Hey there Sign in to join this conversationNew here? Join for free

C4 Differentiation Edexcel

Announcements Posted on
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    Click image for larger version. 

Name:	Untitled.png 
Views:	84 
Size:	16.5 KB 
ID:	136693
    • 13 followers
    Offline

    ReputationRep:
    (Original post by rubadubdub)
    Click image for larger version. 

Name:	Untitled.png 
Views:	84 
Size:	16.5 KB 
ID:	136693

    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A
    • 1 follower
    Offline

    ReputationRep:
    (Original post by rubadubdub)
    Click image for larger version. 

Name:	Untitled.png 
Views:	84 
Size:	16.5 KB 
ID:	136693
    they have k' as a positive constant.

    they then have -k'/sqrtA times sqrt(h)

    so they say that k'/sqrtA is a positive constant k, leaving -k sqrt(h)

    i.e. k and k' are two different constants.
    • 47 followers
    Offline

    ReputationRep:
    (Original post by rubadubdub)
    Click image for larger version. 

Name:	Untitled.png 
Views:	84 
Size:	16.5 KB 
ID:	136693
    They have mentioned that  k=\dfrac{k'}{\sqrt{A}}

    You are confusing  k with  k' , both are different constants.
    • 13 followers
    Offline

    ReputationRep:
    (Original post by rubadubdub)
    Click image for larger version. 

Name:	Untitled.png 
Views:	84 
Size:	16.5 KB 
ID:	136693

    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A





    Uploaded with ImageShack.us
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    (Original post by steve2005)
    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A
    (Original post by just george)
    they have k' as a positive constant.

    they then have -k'/sqrtA times sqrt(h)

    so they say that k'/sqrtA is a positive constant k, leaving -k sqrt(h)

    i.e. k and k' are two different constants.
    (Original post by raheem94)
    They have mentioned that  k=\dfrac{k'}{\sqrt{A}}

    You are confusing  k with  k' , both are different constants.

    Thanks but why is k' equivalent to (-k' . A^-1/2)


    Sorry i understand now thanks
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    (Original post by steve2005)
    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A





    Uploaded with ImageShack.us
    thanks
    • 47 followers
    Offline

    ReputationRep:
    (Original post by steve2005)
    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A





    Uploaded with ImageShack.us
     k = \dfrac{k'}{\sqrt{A}}

k \not= -\dfrac{k'}{\sqrt{A}}

    Probably a typo from you.
    • 13 followers
    Offline

    ReputationRep:
    (Original post by raheem94)
     k = \dfrac{k'}{\sqrt{A}}

k \not= -\dfrac{k'}{\sqrt{A}}

    Probably a typo from you.
    No typo.

    There is nothing wrong with what I have typed.
    • 1 follower
    Offline

    ReputationRep:
    (Original post by steve2005)
    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A





    Uploaded with ImageShack.us

    (Original post by steve2005)
    No typo.

    There is nothing wrong with what I have typed.
     \frac {dh}{dt} = \frac{-k'\sqrt{Ah}}{A} = (\frac{-k'}{\sqrt{A}})\sqrt{h} = -k\sqrt{h}

     where  k = (\frac{k'}{\sqrt{A}})

    I think thats what raheem94 was trying to point out
    • 13 followers
    Offline

    ReputationRep:
    (Original post by just george)
     \frac {dh}{dt} = \frac{-k'\sqrt{Ah}}{A} = (\frac{-k'}{\sqrt{A}})\sqrt{h} = -k\sqrt{h}

     where  k = (\frac{k'}{\sqrt{A}})

    I think thats what raheem94 was trying to point out
    I know what he was trying to say. But what I wrote is correct because k is defined, in this case, to be negative.

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?

    this is what you'll be called on TSR

  2. this can't be left blank
    this email is already registered. Forgotten your password?

    never shared and never spammed

  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. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide the button to the right to create your account

    Slide to join now Processing…

Updated: March 17, 2012
New on TSR

Your favourite film of the year?

For you personally what has been the best 2014 movie

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