Question

When the area in square units of an expanding circle is increasing twice as fast as its radius in linear units

Answer 1

r = 1/π

Explanation:

Here we have

Area of a circle given as

Area = πr²

Where:

r = Radius of the circle

When the area of the circle is expanding twice as fast s the radius we have

`(dA)/(dt) =2 * (dr)/(dt)`

However,

`(dA)/(dt) = (dA)/(dr) * (dr)/(dt)` and

`(dA)/(dr) = (d\pi r^2)/(dr) = 2\pi r`

Therefore, we have

`(dA)/(dt) =2 * (dr)/(dt) = 2\pi r * (dr)/(dt)`

Cancelling like terms

`1= \pi r`

Therefore,
`r = (1)/(\pi )`.

Answer 2

r=1/π

Explanation:

Area of the circle is defined as:

Area = πr²

Derivating both sides

`(dA)/(dr)`=2πr

`(dA)/(dt)` =
`(dA)/(dr)` x
`(dr)/(dt)` = 2πr
`(dr)/(dt)`

If area of an expanding circle is increasing twice as fast as its radius in linear units. then we have :
`(dA)/(dt)` =2
`(dr)/(dt)`

Therefore,

2πr
`(dr)/(dt)` = 2
`(dr)/(dt)`

r=1/π