By approximation what percentage does the volume of a ball change if its surface area is changed by -4%? The volume V and surface area S of a ball of radius r are given by the formulas: v=4/3pir^2 and S=4pir^2
Hello, Tiger!
Is some information missing?
We should have been given one more fact
. . like the Surface Area at a certain instant.
$\displaystyle \text{By approximation what percentage does the volume of a ball change}$
$\displaystyle \text{if its surface area is changed by }\text{-}4\%?$
$\displaystyle V\,=\,\frac{4}{3}\pi r^3,\;\;S\,=\,4\pi r^2$
We have: .$\displaystyle S \:=\:4\pi r^2 \quad\Rightarrow\quad r^2 \:=\:\dfrac{S}{4\pi} \quad\Rightarrow\quad r \:=\:\dfrac{S^{\frac{1}{2}}}{2\sqrt{\pi}} $
Then: .$\displaystyle V \;=\;\dfrac{4\pi}{3}\left(\dfrac{S^{\frac{1}{2}}}{ 2\sqrt{\pi}}\right)^3 \;=\;\dfrac{4\pi}{3}\cdot \dfrac{S^{\frac{3}{2}}}{8\pi^{\frac{3}{2}}} \;=\;\dfrac{1}{6\sqrt{\pi}}\,S^{\frac{3}{2}}$
Take differentials:
. . $\displaystyle \displaystyle dV \;=\;\frac{1}{6\sqrt{\pi}}\cdot\frac{3}{2}S^{\frac {1}{2}}\cdot dS\;=\;\frac{S^{\frac{1}{2}}}{4\sqrt{\pi}}\cdot dS $
We are given: .$\displaystyle dS\:=\:-0.04$
. . Hence: .$\displaystyle \displaystyle dV\;=\;\frac{S^{\frac{1}{2}}}{4\sqrt{\pi}}(\text{-}0.04) \;=\;-\frac{\sqrt{S}}{100\sqrt{\pi}} $
Therefore: .$\displaystyle \displaystyle \Delta V \;\approx\; -\frac{\sqrt{S}}{100\sqrt{\pi}} $
Or....
Surface area changes by $\displaystyle -4\%$
$\displaystyle \Rightarrow\ S(0.96)=4{\pi}\left[0.96r^2\right]=4{\pi}\left[\sqrt{0.96}\;r\right]^2$
New radius is $\displaystyle \sqrt{0.96}\;r$ where r is the original radius.
New volume is $\displaystyle \displaystyle\frac{4}{3}{\pi}\left[\sqrt{0.96}\;r\right]^3=\frac{4}{3}{\pi}r^3\left[0.96^{\frac{3}{2}}\right]$
The percentage change is $\displaystyle \displaystyle\left[1-0.96^{\frac{3}{2}}\right]100\%$