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Another area between curves.

  1. Feb 4, 2007 #1

    tony873004

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    [tex]y = x^2 ,\,\,y = \frac{2}{{x^2 + 1}}[/tex]

    [​IMG]

    It looks obvious from the graph that the intersections are -1 and 1, but just to verify:

    [tex]\frac{2}{{x^2 + 1}} = x^2 \,\, \Leftrightarrow \,\,x^2 \left( {x^2 + 1} \right) = 2\,\, \Leftrightarrow x^4 + x^2 = 2,\,\,x = - 1,\,\,1[/tex]

    So my integration limits are -1 and 1

    Therefore, my area should be
    [tex]A = \int_{ - 1}^1 {\frac{2}{{x^2 + 1}}dx - \int_{ - 1}^1 {x^2 } } dx = 2\int_{ - 1}^1 {\frac{1}{{x^2 + 1}}dx - \int_{ - 1}^1 {x^2 } } dx[/tex]

    Integrating this I get:
    [tex]\left. {2\tan ^{ - 1} x - \frac{1}{3}x^3 } \right]_{ - 1}^1 = \left( {2\tan ^{ - 1} \left( 1 \right) - \frac{1}{3}1^3 } \right) - \left( {2\tan ^{ - 1} \left( { - 1} \right) - \frac{1}{3}\left( { - 1} \right)^3 } \right)[/tex]

    Plugging this into my calculator gives me 1.90170379707065, but the back of the book says pi-2/3 which equals 2.475.

    Just looking at the graph, it seems like the answer would be more than 2, as each of the 4 squares appear slightly more than half shaded.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 4, 2007 #2

    mjsd

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    you probably did something wrong with calculator... everything is fine... I get 2.4749... when I plugged it in
     
  4. Feb 4, 2007 #3

    cristo

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    Alternatively, note that arctan(1)=pi/4, and arctan(-1)=-pi/4, and plug that into the expression.
     
  5. Feb 4, 2007 #4

    Gib Z

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    Why use a calculator at all?

    arc tan 1 = pi/4
    arc tan -1 = -pi/4
    Sub all this in,

    (pi/2 -1/3) - (-pi/2 + 1/3) = pi - (2/3), which is what the book gave.

    You obviously made a mistake with your calculator.

    Edit: Cristo beat me
     
  6. Feb 4, 2007 #5

    tony873004

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    Thanks. Now I'm getting it. I guess I didn't realize how to use the calculator. In other problems, I've been entering stuff like:

    [tex]\sin ^2 \theta [/tex]

    as

    [tex]\left( {\sin \theta } \right)^2 [/tex]

    since the calculator does not have a [tex]\sin ^2 \theta [/tex] button, and I've been getting the right answers.

    So I just assumed that
    [tex]\tan ^{ - 1} x[/tex]

    could be entered as
    [tex]\left( {\tan x} \right)^{ - 1} [/tex]

    But when I actually use the [tex]\tan ^{ - 1} x[/tex] button, I get the right answer.

    Can anyone shed some light on trig functions raised to a power?
     
  7. Feb 4, 2007 #6

    Gib Z

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    Ahh theres your problem mate.

    [tex]\tan ^{ - 1} x[/tex] is not the same as [tex](\tan x)^{-1}[/tex].

    Alot of people get confused with this notation for this.

    When you see a function, f(x), for some reason our notation for its inverse is [tex]f^{-1}(x)[/tex]. Its not the same as 1/(f(x)).

    When you see [tex]\tan ^{ - 1} x[/tex] it means f(x)=tan x, and this function is its inverse. Not the same as 1/(tan x). This is one of the reasons the cofunctions were introduced. 1/tan x= cot x. 1/sin x= csc x and 1/cos x =sec x. That way we can remember, if they wrote tan^-1 x, it cant mean 1/tan x, otherwise theyd have wrote cot.
     
  8. Feb 4, 2007 #7

    cristo

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    It is true that sin2x is used as a shorthand notation for (sinx)2, however sin-1x does not mean (sinx)-1; sin-1x is the inverse function of sine, i.e the function that when composed with with the sine function is the identity function, and thus sin-1(sinx)=x. Note that (sinx)-1 is the reciprocal of sinx, i.e. is 1/sinx.
     
  9. Feb 4, 2007 #8
    This is an example of a rather poor choice of notation, I mean that when we write
    [tex]\sin ^a \theta [/tex]
    we always mean it in the sense of raising the sine function evaluated at theta to the "a"th power as in
    [tex]\left( {\sin \theta } \right)^a [/tex]


    But we also tend to write the inverses functions of the trig functions in the same manner. So as a shorthand for writing
    [tex]\arctan x[/tex]
    we write
    [tex]\tan ^{ - 1} x[/tex]

    I suppose the ambiguity here is avoided by noting that
    [tex]\left( {\tan x} \right)^{ - 1}\ =\ \cot x [/tex]
     
  10. Feb 4, 2007 #9

    tony873004

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    lol... because I'm not a math genius and I just don't remember all that stuff. But every time I use my calculator, I know there's something I'm missing. That's why the teacher laughs when someone asks if we can use a calculator on the test. (we can, and he'd probably accept the decimal answer.)

    This is the toughest part of Calc for me. I find the Calc itself is easy. It's just remembering years of algebra and trig that trip me up.

    Thanks everyone for your help. :smile: More q's tommorow. I'm stuck on a problem with the integrand having absolute value brackets around it, but I'll give it a go tommorow before I give up.
     
    Last edited: Feb 4, 2007
  11. Feb 4, 2007 #10

    tony873004

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    just to clarify:

    [tex]
    \begin{array}{l}
    \sin ^3 \theta = \left( {\sin \theta } \right)^3 \\
    \sin ^2 \theta = \left( {\sin \theta } \right)^2 \\
    \sin ^1 \theta = \left( {\sin \theta } \right)^1 \\
    \sin ^0 \theta = \left( {\sin \theta } \right)^0 \\
    \sin ^{ - 1} \theta \ne \left( {\sin \theta } \right)^{ - 1} \\
    \sin ^{ - 2} \theta = \left( {\sin \theta } \right)^{ - 2} \\
    \sin ^{ - 3} \theta = \left( {\sin \theta } \right)^{ - 3} \\
    \end{array}
    [/tex]
    ???
     
  12. Feb 4, 2007 #11
    Those are all fine, although the last one is just equal to 1.

    That is also correct and sin-1(x)=arcsin(x).

    Yes, I think all of these would again be equal, and the first would be equivalent to csc2(theta) and the second equivalent to csc3(theta).
     
    Last edited: Feb 4, 2007
  13. Feb 4, 2007 #12

    Gib Z

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    As for the absolute value thing, i'll just explain it now because its actually really easy.

    The integral gives the NET area. So When we find the area, and some of the function is negative, we find the areas separately. We Separate the positive sections and negative sections. The negative sections will give a negative integral, so we absolute value it.

    When a part of the function is negative, and we absolute value it, its just all become positive, so it is a reflection of itself in the x axis. So the area is the same!

    So basically, the integral is the net area. The Integral with the absolute value sign is the true Area.
     
  14. Feb 4, 2007 #13

    tony873004

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    Thanks, you saved me the trouble of tex-ing my work and asking here. It's quite intuituve once you think about it.
     
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