Hey there! Sign in to join this conversationNew here? Join for free
Turn on thread page Beta
    • Thread Starter
    Offline

    19
    ReputationRep:
    Name:  answer18.jpg
Views: 16
Size:  203.0 KB

    Attachment 695170695172

    (3i) , so as usual, where did I mess up?
    Attached Images
     
    Offline

    18
    ReputationRep:
    (Original post by ckfeister)
    Name:  answer18.jpg
Views: 16
Size:  203.0 KB

    Attachment 695170695172

    (3i) , so as usual, where did I mess up?
    You had \displaystyle \frac{\mathrm dy}{\mathrm dx} - y^2 = 1, not \displaystyle -y^2\frac{\mathrm dy}{\mathrm dx} = 1 (as you treated it)

    Instead, divide both sides by 1+y^2, to obtain,

    \displaystyle \frac 1 {1+y^2}\frac{\mathrm dy}{\mathrm dx} = 1. Should be relatively clear where this is going.
    Offline

    11
    ReputationRep:
    (Original post by ckfeister)
    Name:  answer18.jpg
Views: 16
Size:  203.0 KB



    (3i) , so as usual, where did I mess up?
    To be honest, I'm struggling to make any sense of that at all. What does the (x dx) (-y^2) bit mean?

    The first step is to separate the variables i.e. formally write:

    \frac{dy}{1+y^2} = dx

    then integrate.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by _gcx)
    You had \displaystyle \frac{\mathrm dy}{\mathrm dx} - y^2 = 1, not \displaystyle -y^2\frac{\mathrm dy}{\mathrm dx} = 1 (as you treated it)

    Instead, divide both sides by 1+y^2, to obtain,

    \displaystyle \frac 1 {1+y^2}\frac{\mathrm dy}{\mathrm dx} = 1. Should be relatively clear where this is going.
    I don't get where this is meant to go, whats the link with tanx?
    Offline

    18
    ReputationRep:
    (Original post by ckfeister)
    I don't get where this is meant to go, whats the link with tanx?
    Are you familiar with how to approach

    \displaystyle \int \frac 1 {1+y^2} \mathrm dy?
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by _gcx)
    Are you familiar with how to approach

    \displaystyle \int \frac 1 {1+y^2} \mathrm dy?
    Nope... first time. I would try sub but I'm sure it won't get tan(x)
    Offline

    18
    ReputationRep:
    (Original post by ckfeister)
    Nope... first time.
    Have you covered integration by substitution? Try substituting y = \tan\theta.
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by ckfeister)
    I don't get where this is meant to go, whats the link with tanx?
    Which module are you doing currently?
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by notnek)
    which module are you doing currently?
    aqa - fp2
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by ckfeister)
    aqa - fp2
    Then you'll need to know how to integrate that using a trig substitution and you should also be aware that is in your fomula book.

    Do you have any ideas which substitution you can use?
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by Notnek)
    Then you'll need to know how to integrate that using a trig substitution and you should also be aware that is in your fomula book.

    Do you have any ideas which substitution you can use?
    x/a tan-1 x/a?
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by ckfeister)
    u = 1+y^2

    du/dy= 2y

    du/2y = dy

    Only one I know, I'm going to watch a video soon on khan academy but this hasn't been on the lessons yet... strange.
    It seems like you haven't covered trig substitutions so you shouldn't be doing this question yet. _gcx has given you the correct substitution above.
    Offline

    18
    ReputationRep:
    (Original post by ckfeister)
    u = 1+y^2

    du/dy= 2y

    du/2y = dy

    Only one I know, I'm going to watch a video soon on khan academy but this hasn't been on the lessons yet... strange.
    Have you covered integration by substitution? Try substituting y = \tan\theta.
    It's not an obvious substitution, but it works quite well as you'll see. Watching a khanacademy video on trig subs, or similar, might be beneficial as you suggested.
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by Notnek)
    It seems like you haven't covered trig substitutions so you shouldn't be doing this question yet. _gcx has given you the correct substitution above.
    (Original post by _gcx)
    It's not an obvious substitution, but it works quite well as you'll see. Watching a khanacademy video on trig subs, or similar, might be beneficial as you suggested.
    Is it x/a tan-1 x/a?
    Offline

    18
    ReputationRep:
    (Original post by ckfeister)
    Is it x/a tan-1 x/a?
    Not quite, you might have made a slip. Can you show your working? (Also, where has a come from? I'd dissuade you from copying from a formula booklet, it's more beneficial to do it yourself first)
    • Thread Starter
    Offline

    19
    ReputationRep:
    (Original post by _gcx)
    Not quite, you might have made a slip. Can you show your working? (Also, where has a come from? I'd dissuade you from copying from a formula booklet, it's more beneficial to do it yourself first)
    C3 - which I learnt there, I'm going to watch khan academy (£350 course cost doesn't cover this)
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 11, 2017
Poll
“Yanny” or “Laurel”
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

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

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

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