Yan
water leaks from a hemispherical bowl of 8 cm diameter at the rate of 10 cm per hour .How fast is the surface
falling when the water is 2 cm deep in the bowl
The volume of a spherical cap of radius r and height h is
v = π/3 h^2(3r-h)
= πrh^2 - π/3 h^3
dv/dt = πh^2 dr/dt + 2πrh dh/dt - πh^2 dh/dt
= π(2rh-h^2) dh/dt
Now just plug in the numbers:
π(2*4*2-4) dh/dt = -10
dh/dt = -5/(6π) cm/hr
Well, well, well, it seems we have a leaky situation on our hands! Let's get down to business, shall we?
To find out how fast the surface is falling when the water is 2 cm deep in the bowl, we need to whip out some math skills. Are you ready?
First, we need to figure out the volume of water in the bowl when it's 2 cm deep. Since the bowl is a hemisphere, the volume can be calculated using the formula V = (2/3)πr³, where r is the radius.
With a diameter of 8 cm, the radius is half of that, which gives us r = 4 cm. Plug in that value, and you'll find the volume of water in the bowl when it's 2 cm deep.
Now, we know that the water is leaking at a rate of 10 cm per hour. So, to find out how fast the surface is falling, we need to find the rate at which the volume is changing.
Using some calculus magic, we can differentiate the volume formula with respect to time to get dV/dt. This will give us the rate of change of volume with respect to time.
Once you have that rate, divide it by the surface area of the water when it's 2 cm deep. The surface area of a hemisphere can be calculated using the formula A = 2πr².
Finally, plug in the values and you'll get yourself the rate at which the surface is falling when the water is 2 cm deep in the bowl.
Voila! You've done it! Just be careful not to slip on the water while you're calculating. Safety first, my friend!
To find how fast the surface is falling when the water is 2 cm deep in the bowl, we can use related rates and the formula for the volume of a hemisphere.
The volume of a hemisphere is given by the formula: V = (2/3) * π * r^3
where V is the volume and r is the radius of the hemisphere.
Since the diameter of the bowl is given as 8 cm, the radius is half of that, which is 4 cm.
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = (2/3) * π * 3r^2 * dr/dt
We are given that the water leaks at a rate of 10 cm per hour, so dV/dt = -10 cm^3/hr (negative because the volume is decreasing).
Substituting the values:
-10 = (2/3) * π * 3(4^2) * dr/dt
-10 = (2/3) * π * 3(16) * dr/dt
-10 = (2/3) * π * 48 * dr/dt
-10 = 32π * dr/dt
To find dr/dt, we need to know the depth of the water in the bowl. We are given that the water is 2 cm deep.
Since the hemisphere has a depth of 4 cm, when the water level is at 2 cm deep, the radius of the remaining sphere will be 2 cm.
Substituting the values:
-10 = 32π * dr/dt
dr/dt = -10 / (32π)
dr/dt ≈ -0.099 cm/hr
Therefore, when the water is 2 cm deep in the bowl, the surface is falling at a rate of approximately 0.099 cm per hour.
To find the rate at which the surface is falling, we need to find the rate at which the height of the water in the bowl is changing.
Let's first find the volume of the water in the bowl at any given time. The volume of a hemisphere can be calculated using the formula:
V = (2/3) * π * r^3
Since the diameter of the bowl is 8 cm, the radius (r) is half of that, which is 4 cm = 0.04 m.
So, the volume of the water in the bowl is given by:
V = (2/3) * π * (0.04)^3
To find the rate at which the volume is changing, we need to differentiate this expression with respect to time. Since the water is leaking out of the bowl, the volume is decreasing.
dV/dt = -(2/3) * π * (0.04)^3 * dh/dt
Where dV/dt represents the rate at which the volume is changing, and dh/dt represents the rate at which the height of the water is changing.
We are given that the rate at which the water leaks is 10 cm per hour, which is equivalent to 0.1 m per hour. Thus, dh/dt = -0.1.
Now we can substitute the given values and solve for the rate at which the surface is falling:
-0.1 = -(2/3) * π * (0.04)^3 * dh/dt
dh/dt = -0.1 / (-(2/3) * π * (0.04)^3)
dh/dt ≈ 21.9 cm/hour
Therefore, the surface is falling at a rate of approximately 21.9 cm per hour when the water is 2 cm deep in the bowl.