Question
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?
Answers
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
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
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
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
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
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
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