An inverted conical tank (with vertex down) is 14 feet across the top and 24 feet deep. If water is flowing in at a rate of 12 ft3/min, find the rate of change of the depth of the water when the water is 10 feet deep.
0.229 ft/min
0.449 ft/min
0.669 ft/min
0.778 ft/min
44.90 ft/min
5 answers
I've looked at the related questions and it still doesn't make much sense to me :(
radius/depth = 7/24
what is radius when depth = 10?
7/24 = r/10
r = 70/24 = 35/12
now the change of volume with depth h is the surface area
area * dh = dV
pi r^2 dh = dV
pi r^2 dh/dt = dV/dt = 12 ft^3/m
pi (35/12)^2 dh/dt = 12
dh/dt = .449 ft/min
what is radius when depth = 10?
7/24 = r/10
r = 70/24 = 35/12
now the change of volume with depth h is the surface area
area * dh = dV
pi r^2 dh = dV
pi r^2 dh/dt = dV/dt = 12 ft^3/m
pi (35/12)^2 dh/dt = 12
dh/dt = .449 ft/min
You could use a fancy formula for volume of a cone and differentiate it to find dV/dh but if you just draw a picture you can see that the area of the surface * dh is the added volume for added dh
you need to figure the volume of a cone measured from the vertex.
Notice for any height h, radius is a function of height, r=7/24 *H
so the volume of water h deep is
V=1/3 PI r^2 h
= 1/3 PI (7/24)^2 h^3
dv/dt= PI ( )^2 h^2 dh/dt
you are given dv/dt, h, solve for dh/dt
Notice for any height h, radius is a function of height, r=7/24 *H
so the volume of water h deep is
V=1/3 PI r^2 h
= 1/3 PI (7/24)^2 h^3
dv/dt= PI ( )^2 h^2 dh/dt
you are given dv/dt, h, solve for dh/dt
oh well, not that hard
V = (1/3)pi r^2 h
r = 7 h/24
so
V = (1/3)pi (7/24)^2 h^3
dV/dh = pi (7/24)^2 h^2 dh
when h = 10
dV = pi (70/24)^2 dh
dV = pi (35/12)^2 dh
V = (1/3)pi r^2 h
r = 7 h/24
so
V = (1/3)pi (7/24)^2 h^3
dV/dh = pi (7/24)^2 h^2 dh
when h = 10
dV = pi (70/24)^2 dh
dV = pi (35/12)^2 dh
2 answers