How fast is the depth of the liquid falling when the level is 6 inches deep in the situation below?A conical funnel is 14 inches in diameter and 12 inches deep. A liquid is flowing out at the rate...

How fast is the depth of the liquid falling when the level is 6 inches deep in the situation below?

A conical funnel is 14 inches in diameter and 12 inches deep. A liquid is flowing out at the rate of 40 cubic inches per second.

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embizze | High School Teacher | (Level 1) Educator Emeritus

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We are given a conical funnel with diameter 14in and height 12in, out of which a liquid is flowing at `40 ("in"^3)/"sec"` . We want to find the rate at which the height is decreasing when the height is 6in.

 

The formula for the volume of a cone is `V=1/3 pi r^2h` where r is the radius of the base, and h is the perpendicular height of the cone.

We need to express r as a function of h: After drawing the picture of the inverted cone and the water at height 6in, we can drop a perpendicular from the center of the base to the apex of the cone. With the original radius of the base (7in) and the radius r of the "base" of the liquid and original height 12in and height of liquid h we have similar right triangles.

`r/h=7/12==>r=(7h)/12`

Now `V=pi/3((7h)/12)^2h=(49pi)/432 h^3`

Then `(dV)/(dt)=(49pi)/432(3h^2)(dh)/(dt)`

When the height of the liquid is 6in we have: `(dV)/(dt)=-40,h=6` so we can solve for `(dh)/(dt)` :

`-40=(49pi)/144h^2*(dh)/(dt)`

`-40=(49pi)/4 * (dh)/(dt)`

`(dh)/(dt)=-160/(49pi)~~-1.039379`

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When the height of the liquid is 6in, the height is changing at approximately -1.04`"in"/"sec"`

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