Finding the approximate change in the perimeter of a circle

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 2K views
a129

Homework Statement


The radius of a circle increases from 3 to 3.01 cm. Find the approximate change in its perimeter.

Here's a link to the actual question, in case you need the answer for 6(a) to solve 6(b)
http://imgur.com/a/nQt6M

Homework Equations


Perimeter of circle = 2πr
Area of circle = πr^2

dy/dx ≈ the limit of δy/δx as δx approaches 0

The Attempt at a Solution


Well I only have a problem in solving 6(b).
We know that δr = 3.01-3 = 0.01cm, and so that means δr is already approaching to 0

From this we can say that dp/dr ≈ δp/δr
So to find the change in the perimeter, (dp/dr)(δr) = δp

My only problem here is to find dp/dr
We know the perimeter of circle is 2πr, so by differentiating this we get dp/dr = 2π. But I don't understand, because now I can't input the value r =3cm into dp/dr since the r isn't there anymore after the equation is differentiated. I feel like the equation would only make sense if the the value dp/dr is equated with the value r=3cm so we know the change of the perimeter at THAT exact point. Since δr= 3.01-3, the diffrence in value is so small that it is ≈ dp/dr when r=3.

My final answer would look like δp= 2π(0.01)
= π/50

It's like it makes sense but at the same time it also doesn't make sense and I feel frustrated by it. Thanks for reading!
 
Physics news on Phys.org
Use the Taylor expansion of perimeter truncated up to the first order.
a129 said:
So to find the change in the perimeter, (dp/dr)(δr) = δp
You seem to already have the formula obtained through the step I suggested above. What prevents you from using this formula?