Turn on thread page Beta
    • Thread Starter
    Offline

    13
    ReputationRep:
    Equation of curve is y=20/x
    A tangent is drawn at a point on the curve
    Show that area of the triangle formed by the tangen and coordinate axes is always 40 at any point
    Offline

    20
    ReputationRep:
    (Original post by FrostedStarzzz)
    Equation of curve is y=20/x
    A tangent is drawn at a point on the curve
    Show that area of the triangle formed by the tangen and coordinate axes is always 40 at any point
    Well, start by differentiating it (you will find it easier to express it as y=20x-1 beforehand). Then find the equation of the tangent in terms of the x-coordinate of the point where it meets the curve.
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by TheMindGarage)
    Well, start by differentiating it (you will find it easier to express it as y=20x-1 beforehand). Then find the equation of the tangent in terms of the x-coordinate of the point where it meets the curve.
    I still don’t understand I differentiate it , what you mean equal tangent in terms of x coordinate at a point where it meets the curve - what I did is 20/x=20x/x^2
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by FrostedStarzzz)
    I still don’t understand I differentiate it , what you mean equal tangent in terms of x coordinate at a point where it meets the curve - what I did is 20/x=20x/x^2
    It's a standard expression to differentiate. \frac{20}{x}=20x^{-1}.

    Now the question says 'at any point' so the best thing to do is to pick a general point on the curve, which would have coordinates (p, \frac{20}{p}), where p \neq 0 for obvious reasons.

    Now you want to find the equation of the line in terms of p which is tangent at that general point.

    Then, you want to see where this line cuts the coordinate axis, and express those points in terms of p again.

    Finally, you want to show that the triangle defined by the y-intercept, the origin, and the x-intercept, has an area of 40 using p and showing that it cancels at some point in the calculation, hence making it negligible.
    Offline

    20
    ReputationRep:
    (Original post by FrostedStarzzz)
    I still don’t understand I differentiate it , what you mean equal tangent in terms of x coordinate at a point where it meets the curve - what I did is 20/x=20x/x^2
    20/x is equal to 20x-1. You need to be able to see how you get this. After that, it's far easier to differentiate it.

    Once you differentiated it, let a be the x-coordinate of a point on the curve. The gradient would be equal to whatever you get when you differentiate, but replacing x with a.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 9, 2017

1,177

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.