Question

Hello , how to do 6(iii)?

Answer 1

Answer:

$\displaystyle \frac{dS}{dt}=\frac{3}{50r}$

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:

$\displaystyle V=\frac{4}{3}\pi r^3$

Therefore:

$\displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

Next, the surface area of a sphere S is given by:

$\displaystyle S=4\pi r^2$

Therefore:

$\displaystyle \frac{dS}{dt}=8\pi r\frac{dr}{dt}$

We can divide both sides by 2:

$\displaystyle \frac{1}{2}\frac{dS}{dt}=4\pi r\frac{dr}{dt}$

We can substitute this into dV/dt. Rewriting:

$\displaystyle \frac{dV}{dt}=r\Big(4\pi r\frac{dr}{dt}\Big)$

So:

$\displaystyle \frac{dV}{dt}=\frac{1}{2}r\frac{dS}{dt}$

Since dV/dt = 0.03 or 3/100:

$\displaystyle \frac{3}{100}=\frac{1}{2}r\frac{dS}{dt}$

Therefore:

$\displaystyle \frac{dS}{dt}=\frac{3}{50r}=\frac{3}{50}\Big(\frac{1}{r}\Big)$

Where k = 3/50.

And we have shown that dS/dt is directly proportional to 1/r.