Just one sec...
Hey! Sign in to get help with your study questionsNew here? Join for free to post

Rate of change of circle inside a square (difficult)

Announcements Posted on
Take our short survey, £100 of Amazon vouchers to be won! 23-09-2016
    • Thread Starter
    Offline

    2
    ReputationRep:
    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??
    Offline

    3
    ReputationRep:
    (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??
    SO the rate of r
    d/dt(r)=6/(2pi)
    The perimeter of the square is 4r
    Offline

    2
    ReputationRep:
    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.
    Offline

    2
    ReputationRep:
    (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.
    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).
    Offline

    2
    ReputationRep:
    (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).
    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".
    Offline

    2
    ReputationRep:
    (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".
    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.
    Offline

    2
    ReputationRep:
    (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.
    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)
    Offline

    2
    ReputationRep:
    (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.
    Wow I don't think you're getting this. It has absolute relevance.

    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.
    Offline

    2
    ReputationRep:
    (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)
    Where in the world has root(2) come from?
    Offline

    2
    ReputationRep:
    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)
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by hassi94)
    Where in the world has root(2) come from?

    (Original post by ztibor)
    SO the rate of r
    d/dt(r)=6/(2pi)
    The perimeter of the square is 4r
    Thanks guys




    (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 right
    Offline

    0
    ReputationRep:
    (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.
    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.
    Offline

    2
    ReputationRep:
    (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
    Well by that I meant if he meant r as a length of the side of the square, not the radius of the circle.
    Offline

    1
    ReputationRep:
    (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)
    I think you need more help than the OP.
    Offline

    2
    ReputationRep:
    (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.
    LOL:confused:
    Offline

    2
    ReputationRep:
    (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)
    Look if you've set the perimeter to = 4x then each side = x.


    Each side is the diameter of the circle and since circumference = pi*diameter then C = pi*x with no root(2)
    Offline

    2
    ReputationRep:
    (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?
    Name:  a76b64967e6a404185e953b.png
Views: 106
Size:  3.6 KB
    Offline

    2
    ReputationRep:
    (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.
    "I've tried to write the circumference as 2pi*r"


    :lolwut:
    Offline

    1
    ReputationRep:
    (Original post by Ilyas)
    ...
    I have uploaded the file you requested, here is the link to it.

    Sorry for posting it here, but i saw that you had blocked visitor messages.
    Offline

    2
    ReputationRep:
    (Original post by JonathanM)
    Urm what?
    Name:  a76b64967e6a404185e953b.png
Views: 106
Size:  3.6 KB
    Okay yeah you've understood this completely incorrectly.

    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
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: May 13, 2012
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.

Poll
Who is going to win Bake Off?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.