Question Coffee is dripping from a conical filter into a cylindrical coffeepot.
ID: 2834276 • Letter: Q
Question
Question
Coffee is dripping from a conical filter into a cylindrical coffeepot. Both the filter cone and the coffeepot have diameter of 6 inches. The cone has a height of 6 inches as well. When the coffee in the cone is 5 inches deep, the coffee is draining at the rate of 10in3/min.
(a). How fast is the level in the pot rising?
Hint: The volume of a cylinder of radius r and height h is V=?r2h.
(b). How fast is the level in the cone falling at the same time?
Hint: The volume of a cone of radius r and height h is V=1/3(?r2h)= ?r2h/3.
Question
(a). The profit P from selling x units of a product is given by P=15+12?x-18/x
(i). Find the average rate of change of P on the interval [1,9].
(ii). Find a formula for the marginal profit, and its numerical value when x=9
(b). A sphere of radius 3inches has its radius changing at a rate of 4 inches per second. How quickly is the volume of the sphere changing?
Explanation / Answer
The change in volume for the filter is -10 in^3/min
The change in volume for the coffee pot is +10 in^3/min
First, we need to describe the radius of the cone as a function of the height
When height = 6, diameter = 6 (radius = 3)
When height = 0, diameter = 0 (radius = 0)
So, h = 2r, or r = (1/2) * h
a)
V = pi * r^2 * h
Since the radius of the coffee pot is unchanging, we can treat it as a constant. r = 6
V = pi * 6^2 * h
V = 36 * pi * h
Derive:
dV/dt = 36 * pi * dh/dt
dV/dt = 10
Solve for dh/dt
10 = 36 * pi * dh/dt
5 = 18 * pi * dh/dt
dh/dt = 5 / (18 * pi)
b)
We want to find the rate at which the height is changing, so let's get the volume of the cone as a function of the height
V = (1/3) * pi * r^2 * h
r = (1/2) * h
V = (1/3) * pi * (1/2)^2 * h^2 * h
V = (1/3) * (1/4) * pi * h^3
V = (pi/12) * h^3
Now derive:
dV/dt = (pi/12) * 3 * h^2 * dh/dt
dV/dt = (pi/4) * h^2 * dh/dt
dV/dt = -10
h = 5
Solve for dh/dt
-10 = (pi/4) * 5^2 * dh/dt
-10 = (pi/4) * 25 * dh/dt
(-10/25) * (4/pi) = dh/dt
dh/dt = -40 / (25 * pi)
dh/dt = -8 / (5 * pi)
Since the question is asking how fast it is falling, then we can't say it's falling at -8/(5pi) inches per minute, but rather that it is falling at 8/(5pi) inches per minute.