A cup has a radius of 2" at the bottom and 6" on the top. It is 10" high. 4 Minutes ago, water started pouring at 10 cubic " per minute. How fast is the water level rising now?

Still not done?

You have 40 in^3 of water in the cup at four minutes after start.
How deep is it?
r = 2 + (6-2)h/10 = 2 + .4 h
v = integral pi r^2 dh from 0 to h
= pi int (4 + 1.6 h + .16 h^2) dh
= pi (4h + .8 h^2 + .0533 h^3)
= 40 at t = 4

40/pi = 12.9 = 4 h + .8 h^2 + .0533 h^3)
find h where v = 3.23
3.23 = h + .2 h^2 + .0133 h^3
guess h = 1 then v =1.21
h = 2.0 , v = 2.90
h = 2.5 , v = 3.96
h = 2.4 , v = 3.74
h = 2.3 , v = 3.51
h = 2.2 , v = 3.31 very close
h = 2.15 , v = 3.21 close enough to 3.23
so do the problem with h = 2.15 deep

I should use some other letter than v because it is v/4pi = 40/4pi = 3.23

actually I think closer to 3.18
3.18 = h + .2 h^2 + .0133 h^3
guess h = 1 then v =1.21
h = 2.0 , v = 2.90
h = 2.5 , v = 3.96
h = 2.4 , v = 3.74
h = 2.3 , v = 3.51
h = 2.2 , v = 3.31
h = 2.15 , v = 3.21
h = 2.12 , v = 3.14
h = 2.13 , v = 3.156
h = 2.14 , v = 3.186 very close to 3.18
so do the problem with h = 2.14 deep

I am pretty sure I showed you how to do the rest before

dv/dt = pi r^2 dh/dt = 40
so
dh/dt = 40/(pi r^2)
but r = 2 + .4 h and at 4 minutes we said h = 2.14
so
r = 2 + .4 (2.14) = 2.86
r^2 = 8.16
dh/dt = 40/(8.16pi) = 1.56 in/min

Thanks again for responding. When we earlier did h=0, the formula was dh/dt=5/2pi=.8 in/min. Intuitively, it makes sense that the rate would reduce as the cup is filling up since the radius is increasing. So, should the answer above have 10 in the numerator instead of 40. Then, the answer would be dh/dt = 10/(8.16pi) = .39 in/min.

What do you think?

To find the rate at which the water level is rising, we need to use the formula for the volume of a frustum of a cone.

The volume of a frustum of a cone is given by the formula:

V = (1/3)πh(h1^2 + h2^2 + h1h2)

Where V is the volume, h is the height, h1 is the radius at the bottom, and h2 is the radius at the top.

In this case, the height (h) is given as 10", the radius at the bottom (h1) is given as 2", and the radius at the top (h2) is given as 6".

Plugging these values into the formula, we get:

V = (1/3)π(10)(2^2 + 6^2 + 2*6)
V = (1/3)π(10)(4 + 36 + 12)
V = (1/3)π(10)(52)
V = (1/3)(10π)(52)
V = (520/3)π

We know that the water is pouring at a rate of 10 cubic inches per minute. So, after 4 minutes, the volume of water that has poured into the cup is:

Volume = 10 cubic inches/minute * 4 minutes
Volume = 40 cubic inches

To find how fast the water level is rising, we need to find the time derivative of the volume with respect to time. Taking the derivative of the volume V with respect to time t gives the rate at which the volume is changing with respect to time:

dV/dt = d((520/3)π)/dt

Since the only variable in the equation is t, we can simplify the derivative:

dV/dt = 0

Therefore, the rate at which the water level is rising is 0 cubic inches per minute. This means that the water level is not rising at all.