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

FP1 - Edexcel - Coordinate systems - Need help with question watch

    • Thread Starter
    Offline

    0
    ReputationRep:
    Name:  photo.jpg
Views: 269
Size:  508.3 KB

    I understand how to work out the normal of the equation but here is where I am confused... When you square root something there is a positive and a negative solution so why have they presumed that y is positive. Isn't there two solutions?
    Offline

    16
    ReputationRep:
    I had this exact same conversation with another user on TSR. Let me try and find that thread...

    While I'm looking though, think of the graph of  y = \sqrt{x} . You see that the range is always positive.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by claret_n_blue)
    I had this exact same conversation with another user on TSR. Let me try and find that thread...
    Thanks a lot! I thought I was being silly and missing something.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by claret_n_blue)
    I had this exact same conversation with another user on TSR. Let me try and find that thread...

    While I'm looking though, think of the graph of  y = \sqrt{x} . You see that the range is always positive.
    I can't make the link as earlier in the book they use the positive and negative of y=...

    Yet for some reason it doesn't apply in this case and I don't understand why.
    Offline

    16
    ReputationRep:
    (Original post by Jackabc)
    Thanks a lot! I thought I was being silly and missing something.
    http://www.thestudentroom.co.uk/show....php?t=2214694

    They basically say that

     (\pm 2)^2 = 4

    however

     \sqrt{4} = 2 .

    This website explains a little more about the "principal square root".
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by claret_n_blue)
    http://www.thestudentroom.co.uk/show....php?t=2214694

    They basically say that

     (\pm 2)^2 = 4

    however

     \sqrt{4} = 2 .

    This website explains a little more about the "principal square root".
    So why is it when x^2=9 you give x=3 or x=-3
    Offline

    16
    ReputationRep:
    (Original post by Jackabc)
    So why is it when x^2=9 you give x=3 or x=-3
    That's because either  (-3) \cdot (-3) = 9 or 3 \cdot 3 = 9 . But  \sqrt{9} = +3 .

    EDIT: The square root symbol itself represents the positive value (hence the graph). But you know that there are two possible values and so we put the negative possible value in ourselves as we're not sure if we want the positive or negative solution.

    What's never been made clear (when it should be) is that some  x such that  x^2 = y is not the same as  \sqrt{y} .
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by claret_n_blue)
    That's because either  (-3) \cdot (-3) = 9 or 3 \cdot 3 = 9 . But  \sqrt{9} = +3 .
    Can you expand on that a bit? That website isn't that clear.

    So x^2=9 so x is equal to the square root of 9. Then if the square root of 9 is 3 then why bother with -3.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by claret_n_blue)
    That's because either  (-3) \cdot (-3) = 9 or 3 \cdot 3 = 9 . But  \sqrt{9} = +3 .

    EDIT: The square root symbol itself represents the positive value (hence the graph). But you know that there are two possible values and so we put the negative possible value in ourselves as we're not sure if we want the positive or negative solution.

    What's never been made clear (when it should be) is that some  x such that  x^2 = y is not the same as  \sqrt{y} .
    That makes sense now. So really we are not thinking the square root of 9, we are thinking what could be multiplied to give 9?
    Offline

    16
    ReputationRep:
    (Original post by Jackabc)
    That makes sense now. So really we are not thinking the square root of 9, we are thinking what could be multiplied to give 9?
    Yep! Spot on

    EDIT: I re-read your post properly. Sorry, if you mean what is multiplied by itself to give you 9, that is the same as taking the square root. When you are taking a square root, you have the possibility to choose from two (the positive or negative one). The convention is to choose the positive one.
    Offline

    1
    ReputationRep:
    I'd suggest implicit differentiation on that specific question (or questions similar to) to avoid that confusion.


    Posted from TSR Mobile
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by claret_n_blue)
    Yep! Spot on
    Name:  photo(1).jpg
Views: 102
Size:  502.0 KB
    As you can see in this question both the positive and negative square root are taken. Why is it in this case only the positive square root can be taken? As t could be positive or negative, t could be negative and y would be below the graph so you would be looking for the negative gradient of the tangent and the graph of the negative square root.
    Attached Images
     
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Joshmeid)
    I'd suggest implicit differentiation on that specific question (or questions similar to) to avoid that confusion.


    Posted from TSR Mobile
    But if you look at my above example they take the negative version and how do I know with example 11, the first example that t is not negative. Would implicit differentiation be reliable then?
    Offline

    1
    ReputationRep:
    (Original post by Jackabc)
    But if you look at my above example they take the negative version and how do I know with example 11, the first example that t is not negative. Would implicit differentiation be reliable then?
    Differentiating implicitly jumps from the first line straight to the 8th line or so, much faster and avoids confusion.

    We have y^{2} = 27x \rightarrow 2y\frac{dy}{dx} = 27 \rightarrow \frac{dy}{dx} = \frac{27}{2y}.

    We can then substitute in y and find the gradient of the tangent at that specific point.


    Posted from TSR Mobile
    • Study Helper
    Offline

    16
    ReputationRep:
    Study Helper
    (Original post by Jackabc)
    But if you look at my above example they take the negative version and how do I know with example 11, the first example that t is not negative. Would implicit differentiation be reliable then?
    YES! I have to go out now but there are "safer" methods of getting the same answer - I'll post back in a bit
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Joshmeid)
    Differentiating implicitly jumps from the first line straight to the 8th line or so, much faster and avoids confusion.

    We have y^{2} = 27x \rightarrow 2y\frac{dy}{dx} = 27 \rightarrow \frac{dy}{dx} = \frac{27}{2y}.

    We can then substitute in y and find the gradient of the tangent at that specific point.


    Posted from TSR Mobile
    I get you, thanks a lot! I will use that in future but could you tell me where using the previous method breaks down if I took the positive and negative square root.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by davros)
    YES! I have to go out now but there are "safer" methods of getting the same answer - I'll post back in a bit
    Thanks a lot. My question now is why wouldn't the previous method work is that t could be negative and so what's is wrong with taking the negative square root of y and then working out the gradient and the normal from there.
    Offline

    1
    ReputationRep:
    (Original post by Jackabc)
    I get you, thanks a lot! I will use that in future but could you tell me where using the previous method breaks down if I took the positive and negative square root.
    Depends on the question, in the second question it refers to both points on the curve and you end up finding equations for the tangent on both, the gradient for one tangent is positive, the gradient of the other is negative as you can see from the image.

    For example 11, we can see that the equation of the normal has a negative gradient(- t), therefore the gradient of the tangent at point P will be positive so we take the positive value of y.


    Posted from TSR Mobile
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Joshmeid)
    Depends on the question, in the second question it refers to both points on the curve and you end up finding equations for the tangent on both, the gradient for one tangent is positive, the gradient of the other is negative as you can see from the image.

    For example 11, we can see that the equation of the normal has a negative gradient(- t), therefore the gradient of the tangent at point P will be positive so we take the positive value of y.


    Posted from TSR Mobile
    I think that's where I am going wrong, how do we know from example 11 that t has a negative gradient? As if t was negative then wouldn't the equation I am supposed to be proving have a positive gradient. It is weird but it seems that it seems to work whatever the case.
    Offline

    1
    ReputationRep:
    (Original post by Jackabc)
    I think that's where I am going wrong, how do we know from example 11 that t has a negative gradient? As if t was negative then wouldn't the equation I am supposed to be proving have a positive gradient. It is weird but it seems that it seems to work whatever the case.
    Ah yeah, stumped me there lol. The square roots have very weird rules.

    I'd just stick to implicit to avoid the confusion.


    Posted from TSR Mobile
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
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.