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 was the water level rising 4 minutes ago? How fast is the water level rising now? What will the rate be when the glass is full?
r = 2 + 4 (h/10) = 2 + .4 h
surface area = pi r^2
surface area*dh = dV
(volume of slice at surface)
pi r^2 dh = DV
pi (4 + 1.6 h + .16 h^2) dh = dV
thus
pi (4 + 1.6 h + .16 h^2) dh/dt = dV/dt
but we know that dV/dt = 10 in^3/min
so
dh/dt = (10/pi)(1/(4 + 1.6 h + .16 h^2)
when h = 0
dh/dt = 10/(4pi) etc
Thanks. I think that I understand the 1st question now. How about the other questions?
You only need solve for h as a function of time
pi(4 + 1.6 h + .16 h^2)dh/dt = dV/dt =10
(4 + 1.6 h + .16 h^2)dh = (10/pi) dt
4 h + .8h^2 + .0533 h^3 = (10/pi) t
1.26 h + .251 h^2 + .0167 h^3 = t
You have to solve this for h and then r when t = 4 min (I would do it by making a few guesses and interpolating)
at the top is easy because then you know that h = 10 and r = 6
OK, I understand now. Thanks for all of your help.
In the 2nd question (rate of water rising now), how do you calculate r?
To find the rate at which the water level is rising, we can use the concept of related rates. We know that the volume of water in the cup is increasing at a rate of 10 cubic inches per minute. We need to find the rate at which the water level is rising.
Let's begin by finding a relationship between the height and the volume of the water in the cup. The volume of a cylinder is given by the formula V = πr^2h, where V is the volume, r is the radius, and h is the height.
Given that the cup has a radius of 2" at the bottom and 6" on the top, the radius of the water at any height h can be expressed using similar triangles. Thus, the radius of the water at height h is given by:
r(h) = (6/10)h
Using this expression for the radius in the volume formula, we can now express the volume of the water as a function of h:
V(h) = π(6/10h)^2h = (36π/100)h^3 = (9π/25)h^3
Next, we'll differentiate V with respect to time (t) to get an expression for the rate at which the water level is rising (dh/dt) in terms of the rate at which the volume is increasing (dV/dt):
dV/dt = (9π/25)(3h^2)(dh/dt)
10 = (9π/25)(3h^2)(dh/dt)
We were given that dV/dt = 10 cubic inches per minute. Now, let's evaluate the expression for dh/dt when the cup is 10" high.
When the cup is 10" high, substituting h = 10 into the equation gives:
10 = (9π/25)(3(10^2))(dh/dt)
10 = (9π/25)(300)(dh/dt)
dh/dt = 10 / [(9π/25)(300)]
Now, let's calculate the value of dh/dt when t = 0 (4 minutes ago) and when the cup is full.
When t = 0 (4 minutes ago), we substitute t = 0 into the equation:
dh/dt = 10 / [(9π/25)(300)]
= 250 / (9π)
≈ 8.84 in/min (rounded to two decimal places)
To find the rate at which the water level is rising now, when the cup is not full, we need to know the current height of the water level. If you provide that information, I can help you determine the rate at which the water level is rising.
Lastly, when the cup is full, the height is equal to 10". Substituting h = 10 into the equation:
dh/dt = 10 / [(9π/25)(300)]
= 250 / (9π)
≈ 8.84 in/min (rounded to two decimal places)
Therefore, when the cup is full, the rate at which the water level is rising is approximately 8.84 inches per minute.