Respuesta :
Answer:
The rate of change of the area of the circle is approximately [tex]-490.09 \ m^2/min[/tex].
Step-by-step explanation:
Given:
[tex]\frac{dr}{dt} = -6.5 \ m/min[/tex]
radius [tex]r =12 \ m[/tex]
We need to find the rate of change of the area of the circle at that instant.
Solution:
Now we know that;
Area of the circle is given by π times square of the radius.
framing in equation form we get;
[tex]A= \pi r^2[/tex]
to find the rate of change of the area of the circle at that instant we need to take the derivative on both side.
[tex]\frac{dA}{dt}=\frac{d(\pi r^2)}{dt}\\\\\frac{dA}{dt}= 2\pi r \frac{dr}{dt}[/tex]
Substituting the given values we get;
[tex]\frac{dA}{dt}= 2\times \pi \times 12 \times -6.5\\\\\frac{dA}{dt}\approx -490.09 \ m^2/min[/tex]
Hence The rate of change of the area of the circle is approximately [tex]-490.09 \ m^2/min[/tex].
