The two cones are aligned on edge to edge so that the total volume of the hourglass becomes 2∗13πr2h
Step-by-step explanation:
a) When the sand has y units left in the upper cone the height of the cone becomes y, the radius remains 1cm.Now let's take the right triangle In the upper coneAnd by similar triangles property we found that the radius of the cone in which the amount of sand is left is y3 unitsNow the volume of the sand in the remaining part of the upper cone isπy32y which gives πy327Now if we apply the same similar triangle property with the cone in the below we find that the radius of the part it has filled up remains the same as y3unitsSo the volume it has filled is (3-y) from the top cone which says that the volume filled on the bottom cone has a height of (3-y)So therefore the volume of the frustum which it has filled up isπ3h(R2+r2+rR) {Is the volume of the frustum with two radii r and R with a height h}i.eπ3(3−y)(12+(y/3)2+(y/3))So now the total volume of the hourglass in terms of y is π3(3−y)(12+(y/3)2+(y/3))+πy327b) The next question tells us that when the height of the sand in the below cone reaches up a height of h centimetreswhich is exactly 3-y in the previous questionso substituting y = h-3 in the previous question we'll find the volume of the hourglass in terms of hπ(h−3)327+π3(6−h)(12+((h−3)/3)2+((h−3)/3)) .c) Total volume of the hourglass isV=V1+V2⇒3π4=πy327+π−π27(3−h)3Putting h=1We have the value of y as⇒3π4=π27y3+π+π278..............(1)⇒y=(54)13Differentiating the first equation with respect to t we have0=π9y2dydt+π9(3−h)2dhdtPutting values that dydt=−0.14inches/sec at h =1 and for ⇒y=(54)13We have0=π9(54)132∗(−0.14)+π922dhdt⇒0.1956π9=4π9dhdt⇒dhdt=0.19564⇒dhdt=0.0489
