Infinitely smaller changes in a related rates question

[NoahsArk]

I was looking at related rates problems, and the problem of finding the rate at which the area of a circle changes with respect to time. In order for the area of a circle to be changing at a constant rate, say per second, it would have to mean that the radius was increasing by a smaller and smaller amount each second. In that case, can the area continue changing by the same amount each second for an infinite amount of time?

 

[jedishrfu]

You could answer this yourself by considering the equation:

##area(t) = A * t = \pi * r^2## where A is some constant rate of change in a area per second units.

solve for r as a function of t.

What do you see for r as t increases? You could plot it via the desmos online calculator at:

www.desmos.com/calculator

 

[NoahsArk]

In the examples given, we already know the change in the radius per second, and are asked to find the instantaneous rate of change of the area at a given time. This part I can do, but was wondering what the physical significance would be of the area changing at a constate rate. If I'm not mistaken, it would have to mean that the radius was increasing by a lesser and lesser amount each second.

 

[jedishrfu]

Yes and I provided you with a means to answer it.

So what do you see?

r is dependent on the square root of time so at one second r is 1 at 2 secs r is 1.414 at 10 secs r is 3.162 and at 100 secs r is 10.

 

[NoahsArk]

I will think about it and write back. Thanks.

 

[jedishrfu]

What’s there to think about?

As time increases, the area increases at a constant rate and r increases as square root of that rate meaning there is no limit for r increasing even the rate of increase gets smaller and smaller to zero.

 

[NoahsArk]

## A * t = \pi * r^2 ##,
## r^2 = \frac {A * t} {\pi} ##
## r = \sqrt \frac {A * t} {\pi} ##
Is this what you meant by solving for r as a function of t?

 

[jedishrfu]

yes, so the equation tells you how r evolves relative to t.

r is proportional to the square root of t

 

[Delta2]

What’s there to think about?

Its not crystal clear that the rate of increase of r is a decreasing function. For this to be crystal clear we have to consider the derivative ##r'(t)## and show that it is indeed a decreasing function. Nothing too hard but just saying...

 

[jedishrfu]

Its not crystal clear that the rate of increase of r is a decreasing function. For this to be crystal clear we have to consider the derivative ##r'(t)## and show that it is indeed a decreasing function. Nothing too hard but just saying...

This was in response to the OP saying he'll think about it and get back that's code for saying I'm moving on to the next problem and will let this one slide.

I was hoping the OP would notice that r is proportional to the square root of t and that there is no upper limit for r just that the change in r gets smaller and smaller as opposed to writing r in terms of t and not thinking any deeper.