Hi guys, I came across this question and just need a little push I think..
If we have a circle inside a square so that the sides of the square lie tangent to the circle, find the rate of change of the perimeter of the square if the rate of change of the circumference of the circle is 6ms^1.
I've tried to write the circumference as 2pi*r
so d/dt (2pi*r) = 6 which obviously isn't right, maybe i'm just not thinking straight..any ideas??
Rate of change of circle inside a square (difficult)

 Follow
 1
 12052012 23:03

 Follow
 2
 12052012 23:11
(Original post by Extricated)
Hi guys, I came across this question and just need a little push I think..
If we have a circle inside a square so that the sides of the square lie tangent to the circle, find the rate of change of the perimeter of the square if the rate of change of the circumference of the circle is 6ms^1.
I've tried to write the circumference as 2pi*r
so d/dt (2pi*r) = 6 which obviously isn't right, maybe i'm just not thinking straight..any ideas??
d/dt(r)=6/(2pi)
The perimeter of the square is 4r 
 Follow
 3
 12052012 23:19
You know it helps to state what topic it is.
You should use connected rates of change.
Label the length of the sides of the square x and the rest should be pretty obvious.
ztibor you didn't do anything lol.
So you have the circumference of the circle
6ms^1 = dC/dt (C= circumference)
Therefore dP/dt = dP/dC *dC/dt
P = Perimeter of square.Last edited by JonathanM; 12052012 at 23:29. 
 Follow
 4
 12052012 23:21
(Original post by JonathanM)
You know it helps to state what topic it is.
You should use connected rates of change.
Label the length of the sides of the square x and the rest should be pretty obvious.
ztibor you didn't do anything lol.
(Except he should've wrote that the perimeter of the square is 8r, I believe). 
 Follow
 5
 12052012 23:24
(Original post by hassi94)
Ztibor has essentially solved the problem so I don't know what you're talking about.
(Except he should've wrote that the perimeter of the square is 8r, I believe).
to "The perimeter of the square is 4r". 
 Follow
 6
 12052012 23:26
(Original post by JonathanM)
How has he solved it? I can't see how he got from "d/dt(r)=6/(2pi)
to "The perimeter of the square is 4r".Last edited by Intriguing Alias; 12052012 at 23:29. 
 Follow
 7
 12052012 23:31
(Original post by hassi94)
He didn't get from one to the other. He got to the first bit, then gave the second bit of information (which okay was wrong but it's easy to make mistakes) and then left the OP to do the simple last bit.
dC/dt = 6
Perimeter = 4x
Circumference = root(2)*pi*x
P = 4(C/pi*root(2))
dP/dC = 4/(pi*root(2))
dP/dt = 24/pi*root(2)Last edited by JonathanM; 12052012 at 23:45. 
 Follow
 8
 12052012 23:39
(Original post by JonathanM)
I don't think the OP understands the question. I understand what the hungarian guy meant now it's just the fact I mistook what he said. And what he said has no relevance to how you solve it anyway. Only the 4r bit.
If d/dt (2pi r) = 6 then d/dt (r) = 6/2pi = 3/pi
And you can logically work out that the square must be of side 2r and so the perimeter is 8r.
Then we can write d/dt(8r) = 24/pi
Now if there's something wrong there, tell me. Otherwise stop commenting that something is wrong or irrelevant just because you don't understand how it's relevant.Post rating:1 
 Follow
 9
 12052012 23:45
(Original post by JonathanM)
I don't think the OP understands the question. I understand what the hungarian guy meant now it's just the fact I mistook what he said. And what he said has no relevance to how you solve it anyway. Only the 4r bit. Top tip for him, a length of a side of a square isn't the radius.
dC/dt = 6
Perimeter = 4x
Circumference = root(2)*pi*x
P = 4(C/pi*root(2))
dP/dC = 4/(pi*root(2))
dP/dt = 24/pi*root(2) 
 Follow
 10
 12052012 23:46
You're answer is wrong the r for the circle is not the same as the r (length of a side) for a square.
Using chain rule I got dP/dt = 24/pi*root(2) 
 Follow
 11
 12052012 23:47
(Original post by hassi94)
Where in the world has root(2) come from?
(Original post by JonathanM)
I don't think the OP understands the question. I understand what the hungarian guy meant now it's just the fact I mistook what he said. And what he said has no relevance to how you solve it anyway. Only the 4r bit. Top tip for him, a length of a side of a square isn't the radius.
dC/dt = 6
Perimeter = 4x
Circumference = root(2)*pi*x
P = 4(C/pi*root(2))
dP/dC = 4/(pi*root(2))
dP/dt = 24/pi*root(2)
lol, it's actually ironic that the only bit that ztibor got wrong (i.e the 4r bit) is what you're claiming is the only bit he's got rightLast edited by Extricated; 12052012 at 23:48. 
 Follow
 12
 12052012 23:48
(Original post by JonathanM)
You're answer is wrong the r for the circle is not the same as the r (length of a side) for a square. 
 Follow
 13
 12052012 23:49
(Original post by Extricated)
lol, it's actually ironic that the only bit that ztibor got wrong (i.e the 4r bit) is what you're claiming is the only bit he's got right 
 Follow
 14
 12052012 23:49
(Original post by JonathanM)
Top tip for him, a length of a side of a square isn't the radius.
dC/dt = 6
Perimeter = 4x
Circumference = root(2)*pi*x
P = 4(C/pi*root(2))
dP/dC = 4/(pi*root(2))
dP/dt = 24/pi*root(2)Post rating:1 
 Follow
 15
 12052012 23:50
(Original post by JonathanM)
Well by that I meant if he meant r as a length of the side of the square, not the radius of the circle. 
 Follow
 16
 12052012 23:52
(Original post by JonathanM)
You're answer is wrong the r for the circle is not the same as the r (length of a side) for a square.
Using chain rule I got dP/dt = 24/pi*root(2)
Each side is the diameter of the circle and since circumference = pi*diameter then C = pi*x with no root(2) 
 Follow
 17
 12052012 23:52
(Original post by F1Addict)
Yeah it isn't r. Its 2r. In words, the length of a side of a square is 2 times the radius of the circle within the square.
Urm what?

 Follow
 18
 12052012 23:53
(Original post by JonathanM)
Well by that I meant if he meant r as a length of the side of the square, not the radius of the circle.

 Follow
 19
 12052012 23:53
(Original post by Ilyas)
...
Sorry for posting it here, but i saw that you had blocked visitor messages. 
 Follow
 20
 12052012 23:54
Original Post:
"we have a circle inside a square so that the sides of the square lie tangent to the circle"
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: May 13, 2012
Share this discussion:
Tweet
Related discussions:
 Difficult Maths/Physics Problems Help Thread
 RESIT Edexcel Biology Unit 1  26th May (6BI01)
 Edexcel GCSE Mathematics A Higher Paper 2016 Unofficial ...
 ocr a f325 revision thread
 Healthy New You: Your Change For Life #3
 OCR C1 19th May 2014
 Maths Ocr 2016 C1
 Edexcel Unit 4: Physics on the Move 6PH04 (11th June 2015)
 AQA Physics PHYA4  20th June 2016 [Exam Discussion ...
 OCR Physics G484  June 2013 Unit 4 (OFFICIAL RETAKE ...
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by: