Water is leaking from a conical funnel at the rate of 5 ent/sec If the radius of the base of thetunnel is 10 cmand altitude is 20 cm. Find the rate at which water level is dropping when it is 5cm from top.wlond kent vertical the petrol is leaking at the rate of

Water is leaking from a conical funnel at the rate of 5 ent/sec If the

1.	Water is leaking from a conical funnel at the rate of 5 ent/sec If the radius of the base of thetunnel is 10 cmand altitude is 20 cm. Find the rate at which water level is dropping when it is 5cm from top.wlond kent vertical the petrol is leaking at the rate of
Answer» dV/dt = 5 cm^3/sec radius(funnel) = 10 cm height(funnel) = 20 cm dh/dt = ? at the point when the water level is dropping 5 cm from the best.volume = π/3(radius)^2(height) 1.) discover the proportion wager. the water and the cone shaped pipe. radius(funnel)/height(funnel) = radius(water)/height(water) give r a chance to be the radius(water) and h be height(water) 10cm/20cm = r/h 1/2(h) = r 2.) the volume of the water is the same as the volume of the conelike pipe. volume = π/3(radius)^2(height) substitute 1/2(h) to the range volume = π/3(1/2h)^2(height) volume = π/3(1/4h^2)(h) volume = πh^3/12 3.) separate the new eq'n shaped as for time. give V a chance to be the volume d/dV(V)dV/dt = d/dh(πh^3/12)dh/dt dV/dt = (πh^2/4)dh/dt connect to the water level, w/c is 15cm (since the water level is dropping when it is 5cm from the best) and the rate of the spilling of the water from the cone shaped channel 5 cm^3/sec = (π*(15 cm)^2/4)(dh/dt) 5 cm^3/sec = 176.7145868 cm^2(dh/dt) - 0.028 cm/sec = dh/dt - > last answer (negative implies that the water is spilling)