Question

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. At what rate is the height of the cone increasing when the height is 2 cm form the base of the cone?​

Answer 1

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 :

`V = (\pi r^2h)/(3)`

Also, h = r/5

r = 5h

`V = (\pi (5h)^2h)/(3)\\\\V = (\pi 25h^3)/(3)`

Differentiating above equation w.r.t h, we get :

`(dV)/(dt) = (d)/(dt)((\pi 25h^3)/(3))\\\\(dV)/(dh) = 25\pi h^2(dh)/(dt)\\\\(dh)/(dt)= ((dV)/(dh))/(25\pi h^2 )\\\\(dh)/(dt)=(10)/(25* 3.14* 2^2)\\\\(dh)/(dt)=0.032\ cm/s`

Hence, this is the required solution.