Page 2 of 2 FirstFirst 12
Results 16 to 27 of 27

Math Help - DE Marathon

  1. #16
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Chris L T521 View Post
    Here's an interesting problem:

    A mass m=1 is attached to a spring with constant k=4; there is no dashpot. The mass is released
    from rest with x(0)=3. At the instant t=2\pi the mass is struck with a hammer, providing
    an impulse p=8. Determine the motion of the mass.
    \frac{d^2x}{dt^2}+4x=8\delta(t-2\pi), \\\ x(0)=3, \\\ \frac{dx}{dt}\bigg|_{x=0}=0

    Now taking the Laplace transform we get

    s^2X-s(3)+4X=8e^{-2\pi s} \iff (s^2+4)X=3s+8e^{-2\pi s}

    X=\frac{3}{2}\left( \frac{2}{s^2+4}\right)+4e^{-2\pi s}\left( \frac{2}{s^2+4}\right)

    Now we take the inverse laplace transfrom to get

    x(t)=\frac{3}{2}\sin(2t)+4\mathcal{U}(t-2\pi)\sin(2[t-2\pi])=\frac{3}{2}\sin(2t)+4\mathcal{U}(t-2\pi)\sin(2t)

    DE Marathon-capture.jpg
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Excellent. I got something similar to that. You can post your question now if you have one.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Chris L T521 View Post
    Excellent. I got something similar to that. You can post your question now if you have one.
    Okay this is similiar to the one above. I think this is a cool problem

    \frac{d^2x}{dt^2}+0.4\frac{dx}{dt}+9.04x=6e^{-\frac{t}{5}}\cos(3t), \\\ x(0)=x'(0)=0
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by TheEmptySet View Post
    Okay this is similiar to the one above. I think this is a cool problem

    \frac{d^2x}{dt^2}+0.4\frac{dx}{dt}+9.04x=6e^{-\frac{t}{5}}\cos(3t), \\\ x(0)=x'(0)=0
    Taking the Laplace Transform of both sides, we get:

    s^2X+.4(sX)+9.04X=6\left.\frac{s}{9+s^2}\right|_{s  \to{.2+s}}\implies (s^2+.4s+9.04)X=6\frac{.2+s}{9+\left(.2+s\right)^2  }

    Therefore, X=6\frac{.2+s}{(s^2+.4s+9.04)(9+(.2+s)^2)}

    Taking the Inverse Laplace Transform, we have the following:

    \begin{aligned}<br />
x(t)&=\mathcal{L}^{-1}\left\{\frac{1}{(s^2+.4s+.04)+9.04-.04}\cdot 6\frac{.2+s}{9+(.2+s)^2}\right\} \\<br />
&=\mathcal{L}^{-1}\left\{\frac{1}{(s+.2)^2+9}\cdot 6\frac{s+.2}{(s+.2)^2+9}\right\} \\<br />
&=\mathcal{L}^{-1}\left\{\left.\frac{1}{s^2+9}\right|_{s\to{s+.2}}  \cdot\left.\frac{6s}{s^2+9}\right|_{s\to{s+.2}}\ri  ght\}<br />
\end{aligned}<br />

    Thus, we see that x(t)=\frac{1}{3}e^{-\frac{t}{5}}\sin(3t)*6e^{-\frac{t}{5}}\cos(3t)

    Thus, \frac{1}{3}e^{-\frac{t}{5}}\sin(3t)*6e^{-\frac{t}{5}}\cos(3t)=\int_0^t \left(\frac{1}{3}e^{-\frac{1}{5}\tau}\sin(3\tau)\right)\left(6e^{-\frac{1}{5}(t-\tau)}\cos(3(t-\tau))\right)\,d\tau

    =2\int_0^t\bigg[e^{-\frac{1}{5}t}\sin(3\tau)\cos(3t-3\tau)\bigg]\,d\tau

    Using the identity \sin(\vartheta)\cos(\varphi)=\frac{1}{2}\left[\sin(\vartheta+\varphi)+\sin(\vartheta-\varphi)\right], we rewrite the integrand as

    e^{-\frac{t}{5}}\int_0^t \left[\sin(3t)+\sin(6\tau-3t)\right]\,d\tau=e^{-\frac{t}{5}}\left.\left[\frac{6}{12}\left(\tau\sin(3t)-\cos(6\tau-3t)\right)\right]\right|_0^t

    =e^{-\frac{t}{5}}\left(t\sin(3t)\right)=te^{-\frac{t}{5}}\sin(3t)

    Thus, \color{red}\boxed{x(t)=te^{-\frac{t}{5}}\sin(3t)}

    Cool problem

    --Chris
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Here is another way to do this. Complete the square on the first term in the denominator to get

