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Math Help - length of curve

  1. #1
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    length of curve

    find the length of the curve C from t=0 to t pi/2 where x(t)=2sin^3t and y(t)=2cos^3t. set up the correct integral and simplify the integrand.

    i think im supposed to use the formula L= sqrt[(dx/dt)^2 + (dy/dt)^2]

    for x' i got 36sin^4tcos^2t
    y' = -36sin^2tcos^4t

    how am i supposed to add(or subtract) them so that it =6sintcost ? that was the answer given for the part inside the integrand. so before rooting im supposed to get 36sin^2tcos^2t but how?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jeph View Post
    find the length of the curve C from t=0 to t pi/2 where x(t)=2sin^3t and y(t)=2cos^3t. set up the correct integral and simplify the integrand.

    i think im supposed to use the formula L= sqrt[(dx/dt)^2 + (dy/dt)^2]

    for x' i got 36sin^4tcos^2t
    y' = -36sin^2tcos^4t

    how am i supposed to add(or subtract) them so that it =6sintcost ? that was the answer given for the part inside the integrand. so before rooting im supposed to get 36sin^2tcos^2t but how?
    L = \int_0^{\pi /2} dt \, \sqrt{ \left ( \frac{dx}{dt} \right ) ^2 + \left ( \frac{dy}{dt} \right )^2 }

    So
    \frac{dx}{dt} = 6sin^2(3t)cos(3t)
    \frac{dy}{dt} = -6cos^2(3t)sin(3t)

    Thus
    L = \int_0^{\pi /2} dt \, \sqrt{ \left ( 6sin^2(3t)cos(3t) \right ) ^2 + \left ( -6cos^2(3t)sin(3t) \right ) ^2 }

    L = \int_0^{\pi /2} dt \, \sqrt{ 36sin^4(3t)cos^2(3t) + 36cos^4(3t)sin^2(3t) }

    L = 6 \int_0^{\pi /2} dt \, \sqrt{ sin^2(3t)cos^2(3t) (sin^2(3t) + cos^2(3t) ) }

    L = 6 \int_0^{\pi /2} dt \,   sin(3t)cos(3t)

    Can you take it from here?

    -Dan
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    i think i might have screwed up on my math typing again...i wanted to say x=2sin^3(t) and y=2cos^3(t). just t not 3t so "sin and cos cubed t" o.o if not then i dont know how to get 3t...

    whats the technique to deferentiate those trigs with cubes. can you go
    2cos^3(t)
    2(cos(t))^3 u=cos(t)
    6(u^2)u'
    -6cos^2(t)sin(t)

    or does the sign switch both times you differentiate? it becomes negative then postive again? or the sign only switches when you differentiate the last cos(t)?
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  4. #4
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    Quote Originally Posted by jeph View Post
    find the length of the curve C from t=0 to t pi/2 where x(t)=2sin^3t and y(t)=2cos^3t. set up the correct integral and simplify the integrand.

    i think im supposed to use the formula L= sqrt[(dx/dt)^2 + (dy/dt)^2]

    ...
    Hello,

    I assume that your functions read like:
    x(t)=2\sin^3(t) and y(t)=2\cos^3(t)

    Then

    \frac{dx}{dt} = 6sin^2(t)cos(t) and
    \frac{dy}{dt} = -6cos^2(t)sin(t)

    If you square both sides of these equations you'll get:

    \left(\frac{dx}{dt} \right)^2= 36sin^4(t)cos^2(t)
    \left( \frac{dy}{dt} \right)^2 = 36cos^4(t)sin^2(t)

    Now calculate the integral:

    L=\int_{0}^{\frac{\pi}{2}} \sqrt{36sin^4(t)cos^2(t) + 36cos^4(t)sin^2(t)} \cdot dt

    = \int_{0}^{\frac{\pi}{2}} \sqrt{36sin^2(t)cos^2(t)\underbrace{\left( \sin^2(t)+\cos^2(t) \right)}} _{= 1}\cdot dt

    = 6 \cdot \int_{0}^{\frac{\pi}{2}} |{sin(t)cos(t)} | \cdot dt

    I'll leave the rest for you .

    (For confirmation only: I've got L = 3)
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  5. #5
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    Quote Originally Posted by jeph View Post
    ...or the sign only switches when you differentiate the last cos(t)?
    Hello,

    have a look here:

    x(t) = 2 \sin^3(t)

    \frac{dx}{dt}= 2 \cdot 3 \cdot \sin^2(t) \cdot(-\cos(t))
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    Quote Originally Posted by earboth View Post
    Hello,

    have a look here:

    x(t) = 2 \sin^3(t)

    \frac{dx}{dt}= 2 \cdot 3 \cdot \sin^2(t) \cdot(-\cos(t))
    shouldnt it be the dy/dt that becomes negative? you differentiate sin(t) right? that just becomes cos(t) not -cos(t)? o.o
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  7. #7
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    Quote Originally Posted by jeph View Post
    shouldnt it be the dy/dt that becomes negative? you differentiate sin(t) right? that just becomes cos(t) not -cos(t)? o.o
    Hello,

    of course you are right - but I hope that you caught the method even though I did an ugly mistake. My apologies.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    L = \int_0^{\pi /2} dt \, \sqrt{ \left ( \frac{dx}{dt} \right ) ^2 + \left ( \frac{dy}{dt} \right )^2 }

    So
    \frac{dx}{dt} = 6sin^2(3t)cos(3t)
    \frac{dy}{dt} = -6cos^2(3t)sin(3t)

    Thus
    L = \int_0^{\pi /2} dt \, \sqrt{ \left ( 6sin^2(3t)cos(3t) \right ) ^2 + \left ( -6cos^2(3t)sin(3t) \right ) ^2 }

    L = \int_0^{\pi /2} dt \, \sqrt{ 36sin^4(3t)cos^2(3t) + 36cos^4(3t)sin^2(3t) }

    L = 6 \int_0^{\pi /2} dt \, \sqrt{ sin^2(3t)cos^2(3t) (sin^2(3t) + cos^2(3t) ) }

    L = 6 \int_0^{\pi /2} dt \, sin(3t)cos(3t)

    Can you take it from here?

    -Dan
    Dan, is there a real reason why you always put dt at the begining of the integral rather than the end? or is it just another one of your personality quirks?
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    Dan, is there a real reason why you always put dt at the begining of the integral rather than the end? or is it just another one of your personality quirks?
    It's a personality quirk. It's also fairly standard in the Physics books I tend to use. There are advantages and disadvantages to its use, but for this level of work I don't see any particular advantage/disadvantage.

    -Dan
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