Water pours out of a conical tank of height 10 feet and radius 4 feet at a rate of 10 cubic feet per minute. How fast is the water level changing when it is 5 feet high?
All three of your questions are pretty much the same idea.
change in volume = surface area * change in width of slice.
the surface area = pi r^2
r at five feet = half of radius at 10 feet so r = 2 feet
Area = pi r^2 = 4 pi ft^2
so
change in volume = (4 pi)(change in height)
so
change in volume/time = 4 pi (change in height/time)
so
10 ft^3/min = 4 pi ft^2 (dh/dt)
so
dh/dt = (10/4pi) ft/min
To find the rate at which the water level is changing, we need to determine how the volume of water in the tank is changing with respect to time.
The volume of a cone can be calculated using the formula: V = (1/3)πr^2h, where V is the volume, r is the radius of the cone's base, and h is the height of the cone.
In this case, the radius of the cone's base remains constant at 4 feet, and the height of the cone is changing. We want to find the rate at which the height is changing when the height is 5 feet.
Let's denote the height of the cone as h(t), where t represents time. We are given that dh/dt (the rate at which the height is changing) is what we need to find when h = 5.
Also, we know that the volume of water pouring out of the tank per minute is 10 cubic feet, so dV/dt (the rate at which the volume is changing) is -10 (negative sign because the volume is decreasing).
We can express the volume of water in terms of the height by substituting r = 4 into the formula for the volume: V = (1/3)π(4^2)h.
Taking the derivative of both sides with respect to time (t), we get:
dV/dt = (1/3)(π)(16)(dh/dt).
Since dV/dt is given as -10 (cubic feet per minute), we have:
-10 = (1/3)(π)(16)(dh/dt).
Simplifying this equation, we find:
dh/dt = -10 / [(1/3)(π)(16)].
Now, we can simply calculate dh/dt using the above equation and determine the rate at which the water level is changing.