Umm, it's Saturday morning, no work, lots of sparetime. An related rates problem. Why not.

We are asked to find the rate at which the water is rising. So that is dy/dt if the height of the water is y. And so we base the y from the bottom of the inverted cone.

Imagine, or draw the figure on paper. Draw a vertical cross-section of an inverted cone that's supposed to be 10cm round at the top and 20 cm tall. So you have an isosceles triangle that is standing on its apex. Draw the altitude from the apex to the midpoint of the 10-cm base at the top. Now you have two congruent rigth triangles, any of which has:

---vertical leg = 20cm

---horizontal leg = 10/2 = 5cm.

Now draw a horizontal line, supposed to be the level of the water rising in the cone, somewhere below the top base. Again, you have another pair of congruent smaller right triangles, each of which has:

---vertical leg = y cm

---horizontal leg = x cm

Volume of the water, V = (pi/3)(x^2)y -------(1)

V is in terms of x and y. For simpler differentiation, we need to express V in terms of one variable only. Since we are interested on dy/dt, we choose V in terms of y only.

So we express x in terms of y.

In the large and small right triangles in the figure, by proportion,

x/y = 5/20

So, x = 5y/20 = y/4

Substitute that into (1),

V = (pi/3)[(y/4)^2]y

V = (pi/48)y^3 ---------(2)

To have dy/dt, we differentiate (2) with respect to time t,

dV/dt = [(pi/16)y^2](dy/dt) ----------(3)

Since dV/dt = 1 cu.cm/sec, and y = (20-5) = 15cm, then dy/dt at that instant is:

Plugging in all known values into (3),

1 = [(pi/16)(15^2)](dy/dt)

1 = (225pi/16)(dy/dt)

dy/dt = 16/(225pi) = 0.022635 cm/sec -------------answer.

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Ooopppss, sorry, the radius at the top is 10cm. I thought I read 10-cm diameter.

Some changes on the computations above then.

x/y = 10/20

x = y/2

V = (pi/3)[(y/2)^2]y = (pi/12)y^3

dV/dt = [(pi/4)y^2](dy/dt)

1 = (pi/4)(225)(dy/dt)

dy/dt = 4/(225pi) = 0.005659 cm/sec ------------final answer.