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

    3
    ReputationRep:
    question:
    can someone tell me why in the mark scheme it does the negative recipcocal thing because it says 'parallel' and not 'perpendicular'. am i missing something??
    Attached Images
     
    Offline

    0
    ReputationRep:
    (Original post by cooldudeman)
    question:
    can someone tell me why in the mark scheme it does the negative recipcocal thing because it says 'parallel' and not 'perpendicular'. am i missing something??
    Its asking for the normal, this is perpendicualr to the tangent so you do the reciprocal
    Offline

    0
    ReputationRep:
    (Original post by hazbaz)
    Its asking for the normal, this is perpendicualr to the tangent so you do the reciprocal
    what the fish, i dont remember any of this... maths is a pisstake
    Offline

    2
    ReputationRep:
    ah, i remember this question on a past paper i have done

    they tell you y=k(root x) so y = kx^1/2

    they tell you that its parallel to the line 2x + 3y = 0 so y=-2/3x

    if lines are parallel gradients are equal, to get the gradient of a curve differentiate.

    so dy/dx = 1/2kx^-1/2

    make it equal to -2/3 (thats the gradient ( y=mx + c) )

    after that it should be easy
    Offline

    2
    ReputationRep:
    The grad of the normal to any curve is given by: \frac{-1}{m}, where m is the gradient of the tangent to the curve at any point.
    Offline

    1
    ReputationRep:
    The normal is parallel to 2x+3y=0. This equation can be rearranged to give y=-2/3x, hence giving the gradient they specify in the MS.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by sohail.s)
    ah, i remember this question on a past paper i have done

    they tell you y=k(root x) so y = kx^1/2

    they tell you that its parallel to the line 2x + 3y = 0 so y=-2/3x

    if lines are parallel gradients are equal, to get the gradient of a curve differentiate.

    so dy/dx = 1/2kx^-1/2

    make it equal to -2/3 (thats the gradient ( y=mx + c) )

    after that it should be easy
    um i just tried that now and i got k=8/3.....
    the answer is 6. i did on the last part, -2/3=k/4
    Offline

    2
    ReputationRep:
    (Original post by cooldudeman)
    um i just tried that now and i got k=8/3.....
    the answer is 6. i did on the last part, -2/3=k/4
    hmm,

    dy/dx = 1/2kx^-1/2

    so 0.5kx^-1/2 = -2/3 (x=4 sub that in)

    k/4 = -2/3 thats what we get ( but this is the gradient of the tangent, for the normal its -1/m )

    so k/4 = 3/2 therefore k = 6
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by sohail.s)
    hmm,

    dy/dx = 1/2kx^-1/2

    so 0.5kx^-1/2 = -2/3 (x=4 sub that in)

    k/4 = -2/3 thats what we get ( but this is the gradient of the tangent, for the normal its -1/m )

    so k/4 = 3/2 therefore k = 6
    its just the reciprocal part that i dont understand because the gradient '-2/3' came from the equation, 2x + 3y = 0 and isnt this the equation of a 'normal' line?
    if it is then, the gradient should already be 'normal'.

    dy/dx gave (or always gives... not sure on that) the gradient of the tangent and -2/3 is the normal gradient so is it like this?:
    (a normal gradient) -2/3= k/4 (a tangent gradient)
    you have to change the normal one into atangent one so -2/3 is 3/2.
    is that why you have to change it?????
    Offline

    2
    ReputationRep:
    (Original post by cooldudeman)
    its just the reciprocal part that i dont understand because the gradient '-2/3' came from the equation, 2x + 3y = 0 and isnt this the equation of a 'normal' line?
    if it is then, the gradient should already be 'normal'.

    dy/dx gave (or always gives... not sure on that) the gradient of the tangent and -2/3 is the normal gradient so is it like this?:
    (a normal gradient) -2/3= k/4 (a tangent gradient)
    you have to change the normal one into atangent one so -2/3 is 3/2.
    is that why you have to change it?????
    dy/dx gives the gradient function of a tangent to a point (x=whatever)

    your misunderstanding it mate, the questions says the NORMAL at point p is parallel to the line 2x + 3y = 0 as far as we're concerned this line is just an ordinary line that has nothing to do with the curve (it could be a normal or a tangent but we're not interested in this- we are just interested in the gradient of the line )

    TO find the normal to a point on a curve you find dy/dx which is the gradient of the tangent and then take the -ve reciprocal to find the gradient of the normal.

    SO if dy/dx = 1/2kx^-1/2 and at that the point the gradient of tangent to that point is 0.5kx^-1/2 = -2/3
    the gradient of the normal would equal -1/ -2/3 which gives 3/2


    Just remember when differentiating you are finding the gradient function of a TANGENT to that point, once you have the function substitute values of x into the function to find the value of the tangent and then -ve reciprocal to find the gradient of the normal

    Please forgive me for any typos
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by sohail.s)
    dy/dx gives the gradient function of a tangent to a point (x=whatever)

    your misunderstanding it mate, the questions says the NORMAL at point p is parallel to the line 2x + 3y = 0 as far as we're concerned this line is just an ordinary line that has nothing to do with the curve (it could be a normal or a tangent but we're not interested in this- we are just interested in the gradient of the line )

    TO find the normal to a point on a curve you find dy/dx which is the gradient of the tangent and then take the -ve reciprocal to find the gradient of the normal.

    SO if dy/dx = 1/2kx^-1/2 and at that the point the gradient of tangent to that point is 0.5kx^-1/2 = -2/3
    the gradient of the normal would equal -1/ -2/3 which gives 3/2


    Just remember when differentiating you are finding the gradient function of a TANGENT to that point, once you have the function substitute values of x into the function to find the value of the tangent and then -ve reciprocal to find the gradient of the normal

    Please forgive me for any typos
    ohh i think i get it nowwww. thanks a lot.
    lol if you dont mind, can you or anyone else help me with the second part?
    i dont know how to find point Q...
    Offline

    1
    ReputationRep:
    (Original post by cooldudeman)
    ohh i think i get it nowwww. thanks a lot.
    lol if you dont mind, can you or anyone else help me with the second part?
    i dont know how to find point Q...
    Q is the point where the normal crosses the x-axis, so where y=0. Just sub that into the normal equation and solve for x.
 
 
 
  • 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
    What newspaper do you read/prefer?
    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.