Question

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

Answer 1

Answer:

r=1/π

Step-by-step explanation:

Area of the circle is defined as:

Area = πr²

Derivating both sides

$\frac{dA}{dr}$=2πr

$\frac{dA}{dt}$  =  $\frac{dA}{dr}$ x $\frac{dr}{dt}$  =  2πr$\frac{dr}{dt}$

If area of an expanding circle is increasing twice as fast as its radius in linear units. then we have : $\frac{dA}{dt}$  =2$\frac{dr}{dt}$

Therefore,

2πr $\frac{dr}{dt}$  =  2  $\frac{dr}{dt}$

r=1/π

Answer 2

Answer:

r = 1/π

Step-by-step 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

$\frac{dA}{dt} =2 \times \frac{dr}{dt}$

However,

$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$ and

$\frac{dA}{dr} = \frac{d\pi r^2}{dr} = 2\pi r$

Therefore, we have

$\frac{dA}{dt} =2 \times \frac{dr}{dt} = 2\pi r \times \frac{dr}{dt}$

Cancelling like terms

$1= \pi r$

Therefore, $r = \frac{1}{\pi }$.