\displaystyle \frac{dS}{dt}=8\pi r\frac{dr}{dt}[/tex]
We can divide both sides by 2:
[tex]\displaystyle \frac{1}{2}\frac{dS}{dt}=4\pi r\frac{dr}{dt}[/tex]
We can substitute this into dV/dt. Rewriting:
[tex]\displaystyle \frac{dV}{dt}=r\Big(4\pi r\frac{dr}{dt}\Big)[/tex]
So:
[tex]\displaystyle \frac{dV}{dt}=\frac{1}{2}r\frac{dS}{dt}[/tex]
Since dV/dt = 0.03 or 3/100:
[tex]\displaystyle \frac{3}{100}=\frac{1}{2}r\frac{dS}{dt}[/tex]
Therefore:
[tex]\displaystyle \frac{dS}{dt}=\frac{3}{50r}=\frac{3}{50}\Big(\frac{1}{r}\Big)[/tex]
Where k = 3/50.
And we have shown that dS/dt is directly proportional to 1/r.
