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Math Help - parametrics!

  1. #1
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    parametrics!

    I need help with this problem!

    An object moving along a curve in the xy-plane has position [x(t), y(t)] with dx/dt=cos(t^2) and dy/dt=sin(t^3). At time t=0, the particle is at position (4,7). Where is the particle when t=2?

    I can get the integrals of both equations, but my problem is that when I try to substitute t=0 to find the constants of integration, it becomes undefined. Help!

    This is what I end up with:
    X=[sin(t^2)]/(2t)+C
    Y=[(-cos(t^3)]/(3t^2)+C

    But then you can't substitute t=0, or am I missing something? Probably am
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  2. #2
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    Quote Originally Posted by runner21 View Post
    I need help with this problem!

    An object moving along a curve in the xy-plane has position [x(t), y(t)] with dx/dt=cos(t^2) and dy/dt=sin(t^3). At time t=0, the particle is at position (4,7). Where is the particle when t=2?

    I can get the integrals of both equations, but my problem is that when I try to substitute t=0 to find the constants of integration, it becomes undefined. Help!

    This is what I end up with:
    X=[sin(t^2)]/(2t)+C
    Y=[(-cos(t^3)]/(3t^2)+C

    But then you can't substitute t=0, or am I missing something? Probably am
    Differentiate your expressions for x(t) and y(t) and see if the match up with the derivatives that were given:

     \frac{d}{dt} (\frac{sin(t^2)}{2t}+C) = \frac{2t.2tcos(t^2)-sin(t^2)(2)}{4t^2}= cos(t^2)-\frac{sin(t^2)}{2t^2} \neq cos(t^2) .

    You haven't integrated correctly.

    Are you sure the given derivatives aren't  \frac{dx}{dt} = cos^2(t) = (cos(t))^2 and  \frac{dy}{dt} = sin^3(t) = (sin(t))^3 ??

    If not then I recommend writing your derivatives as infinite series using the Taylor/Maclaurin series and integrating term by term to the desired accuracy...
    Last edited by mr fantastic; January 6th 2009 at 02:52 AM. Reason: Deleted reference to double posting
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  3. #3
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    Quote Originally Posted by runner21 View Post
    I need help with this problem!

    An object moving along a curve in the xy-plane has position [x(t), y(t)] with dx/dt=cos(t^2) and dy/dt=sin(t^3). At time t=0, the particle is at position (4,7). Where is the particle when t=2?

    I can get the integrals of both equations, but my problem is that when I try to substitute t=0 to find the constants of integration, it becomes undefined. Help!

    This is what I end up with:
    X=[sin(t^2)]/(2t)+C
    Y=[(-cos(t^3)]/(3t^2)+C

    But then you can't substitute t=0, or am I missing something? Probably am
    Hello . Please explain how you got \int \cos\left(x^2\right)~dx=\frac{\sin\left(x^2\right)  }{2x}+C, because I believe there lies your problem.
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  4. #4
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    I think my integrals might be wrong, but I checked what I typed as the question with the problem in my book and its the same. Is there another way to do this problem then without having to integrate, because that seems to be where the problem lies.
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  5. #5
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    Quote Originally Posted by runner21 View Post
    I think my integrals might be wrong, but I checked what I typed as the question with the problem in my book and its the same. Is there another way to do this problem then without having to integrate, because that seems to be where the problem lies.
    I can't really think of any alternative at the moment. Though I may be wrong.

    As I said before, use the Maclaurin series expansions of sin(x) and cos(x) to get expressions for sin(t^3) and cos(t^2), then integrate as many terms as you think you need to to keep your solution relatively accurate:

    \displaystyle sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}... = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)!}


    \displaystyle cos(x) = 1 - \frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}... = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}

    Hence:

    \displaystyle sin(t^3) = t^3 - \frac{t^9}{3!}+\frac{t^{15}}{5!}-\frac{t^{21}}{7!}...=\sum_{n=0}^{\infty} \frac{(-1)^nt^{6n+3}}{(2n+1)!}


    \displaystyle cos(t^2) = 1 - \frac{t^4}{2!}+\frac{t^8}{4!}-\frac{t^{12}}{6!}... =\sum_{n=0}^{\infty} \frac{(-1)^nt^{4n}}{(2n)!}

    You will get your constants of integration correct (they're the same as the coordinates at t = 0!), but your calculation of where the object will be at t=2 will be slightly approximated.
    Last edited by Mush; January 5th 2009 at 10:33 PM.
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    Hello . Please explain how you got \int \cos\left(x^2\right)~dx=\frac{\sin\left(x^2\right)  }{2x}+C, because I believe there lies your problem.
    I used a u-substitution (let u=x^2) and then I took the derivative of that to get du=2x~dx, and because dx is now being multiplied by 2x, I had to divide the entire integral statement by 2x, so that:

    (1/2x)\int \cos\left(u\right)~du=\frac{\sin\left(u\right)}{2x  }+C

    and then I just substituted the x^2 back in for u.
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  7. #7
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    Quote Originally Posted by runner21 View Post
    I used a u-substitution (let u=x^2) and then I took the derivative of that to get du=2x~dx, and because dx is now being multiplied by 2x, I had to divide the entire integral statement by 2x, so that:

    (1/2x)\int \cos\left(u\right)~du=\frac{\sin\left(u\right)}{2x  }+C Mr F: Sacre bleu!! You are treating x as if it was a constant ....!! If you differentiated your 'answer' you would quickly see that it is wrong.

    and then I just substituted the x^2 back in for u.
    ..
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  8. #8
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    Quote Originally Posted by runner21 View Post
    I need help with this problem!

    An object moving along a curve in the xy-plane has position [x(t), y(t)] with dx/dt=cos(t^2) and dy/dt=sin(t^3). At time t=0, the particle is at position (4,7). Where is the particle when t=2?

    I can get the integrals of both equations, but my problem is that when I try to substitute t=0 to find the constants of integration, it becomes undefined. Help!

    This is what I end up with:
    X=[sin(t^2)]/(2t)+C
    Y=[(-cos(t^3)]/(3t^2)+C

    But then you can't substitute t=0, or am I missing something? Probably am
    An integral form of the solution is:

    x(t) = \int_0^t \cos u^2 \, du + 4 \Rightarrow x(2) = \int_0^2 \cos u^2 \, du + 4.

    y(t) = \int_0^t \sin u^3 \, du + 7 \Rightarrow x(2) = \int_0^2 \sin u^3 \, du + 7.

    Neither integral can be found exactly using a finite number of elementary functions. So there are two possibilities:

    1. You have to use technology to get a decimal approximation of each integral.

    2. The expressions given for dx/dt and dy/dt are wrong.
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