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    1) When we are differentiating y in implicit differentiation, why do we have to always put dy/dx at the end of it?

    2) I was just looking at this question from Exam Solutions :
    http://www.examsolutions.net/maths-r...tutorial-1.php

    I didn't really understand why , in this vector question, there are TWO values for a and b.
    If someone could help me I would be really grateful! thanks!
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    (Original post by laurawoods)
    1) When we are differentiating y in implicit differentiation, why do we have to always put dy/dx at the end of it?

    2) I was just looking at this question from Exam Solutions :
    http://www.examsolutions.net/maths-r...tutorial-1.php

    I didn't really understand why , in this vector question, there are TWO values for a and b.
    If someone could help me I would be really grateful! thanks!
    Definition:

    \dfrac{dy}{dx}

    means "the derivative of y with respect to x".

    So when you differentiate a y term with respect to x, you need to put that.

    About the vectors q, it just means that the lines are parallel for 2 values of a and b.
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    (Original post by laurawoods)
    1) When we are differentiating y in implicit differentiation, why do we have to always put dy/dx at the end of it?

    2) I was just looking at this question from Exam Solutions :
    http://www.examsolutions.net/maths-r...tutorial-1.php

    I didn't really understand why , in this vector question, there are TWO values for a and b.
    If someone could help me I would be really grateful! thanks!
    I haven't got time to watch the video right now for 2), but for 1) when you have to do an implicit differentiation you're basically making use of the chain rule - remember when you first learnt the chain rule you wrote y as a function of u and u as a function of x, and found that:

    \dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx}

    With implicit differentiation you normally have something like
    f(y) = g(x)

    so when you differentiate the LHS you can think of f being a function of y and y being a function of x, so the chain rule would tell you that:

    \dfrac{df}{dx} = \dfrac{df}{dy} \dfrac{dy}{dx}
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    (Original post by Indeterminate)
    Definition:

    \dfrac{dy}{dx}

    means "the derivative of y with respect to x".

    So when you differentiate a y term with respect to x, you need to put that.

    About the vectors q, it just means that the lines are parallel for 2 values of a and b.
    thanks davros and indeterminate ...thanks for the answers!
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    (Original post by Indeterminate)
    Definition:

    \dfrac{dy}{dx}

    means "the derivative of y with respect to x".

    So when you differentiate a y term with respect to x, you need to put that.

    About the vectors q, it just means that the lines are parallel for 2 values of a and b.
    Hello there,
    How to integrate this: I would be really grateful if u could help me on this!

    thank you...

    2x/ 2x+3


    thanks!
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    (Original post by laurawoods)
    Hello there,
    How to integrate this: I would be really grateful if u could help me on this!

    thank you...

    2x/ 2x+3


    thanks!
    Well, note that

    2x = (2x+3) - 3

    and that

    \displaystyle a \int \dfrac{f'(x)}{f(x)} \ dx = a \ln|f(x)| + C
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    (Original post by Indeterminate)
    Well, note that

    2x = (2x+3) - 3

    and that

    \displaystyle a \int \dfrac{f'(x)}{f(x)} \ dx = a \ln|f(x)| + C
    hello , i don't see how that could potentially work because if you diff 2x+3 you get just 2 , isn't it so? So it definitely cannot be forming that pattern of diff/ original function , isn't it so? Please correct me...
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    (Original post by laurawoods)
    hello , i don't see how that could potentially work because if you diff 2x+3 you get just 2 , isn't it so? So it definitely cannot be forming that pattern of diff/ original function , isn't it so? Please correct me...
    3= \frac{3}{2} (2)

    Can you do it now?
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    (Original post by Indeterminate)
    3= \frac{3}{2} (2)

    Can you do it now?
    Hello , thanks yes i think I have managed it now! thanks for your time! btw, u got into cambridge last year, right>?
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    (Original post by laurawoods)
    Hello , thanks yes i think I have managed it now! thanks for your time! btw, u got into cambridge last year, right>?
    No as I missed out in STEP III.
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    (Original post by laurawoods)
    thanks davros and indeterminate ...thanks for the answers!
    no problem...it must be National Neg Rep Day today - I see we both got negged for helping you and you got negged for thanking us
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    (Original post by Indeterminate)
    No as I missed out in STEP III.
    hello, sorry to hear that ! what is step 3 if u dont mind me asking ...how does it work?
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    (Original post by laurawoods)
    hello, sorry to hear that ! what is step 3 if u dont mind me asking ...how does it work?
    Info here with a link to the dedicated STEP website:

    http://www.maths.cam.ac.uk/undergrad/admissions/step/
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    (Original post by Indeterminate)
    Info here with a link to the dedicated STEP website:

    http://www.maths.cam.ac.uk/undergrad/admissions/step/
    Oh I thought it was an interview to get into cambridge and oxford?

