Question

The figure shows a circle with radius r cm. Another circle is drawn inside the circle such that its diameter is the radius of the original circle. The process is repeated infinitely.
(a) Find the sum of circumferences of the first n circles and express the answer in terms of r and pi.
(b) If the radius of the first circle is 4 cm, find the sum of circumferences of all circles and express the answer in terns of pi.

Answer 1

Interesting... What's the circumference of the first circle?

Answer 2

Given that its radius is r, of course...

Answer 3

2 pi r

Answer 4

That's right :)
Now, since the diameter of the second circle is the radius of the first, that means its radius is half the radius of the first, correct?

Answer 5

radius of the 2nd circle= r/2 ??

Answer 6

That is correct. So its circumference would be...?

Answer 7

\[\pi r\]

Answer 8

fancy. But let's keep it as...
\[\Large \frac{2\pi r}{2}\]
So that you'll better see the pattern.

Now what about the third circle? Its diameter would be the radius of the second circle, so its diameter is r/2. Means its radius is half of that, which is r/4

What then is the circumference of the third circle?

Answer 9

\[\frac{ 2 \pi r }{ 4 }??\]

Answer 10

Correct :)

Answer 11

See the pattern here?
Circumferences...
\[\large 1st = 2\pi r\\\large 2nd = \frac{2\pi r}{2}\\\large 3rd=\frac{2\pi r}{4}\]etc

Answer 12

The circumferences just keep getting divided by two...

Answer 13

yes

Answer 14

Now... that can only mean that your answer is this....
\[\Large 2\pi r + \frac{2\pi r}{2}+ \frac{2\pi r}{4}+ \frac{2\pi r}{8}....\]\[=\Large 2\pi r\left(\frac12\right)^0+2\pi r\left(\frac12\right)^1+2\pi r\left(\frac12\right)^2+2\pi r\left(\frac12\right)^3...\]

Answer 15

common ratio=1/2 ?

Answer 16

You catch on quick :).
\[\huge \sum_{n=1}^\infty2\pi r \left(\frac12\right)^n\]

Answer 17

Whoops...
\[\huge \sum_{n=\color{red}0}^\infty2\pi r \left(\frac12\right)^n\]
my mistake.

Answer 18

what does ∑ mean?

Answer 19

Sum.
Infinite sum.

Answer 20

\[\large\sum_{n=\color{red}0}^\infty2\pi r \left(\frac12\right)^n= 2\pi r\left(\frac12\right)^0+2\pi r\left(\frac12\right)^1+2\pi r\left(\frac12\right)^2+2\pi r\left(\frac12\right)^3...\]

Answer 21

okay..

Answer 22

Do you not know this form?
Surely you have a way to evaluate infinite sums?

Answer 23

Although, if not, we can always fix that :)

Answer 24

\[4\pi r[1-(\frac{ 1 }{ 2 })^{n}]\]

Answer 25

Nice effort, but you're not supposed to have an 'n' in your answer :)

Answer 26

first n circles....

Answer 27

Ahh... okay :D

Answer 28

And, I suppose you let n go to infinity?

Answer 29

yup

Answer 30

So your final answer would be...?

Answer 31

part a?
that's my final answer, i think

Answer 32

No, that can't be it, because the question specifically says 'in terms of r and pi' so you really shouldn't have an 'n' in that expression.

Answer 33

umm, first n circles, in terms of r and pi. n is also included

Answer 34

Enter... the geometric series...
Where |r| < 1
\[\Large S = a + ar + ar^2 +ar^3 + ar^4 ...\]

Answer 35

Yeah, but eventually, should get rid of that n...

Answer 36

and the answer included n..

Answer 37

silly me. My apologies :D

Answer 38

Now what about part b?

Answer 39

sub. r=4 ?

Answer 40

It doesn't say 'first n circles' though.

Answer 41

all circles...

Answer 42

Yup. To infinity... and beyond. :D

Answer 43

Where this comes in
\[\Large S = a + ar + ar^2 +ar^3 + ar^4 ....\]

Answer 44

Where, in this case,
\[\Large a = 2\pi (4)=8\pi \]

and \[\Large r= \frac12\]

Answer 45

Now, pay attention while we work a little magic...

Answer 46

\[\Large S =\color{red} a + ar + ar^2 +ar^3 + ar^4 ....\]
Bring the lone 'a' to the left side
\[\Large S \color{blue}{-a}= ar + ar^2 +ar^3 + ar^4 ....\]

Answer 47

a/ (1-r)

Answer 48

You do like to take the fun away from things :D

Answer 49

continue then :P

Answer 50

\[\Large \frac{S-a}{r}=a+ar+ar^2+ar^3+ar^4+...\]

Answer 51

\[\Large \frac{S-a}{r}=S\]

Answer 52

\[\Large \frac{S}r -S = \frac{a}r\]\[\Large S\left(\frac1r-1\right)=\frac{a}r\]
\[\Large S\left(\frac{1-r}r\right)=\frac{a}r\]\[\Large S=\frac{a}r\cdot\left(\frac{r}{1-r}\right)\]
\[\Huge \color{blue}{S=\frac{a}{1-r}}\]

Answer 53

Okay, well, I only did that because you were already done and I was bored :P

Answer 54

Nice work @terenzreignz

Answer 55

Thanks... I do like a little 'workout' XD

Answer 56

haha, thanks