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

    20
    ReputationRep:
    The question
    The curve  y=x(x-1)^{2} touches the x-axis at point A. B is the point (2,2)
    on the curve and N is the foot of the perpendicular from B to the x-axis. Prove that the tangent B divides the area between the arc AB, BN and AN in the ratio 11:24

    My problem I have found the area of Area B as shown in my working but the problem is area A I could a different answer please have a look at my working and inform me where I went wrong

    I have attached my working below
    Only my calculations for area B was correct not my area A
    Please help

    thank you
    Attached Images
          
    • Study Helper
    Online

    15
    Study Helper
    (Original post by bigmansouf)
    Only my calculations for area B was correct not my area A
    Please help

    thank you

    Area above the tangent is 11/60

    Area below is 2/5

    Both of which you have.

    Hence ratio 11/60 : 2/5
    or 11/60 : 24/60...

    You just put them together incorrectly.
    • Thread Starter
    Offline

    20
    ReputationRep:
    (Original post by ghostwalker)
    Area above the tangent is 11/60

    Area below is 2/5

    Both of which you have.

    Hence ratio 11/60 : 2/5
    or 11/60 : 24/60...

    You just put them together incorrectly.
    no I'm sorry I didnt find the area for Area A thats the problem I found  A= \frac{13}{12}
    I wrote in the picture that I expected  A=\frac{11}{60} but could  A= \frac{13}{12}
    Also I proceeded to write  \frac{11}{60} \neq \frac{13}{12}
    I am sorry I didnt make it clear but I wanted help in finding Area A as I got the wrong answer.
    Attached Images
     
    • Study Helper
    Online

    15
    Study Helper
    (Original post by bigmansouf)
    no I'm sorry I didnt find the area for Area A thats the problem I found  A= \frac{13}{12}
    I wrote in the picture that I expected  A=\frac{11}{60} but could  A= \frac{13}{12}
    Also I proceeded to write  \frac{11}{60} \neq \frac{13}{12}
    I am sorry I didnt make it clear but I wanted help in finding Area A as I got the wrong answer.
    OK, I just saw you had the correct value.

    In your working for finding the area for A you subtracted the eqn of the tangent from the equation of the curve. This gives you the area between the curve and the tangent. This corresponds with your desired area as long as the tangent is above the x-axis. But for x<8/5 this isn't true.

    You want to work out the area under the curve. Then the area in the triangle (area B). And subtract one from the other to get the area A.
    • Thread Starter
    Offline

    20
    ReputationRep:
    Name:  P_20151030_143547.jpg
Views: 88
Size:  112.3 KB
    Attachment 473551473553

    I have atteched the workings I did but I still got it wrong the book says the equation of the curve is  x^{3}-3x^{2}+4x+8
    Where did I go wrong?
    Attached Images
     
    • Study Helper
    Online

    15
    Study Helper
    (Original post by bigmansouf)
    Name:  P_20151030_143547.jpg
Views: 88
Size:  112.3 KB
    Attachment 473551473553

    I have atteched the workings I did but I still got it wrong the book says the equation of the curve is  x^{3}-3x^{2}+4x+8
    Where did I go wrong?
    Don't understand - what's the original question. Are you sure you've on the right thread?
    • Thread Starter
    Offline

    20
    ReputationRep:
    (Original post by ghostwalker)
    Don't understand - what's the original question. Are you sure you've on the right thread?
    sorry wrong thread i will upload it to the right thread thanks
    • Thread Starter
    Offline

    20
    ReputationRep:
    (Original post by ghostwalker)
    OK, I just saw you had the correct value.

    In your working for finding the area for A you subtracted the eqn of the tangent from the equation of the curve. This gives you the area between the curve and the tangent. This corresponds with your desired area as long as the tangent is above the x-axis. But for x<8/5 this isn't true.

    You want to work out the area under the curve. Then the area in the triangle (area B). And subtract one from the other to get the area A.
    thank you very much i will do this and report back after work
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 31, 2015
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
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.