    Hello , if u don't mind , I wanted some maths help from u :

    U know that in C4 differentiation chapter, we have to know how to do 'connected rates of change" problems. What tips would u give me with regards to them and also are the problems involving cones the hardest we could get on them?
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    (Original post by Indeterminate)
    Info here with a link to the dedicated STEP website:

    http://www.maths.cam.ac.uk/undergrad/admissions/step/
    Also I was wondering about these sorts of questions where they give a different thing as our substitution, for example x=2sinu and stuff like that ...any tips with such problems and what is the aim of giving such difficult things to sub in ...
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    (Original post by laurawoods)
    Oh I thought it was an interview to get into cambridge and oxford?

    Hello , if u don't mind , I wanted some maths help from u :

    U know that in C4 differentiation chapter, we have to know how to do 'connected rates of change" problems. What tips would u give me with regards to them and also are the problems involving cones the hardest we could get on them?
    Well, remember that 'proportional' means that there's a constant k involved, and to extract the maths by reading the question carefully.

    Have a look at Q8 in the first paper (2010) and Q5 in the second (2009).
    Attached Images
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    (Original post by laurawoods)
    Also I was wondering about these sorts of questions where they give a different thing as our substitution, for example x=2sinu and stuff like that ...any tips with such problems and what is the aim of giving such difficult things to sub in ...
    Firstly, don't worry too much as they always give you the substitution, and you only need to figure it out for yourself in FM.

    Secondly, they just want to test you to the maximum extent.

    So, suppose they've asked you to integrate

    \displaystyle \int \dfrac{1}{\sqrt{a^2 - x^2}} \ dx

    using

    x=a \sin u

    First, you would differentiate to get dx in terms of d theta, and then substitute in.
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    (Original post by Indeterminate)
    Firstly, don't worry too much as they always give you the substitution, and you only need to figure it out for yourself in FM.

    Secondly, they just want to test you to the maximum extent.

    So, suppose they've asked you to integrate

    \displaystyle \int \dfrac{1}{\sqrt{a^2 - x^2}} \ dx

    using

    x=a \sin u

    First, you would differentiate to get dx in terms of d theta, and then substitute in.
    Hello there, thanks !
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    (Original post by Indeterminate)
    Firstly, don't worry too much as they always give you the substitution, and you only need to figure it out for yourself in FM.

    Secondly, they just want to test you to the maximum extent.

    So, suppose they've asked you to integrate

    \displaystyle \int \dfrac{1}{\sqrt{a^2 - x^2}} \ dx

    using

    x=a \sin u

    First, you would differentiate to get dx in terms of d theta, and then substitute in.
    Hello, I was just feeling a bit stuck on this question ...I would appreciate it if u could help me on it ... thanks!

    There is this parametrics question : http://www.examsolutions.net/maths-r...tutorial-1.php

    I did the following and just wanted to check with u if I have done it right:

    x=2cost y=sin2t

    dx/dt= -2sint

    so integral sign (sin2t) (-2sint)

    then [sin(3t) / (3) - sint ] and for the limits I got t= 90 degrees and 0 . I subbed them in but got -4/3....what do i do next? am i going the right way is this one of those questions where the limits will swap around because we surely cannot have a negative area!?
    thanks laura
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    (Original post by laurawoods)
    Hello, I was just feeling a bit stuck on this question ...I would appreciate it if u could help me on it ... thanks!

    There is this parametrics question : http://www.examsolutions.net/maths-r...tutorial-1.php

    I did the following and just wanted to check with u if I have done it right:

    x=2cost y=sin2t

    dx/dt= -2sint

    so integral sign (sin2t) (-2sint)

    then [sin(3t) / (3) - sint ] and for the limits I got t= 90 degrees and 0 . I subbed them in but got -4/3....what do i do next? am i going the right way is this one of those questions where the limits will swap around because we surely cannot have a negative area!?
    thanks laura
    Your integral should be

    \displaystyle - 4 \int_{\frac{\pi}{2}}^0 \sin^2 t \cos t \ dt

    which does give 4/3

    Note that x is the larger value when t=0.
 
 
 
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