Water is pouring into a conical vessel 15cm deep and having a radius of 3.75cm across the top. If the rate at which the water rises is 2cm/sec, how fast is the water flowing into the conical vessel when the water is 4cm deep?

Answer 1

by similar triangles, when the water is at height h, the radius of the surface is h/4

v = 1/3 pi r^2 h
v = 1/48 pi h^3

dv/dt = 1/16 pi h^2 dh/dt
dv/dt = 1/16 pi * 4 * 2 = pi/2 cm^3/s

Answer 2

The height of the the container/vessel is 4 cm so we use this to formulate the radius as h/4 in that specific time.

V=1/3 pi (h/4)^2 (h)
V=1/3 ((pi)(h^2))/16 (h)
V=(pi h^3)/48

dv/dt= pi/48 ((3)(4)^2(2cm/sec))
dv/dt= pi/48 (96)
dv/dt= (96 pi)/48
dv/dt= 2 pi cm per sec

CHECKING
2 pi = pi/48 (48) (2cm/sec)
2 pi = (96 pi)/48
2 pi = 2 pi

Answer 3

V(cone)=(π/3)(r^2)(h)----(1)

The cone forms 2 right triangles with its h=15 cm and h=4 cm and radius as its base.

by ratio and proportion of the bigger triangle formed with h=15 and smaller triangle h=4:

r(smaller)=(1/4)h

substitute in equation (1):

V=(π/3)*((1/4)h)^2*(h)

differentiate:

dV/dt=((π/16)*h^2)(dh/dt)----(2)

substitute h=4 and dh/dt=2cm/s in (2)

dV/dt = 2π