Question

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 5/100 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.

Answer 1

a)

we know,

volume of cone = π r2 h /3

where r = radius of cone and h= height of cone

given,

height = 3 inches , radius = 1 inch

now,

at an instant , height = y

to find radius at that instant , use property of similar triangles .

which gives,

3/y = 1/r

r= y/3

so we, got height and radius at that instant .

therefore,

volume at that instant , V =  π r2 h /3 =  π *(y/3)2 *y / 3 =  π y3 /27.

b)

given,

height of sand in bottom cone at that instant = h

therefore, height of empty space in cone = 3-h

to find radius of empty space at that instant , use property of similar triangles .

which gives,

(3-h)/3 = r/1

r= (3-h)/3

therefore, volume of empty space in cone B = V= π r2 h /3 = π *[(3-h)/3]2 (3-h) /3 =π *(3-h)3/27 .

now , volume of sand in cone B = total volume of cone B - volume of empty space

= π r2 h /3 - π *(3-h)3/27

= π 12*3/3 -π *(3-h)3/27 .

= π - π *(3-h)3/27 .

c)

total volume of sand = 3π/4.

total volume of sand = volume of sand in A + volume of sand in B

3 π/4 = π y3 /27 +  π - π *(3-h)3/27

at an instant , h= 1 (given), now find y at that instant

therefore,

3 π/4   = π y3 /27 +  π - π *(3-1)3/27

3 π/4   = π y3 /27 +  π - π *(2)3/27

3 π/4   = π y3 /27 +  π - π *8/27

3 π/4   = π y3 /27 + π 19/27

27* 3 π/4 = π y3 + 19 π

cancelling π

81/4 = y3 +19

y3 = 81/4 -19

y3 = 5/4.

y = 1.077.

now,

volume of uppercone A = V = π y3 /27.

derievating

dV = 3 π y2 dy /27

= π y2 dy /9

dy = 2/100 = 0.02 , y( we got it) = 1.077

therefore,

dV =  π (1.077)2 *0.02 /9

= 0.002578 π

it is true that rate of chane of volume for both the cylinders is equal

therefore,

dV for cone B = 0.002578 π

now,

we have ,

volume of cone B = V = π - π *(3-h)3/27

differentiating,

dV = - π 3*(-1)*(3-h)2 dh/27

at this instant, h=1

0.002578 π = π *(3-1)2 dh /9

0.002578 π = π *22 dh /9

0.002578 π = π *4dh /9

cancelling π

0.002578 = 4dh /9

dh = 9*0.002578/4

= 0.0058 inches per second.

= 58/10000

rate of increase in height h in cone B = 0.0058 inches per second.