Acone (vertex down) with height 10 inches and radius 2 inches is being filled with water at a constant rate of 2 in3/sec. how fast is the surface of the water rising when the depth of the water is 6 inches?

Using implicit differentiation, it is found that the surface of the water is rising at a rate of 0.4775 inches per second.

The volume of a cone of radius r and height h is given by:

Applying implicit differentiation, the rate of change is given by:

Thus:

The surface of the water is rising at a rate of 0.4775 inches per second.

A similar problem is given at link

Step-by-step explanation:

The volume for a cone is

Let's look at the derivative of this to see what we need to fill into it against what we were given to fill into it.  The derivative of the volume with respect to time is

If this is the formula we are to use, we need a radius measure (we have one), a dr/dt (we don't have one), a height (we seem to have 2 of those!), and we are ultimately looking for dh/dt.

Because of the fact that we do not have a dr/dt value along with the fact that we have 2 values of height, we have to find a rewrite for that volume formula where the radius is in terms of the height.  That way we don't need to worry about the rate of change of radius (becasue we weren't given one anyway!).  The reason we were given that the height of the cone is 10 and the radius is 2 is so we can write the radius in terms of the height.  Use proportions to do this:

Cross multiply to get 10r = 2h so

We will now be able to express the volume in terms of height only:

which simplifies down to

This is what we will take the derivative of.  

The 3's cancel, leaving a little bit of a simpler formula:

We were given that dV/dt = 2 and we are to find dh/dt when h = 6.  Filling in what we given then:

or

Multiply by the reciprocal of the fraction on the right and simplify to

or, in decimal form .4420970641in/sec

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step-by-step explanation: