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MTH 174
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Can you please help me with this problem, I don't even know how to set it up properly. If water is rising at .61 cm/sec in a sphere of radius 48 cm then at what net rate in cm^3/sec is water entering the sphere when water depth is 7 cm
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Sure. You can also see that sort of problem in the Class Notes.
Sphere radius is 24 cm.
When the depth of water is the sphere is 7 cm, the surface of the water is 24 cm - 7 cm = 17 cm below the center. The water surface makes a circle. We want to start by finding the area of that circle.
Form a triangle by starting at the center of the sphere. Go straight down until you hit the surface of the water at the center of the circle, forming one leg. Then go straight across the water surface until you hit the sphere, forming a second leg perpendicular to the first. Then go straight back to the center of the sphere, forming the hypotenuse of your right triangle.
Your second leg, which starts at the center of the circle, forms a radius of the circle.
Let a and b be the two legs, c the hypotenuse. Since a^2 + b^2 = c^2, we know that b = sqrt(c^2 - a^2). If b is the second leg then b = sqrt(24^2 - 17^2) cm, roughly 20 cm.
The water surface forms a circle of radius 20 cm. The area of that circle is pi * (20 cm)^3, about 400 pi = 1260 cm^2.
A cylinder .61 cm high, whose base is the circle, has volume V = A * h = 1260 cm^2 * .61 cm = 750 cm^3. So water would have to be flowing in at the rate of 750 cm^3 / second to make the water rise at .61 cm/s.
You might object that the portion of the sphere being filled is getting wider as water rises. That's so. The cylinder wouldn't be as wide as the sphere at any point but its base, so 750 cm^3 wouldn't actually suffice to fill the sphere to a depth of .61 cm. However at the instant the water depth is 7 cm, and for a very small time interval beyond that instant, the sphere doesn't have time to get significantly bigger and the 750 cm^3/s rate would be accurate.
In general if sphere radius is r and water depth y, the circle has radius b = sqrt(r^2 - y^2) and area pi b^2 = pi ( r^2 - y^2 ). Rising at rate y ', in a short time interval `dt the water will rise distance y ' * `dt, sufficient to fill a cylinder of volume `dV = A * h = pi ( r^2 - y^2 ) * (y ' * `dt), so that the rate at which water would rise in the cylinder is `dV / `dt = pi ( r^2 - y^2) * y '. As `dt -> 0 the cylinder differs by less and less from the corresponding portion of the sphere, so that the instantaneous rate of change of volume is in fact equal to `dV / `dt = pi ( r^2 - y^2) * y '.