Question

Water running in a hemispherical bowl of radius 180 cm at the rate of 108000c.c per minute. Find how fast the water level is rising when the depth of the water bowl is 120cm.

Answer 1

Hint: The rate of volume per minute is given. Differentiate the volume of the cap (bowl) formula with respect to time (rate of volume per minute). The value of the rate of volume per minute of the bowl is already given. So equate the differentiation with the value to find the rate of level rise of water per minute.
Volume of the hemispherical bowl (cap) is $\dfrac{\pi }{3}\left( {3r{h^2} - {h^3}} \right)$, where r is the radius of the bowl, h is the height (depth) of the bowl.


Complete step-by-step answer:
We are given that the water is running in a hemispherical bowl of radius 180 cm at the rate of 108000c.c per minute and the depth of the bowl is 120cm.
We have to find the rise in water level.
Volume (v) of the bowl is $\dfrac{\pi }{3}\left( {3r{h^2} - {h^3}} \right)$
We have to differentiate the volume with respect to time to find the rate of volume per minute.
$
  \dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{\pi }{3}\left( {3r{h^2} - {h^3}} \right)} \right) \\
  \dfrac{{dv}}{{dt}} = \dfrac{\pi }{3}\left[ {\dfrac{d}{{dt}}\left( {3r{h^2}} \right) - \dfrac{d}{{dt}}\left( {{h^3}} \right)} \right] \\
  \left( {\because \dfrac{d}{{dt}}{x^n} = n{x^{n - 1}}} \right) \\
  \dfrac{{dv}}{{dt}} = \dfrac{\pi }{3}\left( {3 \times r \times 2h\dfrac{{dh}}{{dt}} - 3{h^2}\dfrac{{dh}}{{dt}}} \right) \\
  \dfrac{{dv}}{{dt}} = \dfrac{\pi }{3}\left( {6rh - 3{h^2}} \right)\dfrac{{dh}}{{dt}} \\
 $
We are already given the value of $\dfrac{{dv}}{{dt}}$, which is 108000c.c per minute.
$
  \dfrac{{dv}}{{dt}} = 108000c{m^3}/\min \\
  \dfrac{{dv}}{{dt}} = \dfrac{\pi }{3}\left( {6rh - 3{h^2}} \right)\dfrac{{dh}}{{dt}} \\
   \to 108000 = \dfrac{\pi }{3}\left( {6rh - 3{h^2}} \right)\dfrac{{dh}}{{dt}} \\
  \therefore \dfrac{{dh}}{{dt}} = \dfrac{{108000}}{{\left( {\dfrac{\pi }{3}\left( {6rh - 3{h^2}} \right)} \right)}} \\
 $
Height (depth) of the bowl is 120cmand radius is 180cm.
$
  \dfrac{\pi }{3}\left( {6rh - 3{h^2}} \right) = \dfrac{\pi }{3}\left( {6 \times 180 \times 120 - 3 \times {{120}^2}} \right) \\
  = \dfrac{\pi }{3}\left( {129600 - 43200} \right) \\
  = \dfrac{\pi }{3}\left( {86400} \right) \\
  = 28800 \times \pi c{m^2} \\
  \dfrac{{dh}}{{dt}} = \dfrac{{108000}}{{\dfrac{\pi }{3}\left( {6rh - 3{h^2}} \right)}} \\
   \to \dfrac{{dh}}{{dt}} = \dfrac{{108000}}{{28800\pi }} \\
   = 1.19cm/\min \\
 $
The rate of rise of water level is 1.19 centimetres per minute.

Note: 1c.c is one cubic centimeter or cube of a centimeter. One cubic centimeter is also equal to one thousandth of one litre. A hemisphere is half of a sphere, formed by cutting the sphere plainly from its center. Be careful with the volumes of the sphere and hemisphere.