     <br />
X=6\frac{.2+s}{(s^2+.4s+9.04)(9+(.2+s)^2)}=6 \left( \frac{s+\frac{1}{5}}{[(s+\frac{1}{5})^2+9]^2}\right)<br />

    Let F(s)=\frac{s}{(s^2+9)^2} \implies F\left(s+\frac{1}{5} \right)=\left( \frac{s+\frac{1}{5}}{[(s+\frac{1}{5})^2+9]^2}\right)

    Note that F(s)=(-1)\left( \frac{1}{2}\right)\frac{d}{ds}\left( \frac{1}{s^2+9}\right)

    So now...


    \mathcal{L}^{-1} \left\{6\cdot F\left( s+\frac{1}{5}\right) \right\}=6\cdot e^{-\frac{1}{5}} \mathcal{L}^{-1} \{F(s)\}=e^{-\frac{1}{5}t}t\sin(3t)
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by TheEmptySet View Post
    Here is another way to do this. Complete the square on the first term in the denominator to get

     <br />
X=6\frac{.2+s}{(s^2+.4s+9.04)(9+(.2+s)^2)}=6 \left( \frac{s+\frac{1}{5}}{[(s+\frac{1}{5})^2+9]^2}\right)<br />

    Let F(s)=\frac{s}{(s^2+9)^2} \implies F\left(s+\frac{1}{5} \right)=\left( \frac{s+\frac{1}{5}}{[(s+\frac{1}{5})^2+9]^2}\right)

    Note that F(s)=(-1)\left( \frac{1}{2}\right)\frac{d}{ds}\left( \frac{1}{s^2+9}\right)

    So now...


    \mathcal{L}^{-1} \left\{6\cdot F\left( s+\frac{1}{5}\right) \right\}=6\cdot e^{-\frac{1}{5}} \mathcal{L}^{-1} \{F(s)\}=e^{-\frac{1}{5}t}t\sin(3t)

    Shoot...I totally forgot this way. You're way seems easier. Anyway, It was a good problem.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Instead of ODEs, why not try a PDE?

    Find the general solution to the boundary value problem for the 1-D Wave Equation:

    \frac{\partial^2y}{\partial^2t}=a^2\frac{\partial^  2y}{\partial^2x}

    y(0,t)=y(L,t)=0; \ (0<x<L, \ t>0)

    y(x,0)=f(x) \ (0<x<L)

    y_t(x,0)=g(x) \ (0<x<L)

    where y(x,t) is the displacement function of a vibrating string with fixed ends, f(x) is the initial position function, and g(x) is the initial velocity function.

    Show all steps.
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Senior Member pankaj's Avatar
    Joined
    Jul 2008
    From
    New Delhi(India)
    Posts
    317
    Quote Originally Posted by Chris L T521 View Post
    Mathstud's Limit Marathon has inspired me to start a Differential Equation Marathon. It will have similar rules : "after someone completes a DE and gets the OK from the DE's poster, he/she posts their own DE."

    There are no restrictions on how you may solve them. But please show all the steps that led you to your solution.

    Here's an easy one:

    Solve \bigg(e^x\sin(y)+\tan(y)\bigg)\,dx+\bigg(e^x\cos(y  )+x\sec^2(y)\bigg)\,dy=0
    I think solution is quite simple if we invoke concept of total derivatives
    siny(e^x)(dx) +(e^x)(cosydy)+ (tany)(dx) + (x)(sec^2(y))dy
    d((e^x)siny) + d(xtanx) = 0
    On integrating,we get
    (e^x)siny + xtanx = c
    Follow Math Help Forum on Facebook and Google+

  9. #24
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Chris L T521 View Post
    Instead of ODEs, why not try a PDE?

    Find the general solution to the boundary value problem for the 1-D Wave Equation:

    \frac{\partial^2y}{\partial^2t}=a^2\frac{\partial^  2y}{\partial^2x}

    y(0,t)=y(L,t)=0; \ (0<x<L, \ t>0)

    y(x,0)=f(x) \ (0<x<L)

    y_t(x,0)=g(x) \ (0<x<L)

    where y(x,t) is the displacement function of a vibrating string with fixed ends, f(x) is the initial position function, and g(x) is the initial velocity function.

