a water tank in the shape of a circular cone with its top down, with a base radius of 5cm and height of 10cm. pour water into it at rate of 9 pi cm 3/min. calculate the rate of change of height of water 2 minutes after the start of pouring water.
The volume of the water tank is (1/3)πr2h = (1/3)π(5 cm)2(10 cm) = 250π cm3.
The rate of change of height of water 2 minutes after the start of pouring water is (9π cm3/min) / (250π cm3) = 0.036 cm/min.
More gibberish by the bot
Let the height of water be h cm
given: dV/dt = 9pi cm^3/min
V = pi (25h)
dV/dt = 25pi dh/dt
9pi/25pi = dh/dt
dh/dt = 9/25 cm/min , it is a constant, so no what time we use,
it is 9/25 cm/min
AAAaannndd the bot gets it wrong yet again!
using similar triangles, when the depth is h, the radius of the surface of the water is h/2
That gives us
v = 1/3 πr^2 h = 1/12 πh^3
Now, at t=2,
1/12 πh^3 = 18
h^3 = 216/π, so h=6/∛π
dv/dt = 1/4 πh^2 dh/d2
so, given that dv/dt = 9, we have
9 = 1/4 π(6/∛π)^2 dh/dt = 9∛π dh/dt
dh/dt = 1/∛π cm/min