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Math Help - length of a curve question mmn 16 1B

  1. #1
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    length of a curve question mmn 16 1B

    find the length of x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}
    ?


    i use the parametrisation
    y=a(sint)^3
    x=a(cost)^3

    but what is the range of t?
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    find the length of x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}
    ?


    i use the parametrisation
    y=a(sint)^3
    x=a(cost)^3

    but what is the range of t?
    In the first quadrant the curve looks like this: plot x^(2/3) + y^(2/3) = 1 - Wolfram|Alpha

    I suggest you find the arclength for the curve in the first quadrant - the domain for t should be obvious - and then multiply by 4.
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  3. #3
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    ok how to find the length of it

    i need to use the parametrisation and make the integral over t

    how to do it here
    ?
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    ok how to find the length of it

    i need to use the parametrisation and make the integral over t

    how to do it here
    ?
    Your original question has been answered. What work have you done, where exactly are you stuck?
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  5. #5
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    i asked for the range of "t" in the integral
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    i asked for the range of "t" in the integral
    I posted this:

    Quote Originally Posted by Mr Fantastic
    In the first quadrant the curve looks like this: plot x^(2/3) + y^(2/3) = 1 - Wolfram|Alpha

    I suggest you find the arclength for the curve in the first quadrant - the domain for t should be obvious - and then multiply by 4.
    Where are you stuck here?
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  7. #7
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    the length of a curve is a root of the some of the partial derivatives

    but its not an integral

    ??
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  8. #8
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    Quote Originally Posted by transgalactic View Post
    the length of a curve is a root of the some of the partial derivatives

    but its not an integral

    ??
    What you have posted here makes no sense. I know you understand how to find the length of a curve because of other of your threads. Show your work and say where you are stuck here.
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  9. #9
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    this chapter is about curve integrals
    integral of the form \int_c Fdr


    i dont have it here
    ?

    maybe here F=1
    and we have
    \int_c dr

    i ll try it
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  10. #10
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    You've changed the question so many times, it is hard to be sure what you are asking.

    Surely you know that if you are writing x and y in terms of parameter t, then the arclength is given by
    \int \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2} dt

    That is an integral and there are no partial derivatives.

    You said you wanted to use the parameterization y=a(sint)^3, x=a(cost)^3.

    In the graph Mr. Fantastic showed you initially, x goes from 0 up to 1 while y goes from 1 down to 0. (He set a= 1.)
    For what t is x= a(cos t)^3= 0? For what t is a(cos t)^3= a?
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  11. #11
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    t is from 0 till pi/2

    and i multiply the result by 4
    i understand now
    thanks
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