Answer :
To solve the problem, we need to find the rate of change of the circumference of a circle when the area of the circle is increasing at a certain rate. Here’s a step-by-step guide on how to do it:
1. Understand the Relationship Between Area and Radius:
The formula for the area [tex]\( A \)[/tex] of a circle is given by:
[tex]\[
A = \pi r^2
\][/tex]
where [tex]\( r \)[/tex] is the radius of the circle.
2. Determine the Radius When the Area is Given:
When the area of the circle is 97 square feet, we can find the radius [tex]\( r \)[/tex] by rearranging the area formula:
[tex]\[
r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{97}{\pi}}
\][/tex]
3. Find the Rate of Change of the Radius ([tex]\( \frac{dr}{dt} \)[/tex]):
We know the area is increasing at a rate of 98 square feet per second ([tex]\( \frac{dA}{dt} = 98 \)[/tex]). Using the relationship:
[tex]\[
\frac{dA}{dt} = 2\pi r \frac{dr}{dt}
\][/tex]
we can solve for [tex]\( \frac{dr}{dt} \)[/tex]:
[tex]\[
\frac{dr}{dt} = \frac{\frac{dA}{dt}}{2\pi r} = \frac{98}{2\pi \cdot \sqrt{\frac{97}{\pi}}}
\][/tex]
4. Find the Rate of Change of the Circumference ([tex]\( \frac{dC}{dt} \)[/tex]):
The formula for the circumference [tex]\( C \)[/tex] of a circle is:
[tex]\[
C = 2\pi r
\][/tex]
Differentiating with respect to time gives:
[tex]\[
\frac{dC}{dt} = 2\pi \frac{dr}{dt}
\][/tex]
Substitute the value of [tex]\( \frac{dr}{dt} \)[/tex] found in step 3 into this formula to get [tex]\( \frac{dC}{dt} \)[/tex].
Putting all of this together, the rate of change of the circumference when the area is 97 square feet is approximately 17.637 feet per second.
1. Understand the Relationship Between Area and Radius:
The formula for the area [tex]\( A \)[/tex] of a circle is given by:
[tex]\[
A = \pi r^2
\][/tex]
where [tex]\( r \)[/tex] is the radius of the circle.
2. Determine the Radius When the Area is Given:
When the area of the circle is 97 square feet, we can find the radius [tex]\( r \)[/tex] by rearranging the area formula:
[tex]\[
r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{97}{\pi}}
\][/tex]
3. Find the Rate of Change of the Radius ([tex]\( \frac{dr}{dt} \)[/tex]):
We know the area is increasing at a rate of 98 square feet per second ([tex]\( \frac{dA}{dt} = 98 \)[/tex]). Using the relationship:
[tex]\[
\frac{dA}{dt} = 2\pi r \frac{dr}{dt}
\][/tex]
we can solve for [tex]\( \frac{dr}{dt} \)[/tex]:
[tex]\[
\frac{dr}{dt} = \frac{\frac{dA}{dt}}{2\pi r} = \frac{98}{2\pi \cdot \sqrt{\frac{97}{\pi}}}
\][/tex]
4. Find the Rate of Change of the Circumference ([tex]\( \frac{dC}{dt} \)[/tex]):
The formula for the circumference [tex]\( C \)[/tex] of a circle is:
[tex]\[
C = 2\pi r
\][/tex]
Differentiating with respect to time gives:
[tex]\[
\frac{dC}{dt} = 2\pi \frac{dr}{dt}
\][/tex]
Substitute the value of [tex]\( \frac{dr}{dt} \)[/tex] found in step 3 into this formula to get [tex]\( \frac{dC}{dt} \)[/tex].
Putting all of this together, the rate of change of the circumference when the area is 97 square feet is approximately 17.637 feet per second.
