Rate of Increase in Area of Circle with Circumference 20pi m

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SUMMARY

The discussion focuses on calculating the rate of increase in the area of a circle when its circumference is 20π meters. The radius is increasing at a constant rate of 0.2 meters per second. Using the formula for the area of a circle, A = πr², and applying the chain rule, the rate of area increase is determined to be dA/dt = 4π m²/s when the radius is 10 meters. The solution emphasizes the importance of including units in the final answer.

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Homework Statement


The radius of a circle is increasing at a constant rate of 0.2 meter per second. What is the rate of increase in the area of the circle at the instant when the circumference of the circle is 20pi meters?


Homework Equations





The Attempt at a Solution


I have no clue how to approach the problem
 
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The usual way to approach such problems is to find a relationship between the two quantities and then apply the chain rule to find the quantity you're interested in.

What is the relationship between the radius of a circle and its area?
Once you have that, try differentiating both sides with respect to time.
 
a=(pi)(r^2)...
how could i differentiate a?.. dA/dt?
 
It's quite simple when you look at it really.
You have dr/dt (rate of increase of radius) = 0.2 where r = radius and t = time (seconds).
you are looking for dA/dt where A = area of circle

A = \pi r^2

\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt} <-- chain rule.

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

Therefore the rule for rate of area increase = 0.4 \pi r

Now you just have to find it at the instant when circumference = 20 \pi

I'm sure you can take it from here easily. I hope I didn't give away too much.
 
Last edited:
Thank you :)

C = 2(pi)r
r = C/2(pi)
r = 20pi/2pi
r = 10
dA/dt = 2(pi)r * dr/dt
dA/dt = 2(pi)r * 0.2
dA/dt = 2(pi)(10) * 0.2
dA/dt = 4pi
is that correct?
 
Looks fine to me. You may get marked down if you don't include units, ie. 4pi m^2 per second.
 

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