Consider a hemispherical bowl of radius 10cm.
(4 points) If the water level in the bowl is h cm deep, find the radius of the surface of the water as a function of h.
(6 points) The water level drops at a rate of 0.1 cm per hour. At what rate is the radius of the water decreasing when the depth is 5 cm?
draw a vertical cross-section of the bowl. Pick a point H on the vertical axis at height h from the bottom. If P is a point on the rim of the bowl, HP = 10. The radius at H is
(10-h)^2 + r^2 = 10^2
r = √(100-(10-h)^2)
To find the radius of the surface of the water as a function of h, we can use the Pythagorean Theorem. Let's call the radius of the water surface R.
Since we have a hemisphere, the radius of the base is the same as the given radius, which is 10cm. The height, h, can be considered as the distance between the top of the hemisphere and the surface of the water.
Using the Pythagorean Theorem, we can write:
R^2 = (10)^2 - h^2
Now, let's differentiate both sides of the equation with respect to t (time):
2R dR/dt = -2h dh/dt
Since we are interested in finding the rate at which the radius is decreasing when the depth is 5 cm, we can substitute h = 5 cm into the equation.
2R dR/dt = -2(5) dh/dt
Simplifying, we have:
R dR/dt = -5 dh/dt
Now, we need to find the rate at which the depth is decreasing, which is given as -0.1 cm per hour. So, we can substitute dh/dt = -0.1 cm/hr into the equation.
R dR/dt = -5(-0.1)
R dR/dt = 0.5 cm/hr
Therefore, the rate at which the radius of the water is decreasing when the depth is 5 cm is 0.5 cm/hr.
To find the radius of the surface of the water as a function of h, we can use the formula for the volume of a hemisphere:
V = (2/3)πr^3
where V is the volume and r is the radius of the hemisphere.
In this case, the volume of the hemisphere will be equal to the volume of water inside the bowl. The volume of water is given by the product of the cross-sectional area and the depth of the water:
V = A * h
The cross-sectional area can be found using the formula for the area of a circle:
A = πr^2
Substituting the expression for A into the equation for V, we get:
V = πr^2 * h
Since the volume remains constant, we can set the two expressions for V equal to each other:
(2/3)πr^3 = πr^2 * h
Simplifying and solving for r, we find:
r = (2/3)h
Therefore, the radius of the surface of the water is given by the function r = (2/3)h.
Now let's move on to the second part of the question. We need to find the rate at which the radius of the water is decreasing when the depth is 5 cm.
The rate at which the water level drops is given as 0.1 cm per hour. This means the rate of change of the depth (dh/dt) is -0.1 cm/hr since the depth is decreasing.
To find the rate at which the radius is decreasing, we can differentiate the equation r = (2/3)h with respect to time t:
dr/dt = (2/3)(dh/dt)
Substituting the given value dh/dt = -0.1 cm/hr, we get:
dr/dt = (2/3)(-0.1) = -0.2/3 = -0.067 cm/hr
Therefore, the rate at which the radius of the water is decreasing when the depth is 5 cm is -0.067 cm/hr.