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C4 Differentiation Edexcel

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(edited 12 years ago)
Reply 1
Avatar for steve2005
steve2005
17
Original post by rubadubdub
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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
Reply 2
Avatar for just george
just george
Original post by rubadubdub
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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.
(edited 12 years ago)
Reply 3
Avatar for raheem94
raheem94
13
Original post by rubadubdub
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They have mentioned that k=k′A k=\dfrac{k'}{\sqrt{A}}

You are confusing k k with k′ k' , both are different constants.
Reply 4
Avatar for steve2005
steve2005
17
Original post by rubadubdub
Untitled.png



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
Reply 5
Avatar for rubadubdub
rubadubdub
OP
10
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=k′A k=\dfrac{k'}{\sqrt{A}}

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



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

Sorry i understand now thanks
(edited 12 years ago)
Reply 6
Avatar for rubadubdub
rubadubdub
OP
10
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
Reply 7
Avatar for raheem94
raheem94
13
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=k′A[br]k≠−k′A[br] k = \dfrac{k'}{\sqrt{A}}[br]k \not= -\dfrac{k'}{\sqrt{A}}[br]

Probably a typo from you.
Reply 8
Avatar for steve2005
steve2005
17
Original post by raheem94
k=k′A[br]k≠−k′A[br] k = \dfrac{k'}{\sqrt{A}}[br]k \not= -\dfrac{k'}{\sqrt{A}}[br]

Probably a typo from you.


No typo.

There is nothing wrong with what I have typed.
Reply 9
Avatar for just george
just george
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.


dhdt=−k′AhA=(−k′A)h= \frac {dh}{dt} = \frac{-k'\sqrt{Ah}}{A} = (\frac{-k'}{\sqrt{A}})\sqrt{h} = -kh\sqrt{h}

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

I think thats what raheem94 was trying to point out :smile:
Original post by just george
dhdt=−k′AhA=(−k′A)h= \frac {dh}{dt} = \frac{-k'\sqrt{Ah}}{A} = (\frac{-k'}{\sqrt{A}})\sqrt{h} = -kh\sqrt{h}

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

I think thats what raheem94 was trying to point out :smile:


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

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