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    Hello, I was just feeling a bit stuck on this question ...I would appreciate it if u could help me on it ... thanks!

    There is this parametrics question : http://www.examsolutions.net/maths-r...tutorial-1.php

    I did the following and just wanted to check with u if I have done it right:

    x=2cost y=sin2t

    dx/dt= -2sint

    so integral sign (sin2t) (-2sint)

    then [sin(3t) / (3) - sint ] and for the limits I got t= 90 degrees and 0 . I subbed them in but got -4/3....what do i do next? am i going the right way is this one of those questions where the limits will swap around because we surely cannot have a negative area!?
    thanks laura
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    (Original post by laurawoods)
    Hello, I was just feeling a bit stuck on this question ...I would appreciate it if u could help me on it ... thanks!

    There is this parametrics question : http://www.examsolutions.net/maths-r...tutorial-1.php

    I did the following and just wanted to check with u if I have done it right:

    x=2cost y=sin2t

    dx/dt= -2sint

    so integral sign (sin2t) (-2sint)

    then [sin(3t) / (3) - sint ] and for the limits I got t= 90 degrees and 0 . I subbed them in but got -4/3....what do i do next? am i going the right way is this one of those questions where the limits will swap around because we surely cannot have a negative area!?
    thanks laura
    you can have negative area, it just shows the area is below the x axis
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    (Original post by advice_guru)
    you can have negative area, it just shows the area is below the x axis
    ok thanks so
    1) are my working outs right?
    2) also , pls can u look at the diagram (in the web link). They are looking for that top shaded area...when we do this method are we working out the full blob's area (right hand side)...
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    (Original post by laurawoods)


    so integral sign (sin2t) (-2sint)

    then [sin(3t) / (3) - sint ]
    Link is not working

    what did you do there

    Are you looking to integrate -2\displaystyle \int \sin 2t \sin t dt
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    (Original post by TenOfThem)
    Link is not working

    what did you do there

    Are you looking to integrate -2\displaystyle \int \sin 2t \sin t dt
    yes and then i got

    sin (3t) / 3 - sint

    Then I subbed in the values of t = 90 degrees and t=0 and ended up with -4/3 .

    On the web link, they have highlighted the top portion (it is an infinity shaped graph and they want us to work out the area of the right hand side blob (top of it above the x axis)) . I thought that when we use this method, we would find out the area of the whole of the right hand side blob (not just the top bit)
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    (Original post by laurawoods)
    yes and then i got

    sin (3t) / 3 - sint

    Then I subbed in the values of t = 90 degrees and t=0 and ended up with -4/3 .

    On the web link, they have highlighted the top portion (it is an infinity shaped graph and they want us to work out the area of the right hand side blob (top of it above the x axis)) . I thought that when we use this method, we would find out the area of the whole of the right hand side blob (not just the top bit)
    I get -\dfrac{4}{3}[\sin^3t]

    But, yes I get -\dfrac{4}{3} too assuming that is 90 at the top and 0 at the bottom

    do you have a link to the question that works
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    Something like this I'd imagine.

    Your integral should have the "lower x value" at the bottom, and "larger x value" at the top, so should be from t=pi/2 at the bottom to t=0 at the top.
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    (Original post by ghostwalker)
    Something like this I'd imagine.

    Your integral should have the "lower x value" at the bottom, and "larger x value" at the top, so should be from t=pi/2 at the bottom to t=0 at the top.
    wow , it is impressive how u produced this diagram! i know u would probably feel a bit angry if I ask this question again :
    but
    1) is this a question where the limits swap around?
    2) so am I right or wrong?
    3) when we do this method and sub in 90 and 0, do we get area of that whole right hand side blob?
    4) is it ok to use 90 and 0 instead of radians version? thanks
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    (Original post by TenOfThem)
    I get -\dfrac{4}{3}[\sin^3t]

    But, yes I get -\dfrac{4}{3} too assuming that is 90 at the top and 0 at the bottom

    do you have a link to the question that works
    hello ghostwalker's diagram is right /...pls can u help me now ! am i right?
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    (Original post by laurawoods)
    wow , it is impressive how u produced this diagram! i know u would probably feel a bit angry if I ask this question again :
    but
    1) is this a question where the limits swap around?
    2) so am I right or wrong?
    3) when we do this method and sub in 90 and 0, do we get area of that whole right hand side blob?
    4) is it ok to use 90 and 0 instead of radians version? thanks
    If ghostwalker is correct then there is no need to swap anything

    The limits would be 90 to 0 rather than 0 to 90

    So the answer is \dfrac{4}{3}
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    (Original post by TenOfThem)
    If ghostwalker is correct then there is no need to swap anything

    The limits would be 90 to 0 rather than 0 to 90

    So the answer is \dfrac{4}{3}
    so is my integration right sin3t / 3 -sint

    ok if 90 is at the top of the integral sign thing surely we get -4/3 isnt it so"
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    (Original post by laurawoods)
    wow , it is impressive how u produced this diagram! i know u would probably feel a bit angry if I ask this question again :
    but
    1) is this a question where the limits swap around?
    2) so am I right or wrong?
    3) when we do this method and sub in 90 and 0, do we get area of that whole right hand side blob?
    4) is it ok to use 90 and 0 instead of radians version? thanks
    I'll start with the last one. In any calculus, differentiation or integration, you will always use radians. It is not OK to use degrees.

    Regarding the first three questions.


    Your area in terms of x and y refering to the top curve on the right is:

    \displaystyle\int_0^2y\;\text{dx  }

    Now that part of the curve is defined by the parameter t, in the interval 0 to pi/2.

    You are effectively just making a substutition. x=2cos theta

    So, for x to be 0 theta is pi/2 and for x to be 2, theta is 0

    So your limits change from 0,2 to pi/2,0 respectively. I.e the upper limit now has the lowest value.

    And this is working out the area of just that top right shaded part.

    PS: Your actual integration was incorrrect, and should be as TenOfThem said.
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    (Original post by laurawoods)
    so is my integration right sin3t / 3 -sint
    I do not know where you got that from

    I told you what I got


    ok if 90 is at the top of the integral sign thing surely we get -4/3 isnt it so"
    But 90 is at the bottom
 
 
 
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