Turn on thread page Beta
    • Thread Starter
    Offline

    9
    ReputationRep:
    Find the Coordinates of a point on the curve y= sin x + cos x at which the tangent is parallel to the line y=x

    So basically you need to find the gradient function of the curve, then equate it to the gradient of the line y=x in order to find the x which wil result in f'(x)=1

    f'(x)= Cos x - Sin x
    Cos x = Sin x + 1
    ( squaring both sides )
    Cos ^2 (x) = Sin^2(x) + 2sinx + 1 => 2sin^2 (x) + 2sin (x) = 0
    2sin(x) [Sin(x) + 1] = 0
    2Sin(x)= 0
    x=ArcSin(0)= 0,180,360
    Or

    x=arcSin(-1)= 270

    (im guessing there should be many points where the x gives a gradient of 1 to the tangent to the curve since the graph is continuous. The textbook answer is (0,1) , is there any specific reason why they took that point? )
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Carlos Nim)
    Find the Coordinates of a point on the curve y= sin x + cos x at which the tangent is parallel to the line y=x

    So basically you need to find the gradient function of the curve, then equate it to the gradient of the line y=x in order to find the x which wil result in f'(x)=1

    f'(x)= Cos x - Sin x
    Cos x = Sin x + 1
    ( squaring both sides )
    Cos ^2 (x) = Sin^2(x) + 2sinx + 1 => 2sin^2 (x) + 2sin (x) = 0
    2sin(x) [Sin(x) + 1] = 0
    2Sin(x)= 0
    x=ArcSin(0)= 0,180,360
    Or

    x=arcSin(-1)= 270

    (im guessing there should be many points where the x gives a gradient of 1 to the tangent to the curve since the graph is continuous. The textbook answer is (0,1) , is there any specific reason why they took that point? )
    Indeed there are infinitely many tangents to y such that they are parallel to y=x, and this is more to do with the fact that y is periodic, not continuous.

    As for why picking that point, it doesn't really justify as you could pick any point on the curve where the gradient is 1 and still answer the question, I guess it's because the closest point to the origin which makes it some sort of a 'simple' point.

    Also, some of your solutions (like x=\pi) are not valid because at those points the gradient is -1. You get these falsely because you squared your equation early on, and squaring makes things positive.
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by Carlos Nim)
    Find the Coordinates of a point on the curve y= sin x + cos x at which the tangent is parallel to the line y=x

    So basically you need to find the gradient function of the curve, then equate it to the gradient of the line y=x in order to find the x which wil result in f'(x)=1

    f'(x)= Cos x - Sin x
    Cos x = Sin x + 1
    ( squaring both sides )
    Cos ^2 (x) = Sin^2(x) + 2sinx + 1 => 2sin^2 (x) + 2sin (x) = 0
    2sin(x) [Sin(x) + 1] = 0
    2Sin(x)= 0
    x=ArcSin(0)= 0,180,360
    Or

    x=arcSin(-1)= 270

    (im guessing there should be many points where the x gives a gradient of 1 to the tangent to the curve since the graph is continuous. The textbook answer is (0,1) , is there any specific reason why they took that point? )
    Unless there is something in the wording of the original question, then one answer is as good as any other.

    However, be aware that squaring an equation can, and in this case does, introduce additional solutions, that do not satisfy the original equation. E.g. x=180; gradient of f(x) is -1.
    • Thread Starter
    Offline

    9
    ReputationRep:
    is there any other way to solve the above equation which won't involve squaring ?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Carlos Nim)
    is there any other way to solve the above equation which won't involve squaring ?
    You can express \cos x - \sin x = R\cos (x+\alpha) form.
    • Thread Starter
    Offline

    9
    ReputationRep:
    (Original post by RDKGames)
    You can express \cos x - \sin x = R\cos (x+\alpha) form.
    ahh yes thx , djd't realise that
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 10, 2018

1,162

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should universities take a stronger line on drugs?
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.