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

Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

Announcements
    Offline

    3
    ReputationRep:
    (Original post by Craig1998)
    30+50e^0.002t> 84
    50e^0.002t>54
    e^0.002t>27/25
    t/500>ln1.08
    t>500ln1.08
    t>38.4

    Therefore t = 39
    Therefore year = 1980+39
    = 2019.

    Oh god I'm failing c4.
    That's correct
    Offline

    2
    ReputationRep:
    (Original post by Craig1998)
    30+50e^0.002t> 84
    50e^0.002t>54
    e^0.002t>27/25
    t/500>ln1.08
    t>500ln1.08
    t>38.4

    Therefore t = 39
    Therefore year = 1980+39
    = 2019.

    Oh god I'm failing c4.
    Yep thats what I got, but the mark scheme says 2018, must be an error on their part.

    Btw this is C3 :cute:, im sure you'll be fine. Thanks!
    Offline

    20
    ReputationRep:
    Q) Why is the gradient of tangent --> 1/(value of dy/dx)

    when https://gyazo.com/1cd6bd9ac94dbec89e08132a94ff86a1
    shows the gradient of tangent is the exact value of dy/dx yet this shows the gradient of tangent
    is https://gyazo.com/a31c58f4855ac1c0d1d8b6177b8a08a2 .

    I understand when you draw a picture you can see gradient of tangent is 1/gradient but say for example
    if they say ' find the equation of tangent of the line. ' then the tangent 'm' value in the equation would be
    the dy/dx value and if they ask 'find the equation of tangent TO A PARTICULAR POINT' then do you do 1/gradient. <--- is that right? :sigh:

    I don't know why this has confused me last minute :lol:
    Online

    21
    ReputationRep:
    wot. Gradient of tangent is dy/dx.
    Offline

    20
    ReputationRep:
    (Original post by Zacken)
    wot. Gradient of tangent is dy/dx.
    That's what I thought. I was confused by what that person did in that workings. :confused:

    Can you explain this to me please.

    Question: https://gyazo.com/dbdd9ce6a296f8c337e633e69cd442b3

    Mark scheme: https://gyazo.com/0a971e1e6c6cd570288ad4b35f6aae62

    I got that 1st line in the mark scheme where I dy/dx: 3sqroot(2) . cos (theta+ alpha)

    and dy/dx = 1/2 so why can't I do 3sqroot(2) . cos(theta+alpha) = 1/2?

    How did they get -2 or 2? Is it from the graph or something I missed? :erm:
    Online

    21
    ReputationRep:
    (Original post by XxKingSniprxX)
    ...
    ...you could have given us the question in the first place.

    This is of the form x = f(y). So the gradient of the tangent is given by dx/dy.

    In most questions, the curve is defined as y = f(x) so the gradient of the tangent is dy/dx.

    In the question you've linked, you've found x = f(y), so by differentiating you get dx/dy = f ' (y).

    You want dy/dx. So dx/dy = 1/(dy/dx).
    Offline

    20
    ReputationRep:
    (Original post by Zacken)
    ...you could have given us the question in the first place.

    This is of the form x = f(y). So the gradient of the tangent is given by dx/dy.

    In most questions, the curve is defined as y = f(x) so the gradient of the tangent is dy/dx.

    In the question you've linked, you've found x = f(y), so by differentiating you get dx/dy = f ' (y).

    You want dy/dx. So dx/dy = 1/(dy/dx).
    :facepalm: I've been reading the question as y = f(x) ---> L

    Thanks, I understood it easy once I realised it was x = f(y)
    Online

    21
    ReputationRep:
    (Original post by XxKingSniprxX)
    :facepalm: I've been reading the question as y = f(x) ---> L

    Thanks, I understood it easy once I realised it was x = f(y)
    Coolio.
    Offline

    3
    ReputationRep:
    (Original post by XxKingSniprxX)
    Q) Why is the gradient of tangent --> 1/(value of dy/dx)

    when https://gyazo.com/1cd6bd9ac94dbec89e08132a94ff86a1
    shows the gradient of tangent is the exact value of dy/dx yet this shows the gradient of tangent
    is https://gyazo.com/a31c58f4855ac1c0d1d8b6177b8a08a2 .

