1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculus problem find the equation of the line given two points

  1. Mar 14, 2012 #1
    Consider the curve f(x) = 1/x
    Consider two points on f(x): Pa and Qa, where the x-coordinate of Pa is a, and the x-coordinate of Qa is a+1.
    Let La be the line connecting Pa and Qa

    1.) Find the equation for La
    2.) Find a formula that expresses A(a) = the area between f(x) and La
    3.) Determine lim(a-->∞) A(a) and lim(a-->0) A(a)
    4.) Does A(a) have an extremum for 0 < a? If yes, determine if it's a minimum or maximu (or neither?), and what value of A(a) is at the extremum.

    Extra: Redo the problem, where the x-coordinate of Qa is a^2


    Okay, so I know that the coordinate of Pa is (a, 1/a) and I know that the coordinate of Qa is (a+1, 1/a+1). So, you have the equation for the line y=mx+b. So, the slope would be [(1/a+1)-(1/a)] / [(a+1)-(a)], which would simplify down to m = -(1) / (a^2+a). Correct? How do I go on from here?
     
    Last edited: Mar 14, 2012
  2. jcsd
  3. Mar 14, 2012 #2
    Remember back to your Pre-Calc days... you have the slope and a point on the line. Is there a formula that can combine the two to get the equation of a line?
     
  4. Mar 14, 2012 #3
    Right, well I know the equation y-y1=m(x-x1).
    So, I would have: y-(1/a)=(-1 / a^2+a)*(x-a) ---> y=(-x / a^2+a)+(a / a^2+a)+(1/a).
    Right? Or... y=(-x / a^2+a)+(2a+1 / a(a+1))
    Which would imply that:
    y = (2a+1-x)/(a^2+a)
    Right?
     
  5. Mar 14, 2012 #4
    Just be careful with your equations when posting online. Try to remember to use brackets. I know that you intended for your final equation to look like

    [itex]y = \displaystyle\frac{-x}{a^{2}+a} + \frac{2a+1}{a^{2}+a}[/itex]


    And I just saw you updated your post, your final answer is fine (the brackets are good) but it would be better to split up the x term and the constant term (it'll be easier to integrate that way). The equation you had before was good, just be careful about parentheses.
     
  6. Mar 14, 2012 #5
    Okay...
    So, when setting up this area equation... would it simply be the the integral of -(x / a^2+a)+(2a+1 / a^2+a) dx - integral of 1/x dx [from x=a to x=a+1]?

    Okay, doing that, I get that the function for the area between La and f(x) is:

    A(a) = (2a+1 / 2a^2+2a) - ln(a+1) - ln(a)

    Is this correct?
     
    Last edited: Mar 14, 2012
  7. Mar 14, 2012 #6
    That is how you set it up, I don't think your final answer is correct though, let me check...

    I think you might have forgotten to distribute a negative sign when taking the integral, the natural logs aren't right. Try again and see what you come up with.
     
    Last edited: Mar 14, 2012
  8. Mar 14, 2012 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Use brackets! Is A(a) = (2a+1)/(2a^2+2a) - ln(a+1)-ln(a), or is it what you actually *wrote*, which would be [tex] A(a) = 2a + \frac{1}{2a^2} + 2a - \ln(a+1) - \ln(a)[/itex]?

    RGV
     
  9. Mar 14, 2012 #8
    Yes, A(a) = (2a+1)/(2a^2+2a) - ln(a+1) - ln(a)

    or is it

    A(a) = (2a+1)/(2a^2+2a) - ln(a+1) + ln(a) ?
     
  10. Mar 14, 2012 #9
    The second answer is what I got, and you can simplify that further if you like (properties of logs).

    You still seem unsure about the sign of the last logarithm. How come?
     
  11. Mar 14, 2012 #10
    I don't know... I just copied that from my first try.
    Anyway, would the reduced equation be:
    A(a) = (2a+1)/(2a^2+2a) + ln[(a)/(a+1)] ?
     
  12. Mar 14, 2012 #11
    Yes, that's the right answer, but if you are unsure, you should go back and redo your work so you know why that is the right answer; i.e. convince yourself it is the right answer.
     
  13. Mar 14, 2012 #12
    No, I see what I did originally: I didn't carry the minus sign over.
    Okay, so now that we have a function for A(a), how do the limits work?

    We have:

    (a) lim[a→∞]
    (b) lim[a→0]

    I got

    lim[a→∞] = 0
    lim[a→0] = ∞

    Is that correct?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook