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Really Challenging Area Between Curves Problem

  1. Dec 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose C:y=f(x) with f a twice-differentiable function such that f''(x)> 0 for each x on the closed interval [0,a] where a is a positive constant. Suppose T is the tangent line to C at a point P= (r,f(r)) on C where r is in the open interval (0,a). Let A be the area of the plane region bounded by the y-axis, the vertical line K: x=a, the curve C, and the tangent line T.

    Express A as a function of r.

    3. The attempt at a solution

    The curve is concave up on the closed interval [0,a]. After sketching the curve, drawing a tangent line at a point P on the interval between 0 and a that lies below the curve, I realized that the area A lies between the curve C, and the tangent line T. The area A could be the integral of C:y=f(x) subtracting the area under the tangent line. A friend and I then thought of interpreting the area below the tangent line T to be a trapezoid (f(0)+f(a))/2. Although the area of the trapezoid doesn't use x as a variable and seems to work, I can't seem to find an expression to substitute r for x in the integral for the area of the curve C. This problem has been bugging me for the past few days, I would sleep a lot better if I can solve it! I would really appreciate it if someone could help me out.
     
    Last edited: Dec 11, 2013
  2. jcsd
  3. Dec 12, 2013 #2
    How about ##e^x##? Won't that work for starters? Worry about the others later if we can just do that one. Then from what you said, I constructed an example for r=1 in the plot below and looks like we need the area of the blue right? If that's so, then it's not hard to find an equation of that tangent line as a function of r right? Once we get that line and we know the function, then we can compose the integral for the two shaded regions as a function of r.

    attachment.php?attachmentid=64717&stc=1&d=1386844112.jpg
     

    Attached Files:

    Last edited: Dec 12, 2013
  4. Dec 12, 2013 #3
    Thanks for replying! I think I was able to solve for the area for e^x. I integrated f(x)= e^x from 0 to a and got F(x)= e^a-1. I then solved for the tangent line. I think it is f(x)- e^r=(e^r)(x-r). I then have to integrate the difference right? After integrating the difference from 0 to a, I got Area= (e^a) - [(e^r)(a^2)]/[2] - re^r(a) + ae^r - 1. Does this look right?

    I also have to prove why the area is minimized when r=a/2.
     
    Last edited: Dec 12, 2013
  5. Dec 12, 2013 #4

    LCKurtz

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    I don't think working with ##e^x## is very instructive. You need to start by writing the integral for the area in the general case. You can write the equation of the tangent line at ##(r,f(r))## since you also have the slope ##f'(r)## there. Do that and show us the integral you set up for the area.
     
  6. Dec 12, 2013 #5
    Ok, after integrating the function f(x) and the tangent line y= f'(r)(x-r)+f(r) on [0,a], I got F(a)-F(0)-[f'(r)a^2]/2 - f'(r)ra + f(r)a. I got this from integrating both the function and the tangent line and setting up a difference between them. Is this correct? If it is, how would I minimize this. The r in my Area function is not a variable, how can I differentiate and use the Extreme Value Theorem on it? Thanks! Please reply ASAP. My test is tomorrow and this is the last thing I need to study.
     
  7. Dec 13, 2013 #6

    LCKurtz

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    Please show your work.

    You can't integrate the area because you don't have a formula for ##f(x)##. You need to do what I suggested. Set up the integral and show us.
     
  8. Dec 13, 2013 #7
    Is this correct?

    [itex]\int[/itex]f(x) - f'(r)(x-r)+f(r).

    I solved for the tangent line by solving for y-f(r)=f'(r)(x-r). This is just the generic point slope formula.

    I can't find a way to rewrite x in terms of a and r in the integral.
     
  9. Dec 13, 2013 #8

    LCKurtz

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    You have a sign wrong. The integral should be$$
    A=\int_0^a f(x)-f'(r)(x-r) - f(r)~dx$$which is a function of ##r##. Now calculate ##A'(r)## assuming you can differentiate under the integral sign, and remembering you are differentiating with respect to ##r##, not ##x##, and set it equal to ##0## looking for a critical point. So you will have integral(something)dx = 0. Now work out that integral remembering the variable is dx and see if you can solve the problem.

    It was very late last night when I returned home and after looking at your first method, I think it would work if you do the algebra correctly.
     
    Last edited: Dec 13, 2013
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