# Thread: Differentiation to Rate of Change

1. ## Differentiation to Rate of Change

1) The total surface area of a hemisphere is decreasing at a rate of 2cm^2/s. Find the rate of change of the volume of the hemisphere when its radius is 1/2cm.

2) Sand is poured into a right circular cone at the rate of 18cm^3/s, of which the height is three-quarters its radius. Calculate the instant when the radius is 6cm, the rate of change of
ii) the area of the area.

Sorry, I wanted to ask why for the first question I have to express volume in terms of surface area and then use the chain rule, dV/dt = dV/dA * dA/dt. How come I cannot find the dr/dt first by using dA/dt = dA/dr * dr/dt. After i got the dr/dt, I use dV/dt = dV/dr * dr/dt. But the answer is different.

However for the second question part ii, i do not need to express volume in terms of area. I just have to use dA/dr * the part (i) answer dr/dt. I am a bit confused when i should express something in terms of something. Thanks.

2. Originally Posted by fantasylo
1) The total surface area of a hemisphere is decreasing at a rate of 2cm^2/s. Find the rate of change of the volume of the hemisphere when its radius is 1/2cm.

2) Sand is poured into a right circular cone at the rate of 18cm^3/s, of which the height is three-quarters its radius. Calculate the instant when the radius is 6cm, the rate of change of
ii) the area of the area.

Sorry, I wanted to ask why for the first question I have to express volume in terms of surface area and then use the chain rule, dV/dt = dV/dA * dA/dt. How come I cannot find the dr/dt first by using dA/dt = dA/dr * dr/dt. After i got the dr/dt, I use dV/dt = dV/dr * dr/dt. But the answer is different.

However for the second question part ii, i do not need to express volume in terms of area. I just have to use dA/dr * the part (i) answer dr/dt. I am a bit confused when i should express something in terms of something. Thanks.
Q1.

Hi fantasylo,

maybe check your calculations against the following....

surface area of hemisphere

$SA=2{\pi}r^2+{\pi}r^2=3{\pi}r^2$

rate of change of surface area

$\displaystyle\huge\frac{d}{dt}\left(3{\pi}r^2\righ t)=3{\pi}\frac{dr}{dt}\frac{d}{dr}r^2=6{\pi}r\frac {dr}{dt}=-2\Rightarrow\ \frac{dr}{dt}=-\frac{1}{3{\pi}r}$

volume of hemisphere

$v=\displaystyle\huge\frac{2}{3}{\pi}r^3$

rate of change of volume (i)

$\displaystyle\huge\frac{dv}{dt}=\frac{dv}{dr}\frac {dr}{dt}=\frac{2}{3}{\pi}3r^2\frac{dr}{dt}$

$=\displaystyle\huge\ 2{\pi}r^2\frac{dr}{dt}=-\frac{2{\pi}r^2}{3{\pi}r}=-\frac{2}{3}r$

rate of change of volume (ii)

$\displaystyle\huge\frac{2}{3}{\pi}r^3=\left(\frac{ 2r}{9}\right)3{\pi}r^2\Rightarrow\ v=\frac{2}{9}rA=product,\ since\ r\ and\ A\ are\ variables$

$\displaystyle\huge\frac{dv}{dt}=\frac{dv}{dA}\frac {dA}{dt}=-2\frac{dv}{dA}=-\frac{4}{9}\frac{d}{dA}(rA)$

$=\displaystyle\huge\ -\frac{4}{9}\left(r+A\frac{d}{dA}r\right)=-\frac{4}{9}\left[r+3{\pi}r^2\frac{1}{\left(\frac{dA}{dr}\right)}\ri ght]$

$=\displaystyle\huge\ -\frac{4}{9}\left(r+\frac{3{\pi}r^2}{6{\pi}r}\right )=-\frac{4}{9}\left(r+\frac{r}{2}\right)=-\frac{4}{9}\left(\frac{3}{2}r\right)$

$=\displaystyle\huge\ -\frac{6}{9}r=-\frac{2}{3}r$