Calculus related rates problem?

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a cone shaped coffee filter of radius 6cm. and depth 10 cm contains water, which drips out through a hole at the bottom at a constant rate of 1.5 cm^3 per second. a) if the filter ...show more
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a)

volume for a perfect cone is the volume for what it would be as a cylinder*1/3.

pi(6^2)10 = The cone-shaped coffee filter's volume in cubic centimeters

1.5 cubic centimeters drain per second....


b)

assuming the coffee machine is held perpendicular to the center of earth's gravity, you can use the many properties of right triangles.

Rough diagram of cone, coffee filter cut-away view.

< radius (half cone)>
-....................
--..................
---.................
----...............
-----............ /\
------...........10cm
-------...........\/
--------.........
---------.......
----------.....

10cm is the maximum value for h. any cutaway right triangle will be similar to the one created if looking at the 2-D cross section of the filter cut/divided vertically done the middle.

6(h/10)= the radius of a filter cylinder at any h. I recommend that you use this to find the equation for the volume.


c)

The question is a bit of a contradiction. when the water level is at 8cm time isn't passing, otherwise it would not be 8cm. I could find how much time passes between a range of depths such as 8.25cm to 7.75cm. The rate is a constant 1.5 cubic centimeters volume per second, regardless of how full it is. The level falls at the total volume of water (see part-b to help find the exact numerical value) - 1.5 cubic centimeters of water per second (see part-b again to find the depth/height lost at a rate of 1.5cm^3 per second when h=8).
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