Question

Ice cream is being put into a cone at a rate of 4 cm3/s. The cone has a radius of 2 cm and a height of 8 cm.

At what rate is the height of ice cream rising when it is half full?

Be careful- it is not half full when the height of the ice cream is 4 cm!

Answer 1

0.5052 cm per sec ( approx )

Explanation:

∵ The volume of a cone is,

`V=(1)/(3)\pi (r)^2h`

Where,

r = radius,

h = height,

Here, r = 2 and h = 8,

Thus, the volume of the cone is,

`V=(1)/(3)\pi (2)^2 8=(32\pi )/(3)`

When the cone is half filled,

Volume would be,

`V_1=(V)/(2)=(16\pi )/(3)`

Now,
`(r)/(h)=(1)/(4)\implies r=(h)/(4)`

`V_1=(1)/(3)\pi ((h)/(4))^2 h=(\pi h^3)/(48)----(1)`

`\implies (\pi h^3)/(48) = (16\pi )/(3)`

`\implies h \approx 6.35`

Differentiating equation (1) with respect to t ( time )

`(dV_1)/(dt)=(\pi h^2)/(16)(dh)/(dt)`

We have,
`(dV_1)/(dt)=4\text{ cube cm per s},h=6.35\text{ cm}`

`4=(\pi (6.35)^2)/(16)* (dh)/(dt)`

`\implies (dh)/(dt)=(64)/(6.35^2 \pi)\approx 0.5052\text{ cm per sec}`