Respuesta :
a) The volume is (π/27)y³
b) the volume is π -(π/27)(3 -h)³.
c) the rate at which the sand in the lower cone is increasing is 0.0116 inches per second
a) since we know The volume of a cone is: V = 1/3πr²h, for the problem here we have h = y and r = y/3, so the volume is : (1/3)π(y/3)²(y) = (π/27)y³
b) Now, we know The empty volume of the lower cone is the same as the volume of the upper cone with y=3, so V = (π/27)(3³) = π , as the sand height is h in the lower cone, so the height of the empty space is (3-h), the volume of the empty space is : (π/27)(3-h)³ and the volume will be the difference between that and the cone volume: π -(π/27)(3 -h)³. c) since the ratio of rates of change in height will be inversely proportional to the surface areas of the sand in each section of the cone, the ratio will be proportional to the square of the distance from the sand surface to the cone tip. For the height of the sand in the lower cone of 1 inch high, its volume is: π -(π/27)(3 -1)³ = π -(8π/27) = 19π/27, then the volume of the sand in the upper cone is: 3π/4 -19π/27 = 5π/108, the height of the sand in the upper cone is then
(π/27)h³ = π(5/108)
h³ = 5/4 ,h = ∛(5/4), the ratio of the surface areas in the two cones are The ratio of surface areas : (2 / ∛(5/4))², the rate of decrease is 4/100, rate of increase of sand in lower cone is (4/100)/((2 / ∛(5/4))²) =(∛12.5)/200 ≈ 0.0116
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