an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is tan – ( . ). Water is poured into it at a constant rate of cubic metre per hour.
Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is m. Solution Let r , h and α be as in Fig . . Then tan h α = So α = tan h − .
But α = tan – ( . ) (given) h = . r = h Let V be the volume of the cone. Then V = h h r h h Therefore V dt = h dh dh ⋅ (by Chain Rule) dh h dt Now rate of change of volume, i.e., V dt = m /h and h = m.
Therefore = ( ) dh ⋅ dh dt = m/h π = Thus, the rate of change of water level is m/h Example A man of height metres walks at a uniform speed of km/h away from a lamp post which is metres high. Find the rate at which the length of his shadow increases. Fig . Solution In Fig .
, Let AB be the lamp-post, the lamp being at the position B and let MN be the man at a particular time t and let AM = l metres. Then, MS is the shadow of the man. Let MS = s metres. Note that ∆ MSN ~ ∆ ASB MS AS = MN AB AS = s (as MN = and AB = (given)) Thus AM = s – s = s .
But AM = l So l = s Therefore dl dt = ds Since dl dt = km/h. Hence, the length of the shadow increases at the rate km/h. Example Find intervals in which the function given by f ( x ) =