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
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
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
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
To find the rate of change of the depth of the water, we can use related rates. We know the rate at which water is flowing into the tank, and we want to find the rate of change of the depth of the water.
Let's denote the depth of the water as h and the rate of change of the depth as dh/dt.
We are given that water is flowing into the tank at a rate of 12 ft3/min. This means that the volume of water in the tank is increasing at a rate of 12 ft3/min.
The volume of a cone can be calculated using the formula V = (1/3)πr²h, where r is the radius of the top of the cone and h is the height of the cone.
Given that the tank is inverted and has a radius of 7 ft (since it is 14 feet across the top), we can write the equation for the volume of the cone as V = (1/3)π(7)²h = (49/3)πh.
Now, let's find the derivative of the volume with respect to time, dV/dt, and the derivative of the volume with respect to height, dV/dh.
dV/dt = (49/3)π(dh/dt) (1)
We are given that dV/dt = 12 ft3/min, so we can substitute this value into equation (1):
12 = (49/3)π(dh/dt)
Now, let's solve for dh/dt by isolating it:
dh/dt = (12 * 3) / (49π) = 0.229 ft/min
Therefore, the rate of change of the depth of the water when it is 10 feet deep is approximately 0.229 ft/min.
So the correct answer is 0.229 ft/min.