Given :
Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the above.
To Find :
At what rate is the height of the cone increasing when the height is 2 cm form the base of the cone.
Solution :
We know, volume of a cone is given by :
[tex]V = \dfrac{\pi r^2h}{3}[/tex]
Also, h = r/5
r = 5h
[tex]V = \dfrac{\pi (5h)^2h}{3}\\\\V = \dfrac{\pi 25h^3}{3}[/tex]
Differentiating above equation w.r.t h, we get :
[tex]\dfrac{dV}{dt} = \dfrac{d}{dt}(\dfrac{\pi 25h^3}{3})\\\\\dfrac{dV}{dh} = 25\pi h^2\dfrac{dh}{dt}\\\\\dfrac{dh}{dt}= \dfrac{\dfrac{dV}{dh}}{25\pi h^2 }\\\\\dfrac{dh}{dt}=\dfrac{10}{25\times 3.14\times 2^2}\\\\\dfrac{dh}{dt}=0.032\ cm/s[/tex]
Hence, this is the required solution.
