Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    13
    ReputationRep:
    "Point P lies on the curve.
    The normal to the curve at P is parallel to the line 2y + x = 0
    Find the x coordinate of P."

    I get that the gradient of the line is -1/2 but surely the gradient of the normal should be -1/2 as well if it is parallel to this line?

    I really don't understand the wording of the question. The mark scheme says that the gradient you should use is 2 (which I would understand why in any other case) but the question says the normal is parallel to that line so the gradient of the normal should be the same i.e -1/2!?
    If the gradient was to be 2 then you'd be using the tangent at P but how would you know in the exam to use the tangent if the question hadn't said to?
    Offline

    2
    ReputationRep:
    What question?
    Offline

    17
    ReputationRep:
    'The curve' isn't very specific...
    Offline

    2
    ReputationRep:
    In the question, you're given the f'(x) function. The f'(x) gives you the gradient of the tangent to a point. If you know the gradient of the normal is -1/2, then you know that the tangent gradient is 2. You can then solve 2=f'(x)
    Online

    22
    ReputationRep:
    (Original post by jessyjellytot14)
    "Point P lies on the curve.
    The normal to the curve at P is parallel to the line 2y + x = 0
    Find the x coordinate of P."

    I get that the gradient of the line is -1/2 but surely the gradient of the normal should be -1/2 as well if it is parallel to this line?

    I really don't understand the wording of the question. The mark scheme says that the gradient you should use is 2 (which I would understand why in any other case) but the question says the normal is parallel to that line so the gradient of the normal should be the same i.e -1/2!?
    If the gradient was to be 2 then you'd be using the tangent at P but how would you know in the exam to use the tangent if the question hadn't said to?
    You really need to include the full question next time, I wouldn't have known what you were asking had I not already answered this very question previously.

    The derivative of a function, \frac{dy}{dx} gives you the gradient of the tangent at the point x on the curve y by definition.

    So, for example: the curve y=x^2 has gradient function/derivative \frac{dy}{dx} = 2x. That, is the gradient of the tangent at any point x is 2x. The gradient of the tangent to the curve at x=3 is 2(3) = 6; this is the gradient of the tangent to the curve.

    Secondly:

    The normal and tangent at a point on a curve are mutually perpendicular to one another.

    Hence their gradients satisfy m_N m_T = -1.

    So, wrapping all this up:

    If you are given or you found the gradient of the normal to that point, you then want to find the gradient of the tangent at that point using the perpendicularity argument. This then lets you say that \frac{dy}{dx} = m_T = -\frac{1}{m_N} at the point P.

    It is incorrect to say that \frac{dy}{dx} = m_N at P, as dy/dx does not measure the gradient of the normal, it measures the gradient of the tangent. So if you want to find P, you need to use the gradient of the tangent.
    Offline

    3
    ReputationRep:
    The way I intepret what you have written would lead me to think that the gradient of the normal would be -1/2 aswell.
    Online

    22
    ReputationRep:
    (Original post by Bern Herkins)
    The way I intepret what you have written would lead me to think that the gradient of the normal would be -1/2 aswell.
    That's because the gradient of the normal is -1/2, there are no two interpretations of it.
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by Zacken)
    You really need to include the full question next time, I wouldn't have known what you were asking had I not already answered this very question previously.

    The derivative of a function, \frac{dy}{dx} gives you the gradient of the tangent at the point x on the curve y by definition.

    So, for example: the curve y=x^2 has gradient function/derivative \frac{dy}{dx} = 2x. That, is the gradient of the tangent at any point x is 2x. The gradient of the tangent to the curve at x=3 is 2(3) = 6; this is the gradient of the tangent to the curve.

    Secondly:

    The normal and tangent at a point on a curve are mutually perpendicular to one another.

    Hence their gradients satisfy m_N m_T = -1.

    So, wrapping all this up:

    If you are given or you found the gradient of the normal to that point, you then want to find the gradient of the tangent at that point using the perpendicularity argument. This then lets you say that \frac{dy}{dx} = m_T = -\frac{1}{m_N} at the point P.

    It is incorrect to say that \frac{dy}{dx} = m_N at P, as dy/dx does not measure the gradient of the normal, it measures the gradient of the tangent. So if you want to find P, you need to use the gradient of the tangent.
    Oh okay so basically to find the x coordinate of a point on a curve you always equate dy/dx to the gradient of the tangent at that point?
    Online

    22
    ReputationRep:
    (Original post by jessyjellytot14)
    Oh okay so basically to find the x coordinate of a point on a curve you always equate dy/dx to the gradient of the tangent at that point?
    Yes.
    Offline

    15
    ReputationRep:
    (Original post by jessyjellytot14)
    Oh okay so basically to find the x coordinate of a point on a curve you always equate dy/dx to the gradient of the tangent at that point?
    Sometimes tangents may involve use of the discriminant
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by IrrationalRoot)
    'The curve' isn't very specific...
    I didn't include the rest of the question because I know how to solve it, but i didn't know which gradient to use.. and the part of the question I included is all you need to work that out.
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by Xenon17)
    Sometimes tangents may involve use of the discriminant
    Equal roots?
    Offline

    15
    ReputationRep:
    (Original post by jessyjellytot14)
    Equal roots?
    Yep
 
 
 
  • 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
    Brussels sprouts
    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
  • 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.