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Math Help - Integration question involving parametric equations

  1. #1
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    Integration question involving parametric equations

    The curve C has parametric equations x=\ln(t+2), y=\frac{1}{(t+1)},  t >-1

    The finite region R between the curve C and the x-axis is bounded by the lines with equations x = ln 2 and x = ln 4.

    a)Show that the area of R is given by the integral \int_{0}^{2}\frac{1}{(t+1)(t+2)}\,dt (4 marks)

    b)Hence find an exact value for this area (6 marks)

    c)Find a cartesian equation of the curve C, in the form y=f(x) (4 marks)

    d)State the domain of values for x for this curve (1 mark)


    I came across this question on a past exam paper i was using to revise from and although i attempted it I am still quite unsure of how to tackle this question or a similar question if it comes up.

    I would be grateful if somebody could show me how to go about answering this question and give me some guidelines of how to think about a question like this in the future. Thanks to all!
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  2. #2
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    The thing to remember for these types of questions is that the region you require is

    \int^a_b y\, dx

    = \int^a_b y \frac{dx}{dt} dt

    by the chain rule.

    so for you we have

    \int^a_b \frac{1}{(t+1)} \frac{dx}{dt} dt = \int^a_b \frac{1}{(t+1)} \frac{1}{(t+2)} dt

    Now you need to find your values of a and b, well they are when x = ln(4) and x = ln(2), look at your expression for x and you see that these occur when t = 2 and 0 respectively.

    So putting it all together we have

    R = \int^2_0 \frac{1}{(t+1)(t+2)} \, dt

    as required, which you can solve using partial fractions
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Big_Joe View Post
    The curve C has parametric equations x=\ln(t+2), y=\frac{1}{(t+1)},  t >-1

    The finite region R between the curve C and the x-axis is bounded by the lines with equations x = ln 2 and x = ln 4.

    a)Show that the area of R is given by the integral \int_{0}^{2}\frac{1}{(t+1)(t+2)}\,dt (4 marks)
    Stonehambey addressed this

    b)Hence find an exact value for this area (6 marks)
    he also addressed this. use partial fractions...or as an alternative, algebra

    Note that \frac 1{(t + 1)(t + 2)} = \frac {t + 2 - (t + 1)}{(t + 1)(t + 2)} = \frac {t + 2}{(t + 1)(t + 2)} - \frac {t + 1}{(t + 1)(t + 2)} = \frac 1{t + 1} - \frac 1{t + 2}

    this is where partial fractions would have gotten you. now, integrate term by term

    c)Find a cartesian equation of the curve C, in the form y=f(x) (4 marks)
    look at x = \ln (t + 2)

    \Rightarrow t + 2 = e^x

    \Rightarrow t + 1 = e^x - 1

    thus, we can replace the t + 1 in the y(t) function to get the y(x) function we require.

    d)State the domain of values for x for this curve (1 mark)
    i suppose you can handle this, bearing in mind how our curve is defined and the answer to part (c)
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