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Stationary Points and finding variables?

  1. Nov 4, 2011 #1
    Stationary Points and finding variables? :)

    1. The problem statement, all variables and given/known data
    The curve with the equation y=ax^2 + bx+c has a stationary point at (1,2). When x=0, the slope of the curve is 45 degrees. Find a, b, c.


    2. Relevant equations
    I'm not sure how to create an equation that involves the 45 degrees


    3. The attempt at a solution
    There are three equations to solve for the three variables, but so far I only have 2 equations:

    2= a+b+c
    and
    2a+b=2

    DW, SOLVED IT! :)
     
    Last edited: Nov 4, 2011
  2. jcsd
  3. Nov 5, 2011 #2

    Simon Bridge

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    Re: Stationary Points and finding variables? :)

    Well done!

    Sometimes explaining the problem to someone else will lead to a solution :)
    Want to share how you did it?
     
  4. Nov 5, 2011 #3
    Re: Stationary Points and finding variables? :)

    the derivative of a function is its gradient, which is also defined as rise/run. In trig, rise/run = tanθ!

    therefore tan45°=1

    And then just use one of the aforementioned methods of substitution into y'.

    :)
     
  5. Nov 5, 2011 #4

    HallsofIvy

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    Re: Stationary Points and finding variables? :)

    Yes. The fact that the function has a stationary point at (1, 2) tells you two things:
    First that the value of the function at x= 1 is 2: [itex]a(1)^2+ b(1)+ c= a+ b+ c= 2[/itex].
    Second that the derivative there is 0: 2a(1)+ b= 2a+ b= 0.

    The fact that the tangent line, at x= 0, is at 45 degrees to the x-axis tells you that the slope of the tangent line there is tan(45)= 1: 2a(0)+ b= 1.

    Your three equations are b= 1, 2a+ b= 0 so 2a+ 1= 0, and a+ b+ c= 2 so a+ c+ 1= 2.
     
  6. Nov 5, 2011 #5

    Simon Bridge

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    Re: Stationary Points and finding variables? :)

    Outstanding Dramacon: I figured I'd give you the glory of explaining it :) There is a chance someone else will google to your problem when they are looking for a solution to theirs. Now you've just helped them.

    HallsofIvy, is a veteran, and has included the complete solution.

    I prefer the "rise over run" version since it uses what "slope" means in relation to the gradient ... though the trig is a useful addition so this is totally not a criticism. The tangent of an angle is also rise over run - of course.
     
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