Answer:
r=1/π
Step-by-step explanation:
Area of the circle is defined as:
Area = πr²
Derivating both sides
[tex]\frac{dA}{dr}[/tex]=2πr
[tex]\frac{dA}{dt}[/tex] = [tex]\frac{dA}{dr}[/tex] x [tex]\frac{dr}{dt}[/tex] = 2πr[tex]\frac{dr}{dt}[/tex]
If area of an expanding circle is increasing twice as fast as its radius in linear units. then we have : [tex]\frac{dA}{dt}[/tex] =2[tex]\frac{dr}{dt}[/tex]
Therefore,
2πr [tex]\frac{dr}{dt}[/tex] = 2 [tex]\frac{dr}{dt}[/tex]
r=1/π
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
[tex]\frac{dA}{dt} =2 \times \frac{dr}{dt}[/tex]
However,
[tex]\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}[/tex] and
[tex]\frac{dA}{dr} = \frac{d\pi r^2}{dr} = 2\pi r[/tex]
Therefore, we have
[tex]\frac{dA}{dt} =2 \times \frac{dr}{dt} = 2\pi r \times \frac{dr}{dt}[/tex]
Cancelling like terms
[tex]1= \pi r[/tex]
Therefore, [tex]r = \frac{1}{\pi }[/tex].