    I understand when you draw a picture you can see gradient of tangent is 1/gradient but say for example
    if they say ' find the equation of tangent of the line. ' then the tangent 'm' value in the equation would be
    the dy/dx value and if they ask 'find the equation of tangent TO A PARTICULAR POINT' then do you do 1/gradient. <--- is that right? :sigh:

    I don't know why this has confused me last minute :lol:
    I would expect someone getting 95+ to not be thrown off by this of all things.
    Offline

    20
    ReputationRep:
    (Original post by yesyesyesno)
    I would expect someone getting 95+ to not be thrown off by this of all things.
    Haha.. we all have our :shot: moments.
    Offline

    2
    ReputationRep:
    anyone have the AQA C3 paper? ive heard it was a beast of a paper. I wanna see how hard it really was
    Offline

    20
    ReputationRep:
    How do I go from my alternative answer to getting the quadratic equation they formed in this question?

    https://gyazo.com/861095830fbde070b20076a796aeba94
    Offline

    2
    ReputationRep:
    (Original post by XxKingSniprxX)
    How do I go from my alternative answer to getting the quadratic equation they formed in this question?

    https://gyazo.com/861095830fbde070b20076a796aeba94
    That's how you do it. But to be honest I don't know why your way won't work. Always hated that about logs. For some reason I can only get one way to work. Rest give me wrong answers even if they seem right.

    Posted from TSR Mobile
    Attached Images
     
    Offline

    3
    ReputationRep:
    (Original post by XxKingSniprxX)
    How do I go from my alternative answer to getting the quadratic equation they formed in this question?

    https://gyazo.com/861095830fbde070b20076a796aeba94
    You can't antilog like that, you have to use the log rule, from C2: -log(x) is log(1/x)

    Once you've both logs on the side, you can anti-log it.
    Offline

    2
    ReputationRep:
    (Original post by SaadKaleem)
    You can't antilog like that, you have to use the log rule, from C2: -log(x) is log(1/x)

    Once you've both logs on the side, you can anti-log it.
    Do you mind explaining further please. I don't understand. I get how to do it but I don't get why the way he did it first doesn't work. Please and thanks.

    Posted from TSR Mobile
    Offline

    20
    ReputationRep:
    (Original post by ImNervous)
    Do you mind explaining further please. I don't understand. I get how to do it but I don't get why the way he did it first doesn't work. Please and thanks.

    Posted from TSR Mobile
    You can just anti log from the 1st line since its already in logs. Your way is much more working out involved.

    (Original post by SaadKaleem)
    You can't antilog like that, you have to use the log rule, from C2: -log(x) is log(1/x)

    Once you've both logs on the side, you can anti-log it.
    :zomg: --> :facepalm: I can't believe I'm forgetting basic maths. :L

    Offline

    2
    ReputationRep:
    (Original post by XxKingSniprxX)
    You can just anti log from the 1st line since its already in logs. Your way is much more working out involved.



    :zomg: --> :facepalm: I can't believe I'm forgetting basic maths. :L

    Ooh wow. I'm an idiot. Lol

    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    Who else almost made vectors their b*tch here?


    Posted from TSR Mobile
    Offline

    20
    ReputationRep:
    (Original post by ImNervous)
    Ooh wow. I'm an idiot. Lol

    Posted from TSR Mobile
    :five: at the stupid maths mistakes.
    Offline

    3
    ReputationRep:
    (Original post by кяя)
    Who else almost made vectors their b*tch here?


    Posted from TSR Mobile
    Vectors is my b*tch now ;D

    (Original post by XxKingSniprxX)
    :five: at the stupid maths mistakes.
    These silly arithmetic mistakes always get you and you lose free marks, costed me my 200 UMS in C12 >.<
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    How are your GCSEs going so far?
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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