Posted by Rachel | Posted in Coffee and Accessories | Posted on 09-10-2009
Tags: coffee filter cone 6
Calculus related rates problem?
I’m stuck with this homework problem. There’s a diagram of a funnel filled with coffee above a pot of coffee. It says:
Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of 10 in^3/min.
The diagram shows the diameter of the cone at 6″, the height of the funnel at 6″, and the base of the coffee pot at 6′.
(a) How fast is the level in the pot rising when the coffee in the fonce is 5 in. deep?
(b) How fast is the level in the cone falling at that moment?
Thanks!!
Do you mean the base of the pot is 6″ too, or really 6′ ? I don’t understand how this is physically set up. Is it a funnel over and outside of a cylinder? Are you asking how fast the level in the pot rises if the level of coffee in the FUNNEL is 5 inches?
I’ll at least show you how to get an expression for the rate that the height in the cone changes:
Notice that the radius of the cone and the height of the cone are at right angles. When you combine this with the side of the cone, you get a right triangle. This right triangle has its vertical leg at 6″ (the height of the cone), and its horizontal leg at a width of 3″ (the cone’s radius, or half of the width of the cone).
Let’s say the coffee in the cone is h inches deep, and the radius of the surface is r. You can make another right triangle here, and it will be similar to the other right triangle. So h/r = 6/3, no matter what the values are of h and r. This means h = 2r, or r = h/2.
The volume of a cone is (1/3)pi r^2 h. Since r = h/2, this means V = (2/3) pi (h/2)^2 h = (2pi / 12) h^3 = (pi/6)h^3. The height decreases as more coffee runs out of the cone. That means the height is a function of time, so the derivative of V with respect to time is:
dV/dt = (pi/2)h^2 * (dh/dt)
We know that the volume is changing at -10in^3 / min (its negative, because it’s DECREASING at that rate as it leaves the funnel), so set this equal to -10 to get
-10 = (pi/2) h^2 (dh/dt)
dh/dt = -20 / (pi h^2)
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