The rate of increase in the area of the circle at the instant when the circumference of the circle is 20π m is 12.56 [tex]\rm m^2/sec[/tex] .
Given :
Radius of a circle is increasing at a constant rate of 0.2 m/sec.
Circumference of the circle at an instant is [tex]20 \pi[/tex] m.
Solution :
We know that area of the circle is,
[tex]\rm A = \pi\times r^2[/tex] --- (1)
Now differentiate equation (1) with respect to time t,
[tex]\rm \dfrac{dA}{dt}=2\pi r \dfrac{dr}{dt}[/tex]
Given that
[tex]\rm \dfrac{dr}{dt}=0.2\;m/sec[/tex]
so,
[tex]\rm \dfrac{dA}{dt}= 0.4\times \pi\times r[/tex] ------ (2)
We also know that circumference of the circle is,
[tex]\rm C=2 \pi r[/tex] --- (3)
At [tex]\rm C = 20 \pi[/tex] from equation (3) we get,
[tex]2\pi r = 20 \pi[/tex]
[tex]\rm r =10\;m[/tex]
Now put the value r in equation (2) we get
[tex]\rm \dfrac{dA}{dt}=0.4\times \pi \times 10[/tex]
[tex]\rm \dfrac{dA}{dt} = 12.56\;m^2/sec[/tex]
The rate of increase in the area of the circle at the instant when the circumference of the circle is 20π m is 12.56 [tex]\rm m^2/sec[/tex] .
For more information, refer the link given below
https://brainly.com/question/16418397?referrer=searchResults
This question involves the concepts of circumference, area, and derivative.
The rate of increase in area is "12.57 m²/s".
The area of a circle is given as:
[tex]A = \pi r^2[/tex]
Taking the derivative with respect to time (t) on both sides, we get:
[tex]\frac{dA}{dt}= \pi\frac{d(r^2)}{dt}\\\\\frac{dA}{dt}= 2\pi r\frac{dr}{dt}\\[/tex]
where,
[tex]\frac{dA}{dt}[/tex] = rate of change in area = ?
[tex]2\pi r[/tex] = Circumference of the circle = 20π m
[tex]\frac{dr}{dt}[/tex] = rate of increase in radius = 0.2 m/s
Therefore,
[tex]\frac{dA}{dt}=(20\pi\ m)(0.2\ m/s)\\\\\frac{dA}{dt}=12.57\ m^2/s[/tex]
Learn more about the area of the circle here:
https://brainly.com/question/22275589?referrer=searchResults
The attached picture shows the area of circle.
