Water is running in a hemispherical bowl of radius 180 ~cm at the rate of 108 cubic decimetres per minute. How fast the water level is rising when depth of the water level in the bowl is 120 ~cm ? (1 decimeter =10 ~cm )

Step-by-step solution

Radius of hemispherical bowl ( r )=180 ~cm Rate of flow ( dV dt )=108 dm ^3 / min aligned & =108000 ~cm ^3 / min & = 108000 60 ~cm ^3 / sec & =1800 ~cm ^3 / sec aligned Let depth of water in bowl be x . Volume of water in hemispherical. aligned & bowl ( V )= 3 x^2(3 r -x) & V = 3 x^2(3 180-x) aligned V =180 x^2- 3 x^3 Differentiating w.r.t. x , we get dV dt =360 x d x dt - x^2 d x dt . dV dt |_ x=120 = d x dt (360 120-120^2 ) 1800= d x dt (360 120-120^2 ) 15= d x dt (360 -120 ) d x dt = 15 240 = 1 16 cm / sec