# A conical icicle is melting at a rate of 10.0 cm^3/h. At 10:00 am, the icicle is 30 cm long and 8 cm in diameter at its widest point. The ...icicle keeps the same proportions as it melts. Determine...

A conical icicle is melting at a rate of 10.0 cm^3/h. At 10:00 am, the icicle is 30 cm long and 8 cm in diameter at its widest point. The ...

icicle keeps the same proportions as it melts. Determine the rate at which its length is decreasing at 5:00 pm.

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(1) At 10:00 am `r=4,h=30,V=160pi`

(2) At 5:00pm `V=160pi-70~~432.65` .

(3)Since the ratio of radius to height is constant, `r/h=4/30 => r=(2h)/15`

(4) `V=1/3pir^2h=1/3pi((2h)/15)^2h=(4pi)/675h^3` .

(5) At 5:00pm we have `432.65=((4pi)/675)h^3=>h~~28.54`

(6) We want to find `(dh)/(dt)` . Now:

`(dV)/(dt)=(4pi)/675*3h^2(dh)/(dt)` Plugging in known values we get:

`-10=(4pi)/222(28.54)^2(dh)/(dt)`

Thus `(dh)/(dt)~~(-10(225))/(4pi(28.54)^2)~~-.22`

**So the length of the icicle is getting shorter by .22cm/h at 5:00pm**

Since the conical icicle melts at a rate of `10 (cm^3)/h =gt (dV)/(dt) =10 =gt dV = 10dt =gt V = 10t + c`

At t = 0 => V = 0 => c = 0

The time passed between 10:00 a.m. and 5:00 p.m. is of 7 hours => the lost volume is `V_1 = 10*7 = 70 cm^3`

Let's calculate the volume of the icicle at 10:00 a.m.

`V = pi*r^2*h/3 = pi*16*30/3 = 160pi cm^3`

After 7 hours, the volume remained is:

Since the icicle keeps the same proportions as it melts =>`r/h = r_1/h_1 = 4/30 = 2/15 =gt r_1 = 2h_1/15`

The volume of the icicle at 5:00 p.m. is:

`432 = (pi*(2h_1/15)^2*h_1)/3`

`1296 = (4/225)*pi*(h_1)^3`

`(h_1)^3 = 23216 =gt h_1 = root(3)(23216) =gt h_1 = 28.52 `

`cm.` **The length the icicle reaches at 5:00 p.m. is of 28.52 cm.**