Question

If a snowball melts so that its surface area decreases at a rate of 4 cm2/min, find the rate at which the diameter decreases when the diameter is 9 cm.

Answer 1

Main Answer:

Step-by-step explanation:

Given that,

`(dA)/(dt)` = -4

The surface area of the snow ball = 4π
`r^(2)`

Let the surface area = A

If A denotes the surface area and D the diameter then,

A = 4π
`r^(2)` = 4π
`((D)/(2) )^(2)`

⇒ A = π
`D^(2)`

Differentiate with respect to r,

`(dA)/(dD)` = 2πD

By using chain rule,

`(dA)/(dD)` =
`((dA)/(dt))`
`((dt)/(dD))` =
`((dA)/(dt))/((dD)/(dt))`

⇒ 2πD =
`-(4)/((dD)/(dt))`

∴
`(dD)/(dt)` =
`-(4)/(2\pi\\D)`

when D = 9,

`(dD)/(dt)` =
`-(4)/(18\pi)`

∴
`(dD)/(dt)` =
`-(2)/(9\pi)`

so, the diameter is decreasing at a rate of
`(2)/(9\pi)` .