(old final exam problem-sp2009) an hourglass is made up of two glass cones connected at their tips. both cones have radius 1 inch and height 3 inches. when the hourglass is flipped over, sand starts falling to the lower cone. (a) when the sand remaining in the upper cone has height y inches, its volume a in terms of y is . (b) when the sand in the lower cone has reached a height of h inches, its volume b in terms of h is . (hint: b is the volume of the bottom cone minus the volume of the empty space above the sand.) (c) assume the total volume of sand in the hourglass is 3π/4 cubic inches. also, assume the height of the sand in the upper cone is decreasing at a rate of 4/100 inches per second. at the instant when the sand in the lower cone is 1 inch high, the height of the sand in the lower cone is increasing at a rate of inches per second.

  (a)  A = (π/27)y³

  (b)  B = π -(π/27)(3 -h)³

  (c)  0.0116 in/s

Step-by-step explanation:

(a) The volume of a cone is given by the formula ...

  V = 1/3πr²h

For this part of the problem, the volume is designated A and we have h = y and r = y/3, so the volume is ...

  A = (1/3)π(y/3)²(y)

  A = (π/27)y³

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(b) The empty volume of the lower cone (V) is the same as the volume of the upper cone with y=3, so is ...

  V = (π/27)(3³) = π . . . . . cubic inches

When the sand height is h in the lower cone, the height of the empty space is (3-h), so the volume of the empty space is ...

  empty volume = (π/27)(3-h)³

and the volume B is the difference between that and the cone volume:

  B = cone volume - empty volume

  B = π -(π/27)(3 -h)³

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(c) We want the rate of change of height in the lower cone at a particular height when the rate of change of height in the upper cone is a particular value. We could sum the volumes and differentiate to find the rate of change of one height with respect to the other. (We would still need to find the height in the upper cone.) Or, we can cut to the chase.

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. In turn, that ratio will be proportional to the square of the distance from the sand surface to the cone tip.

When the sand in the lower cone is 1 inch high, its volume is ...

  B(1) = π -(π/27)(3 -1)³ = π -(8π/27) = 19π/27

Then the volume of the sand in the upper cone is ...

  A = 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) . . . . . . . height of sand in upper cone

The sand surface in the lower cone is 1 inch above the base, so 2 inches from the apex. The ratio of surface areas in the two cones is then ...

  lower sand area / upper sand area = (2 / ∛(5/4))²

The height of the sand in the lower cone will be increasing at a rate inversely proportional to this and proportional to the rate of decrease of height in the upper cone, given as 4/100 in/s.

  rate of increase of sand in lower cone = (4/100)/((2 / ∛(5/4))²)

  = (∛12.5)/200 ≈ 0.0116 . . . . inches per second

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

$12. that is my answer.