    Show all steps.
    The best way to solve this would be to set up two boundary value problems [where each one has one nonhomogeneous condition], and then solve the problems using the technique of Separation of Variables:

    \begin{array}{cc}\text{Problem A} & \text{Problem B}\\<br /> <br />
\begin{aligned} y_{tt}&=a^2y_{xx}\\<br />
y(0,t)&=y(L,t)=0 \\<br />
y(x,0)&=f(x)\\<br />
y_t(x,0)&=0\end{aligned} & \begin{aligned} y_{tt}&=a^2y_{xx}\\<br />
y(0,t)&=y(L,t)=0 \\<br />
y(x,0)&=0\\<br />
y_t(x,0)&=g(x)\end{aligned}\end{array}

    Problem A:

    If we let y(x,t)=X(x)T(t), then the PDE becomes:

    XT''=a^2X''T

    Now separate the variables:

    \frac{X''}{X}=\frac{T''}{a^2T}

    The two functions \frac{X''}{X} and \frac{T''}{a^2T} agree \forall x,t if they both are equal to the same constant. We now let :

    \frac{X''}{X}=\frac{T''}{a^2T}=-\lambda

    What we now have is an eigenvalue problem. We can now solve two ordinary DEs:

    \begin{aligned} X''+\lambda X&=0 \\ T''+\lambda a^2T&=0\end{aligned}

    Since the endpoint conditions y(0,t)=X(0)T(t)=0 and y(L,t)=X(L)T(t)=0 require that X(0)=X(L)=0 if T(t) is non-trivial. Thus, X(x) must satisfy the eigenvalue problem

    X''+\lambda X=0, ~~~X(0)=X(L)=0

    If \lambda=0, then the eivenvalue problem becomes:

    X''=0

    When solved, the general solution is X=Ax+B. When we apply the endpoint conditions, we have A=B=0. This gives us the trivial solution X(x)\equiv 0. Thus, \lambda=0 is NOT an eigenvalue.

    If \lambda<0, we will let \lambda=-\alpha^2 \ (a>0). The eigenvalue problem now becomes:

    X''-\alpha^2X=0

    When solved, the general solution is:

    X=c_1e^{\alpha x}+c_2e^{-\alpha x}=A\cosh(\alpha x)+B\sinh(\alpha x); ~ (A=c_1+c_2, ~ B=c_1-c_2)

    where

    \begin{aligned} \cosh(\alpha x)&=\frac{1}{2}\left(e^{\alpha x}+e^{-\alpha x}\right) \\ \sinh(\alpha x)&=\frac{1}{2}\left(e^{\alpha x}-e^{-\alpha x}\right)\end{aligned}

    Applying the endpoint condition X(0)=0, we have X(0)=A\cosh(0)+B\sinh(0)=A=0 so that X(x)=B\sinh(\alpha x). When we apply the other endpoint X(L)=0, we have X(L)=B\sinh(\alpha x)=0. Thus, B=0 since \alpha \neq 0 and \sinh(\alpha x)=0 for x=0. Thus the only solution is the trivial solution X(x)\equiv 0. As a result, there are no negative eigenvalues.

    Now, when \lambda>0, we let \lambda=\alpha^2,~(\alpha>0). The eigenvalue problem then becomes:

    X''+\alpha^2X=0

    The general solution has the form:

    X=A\cos(\alpha x)+B\sin(\alpha x)

    Applying the endpoint condition X(0)=0, we have X(0)=A\cos(0)+B\sin(0)=A=0 so that X(x)=B\sin(\alpha x). But if we apply the other condition X(L)=0, we have X(L)=B\sin(\alpha L). This can occur when B\neq 0 when \alpha L is a postive multiple of \pi:

    \pi,~2\pi,~3\pi,...,~n\pi~~~~\left(n\in\mathbb{Z}^  +\right)

    Therefore

    \alpha L=\pi,~2\pi,~3\pi,...,~n\pi

    Solving for \alpha, we get:

    \alpha=\frac{\pi}{L},~\frac{2\pi}{L},~\frac{3\pi}{  L},...,~\frac{n\pi}{L}

    Since \lambda=\alpha^2, we can say that:

    \lambda=\frac{\pi^2}{L^2},~\frac{4\pi^2}{L^2},~\fr  ac{9\pi^2}{L^2},...,~\frac{n^2\pi^2}{L^2}

    We now can see that we have an infinte sequence of eigenvalues defined by:

    \lambda_n=\frac{n^2\pi^2}{L^2}~~~~\left(n\in\mathb  b{Z}^+\right)

    The associated eigenfunction is

    X_n=\sin\left(\frac{n\pi}{L}\right),~~~~\left(n\in  \mathbb{Z}^+\right)

    -------
    ***

    The homogeneous initial condition

    y_t(x,0)=X(x)T'(0)=0

    Implies that T'(0)=0. Thus, the solution T_n(t) that is associated with the eigenvalue \lambda_n=\frac{n^2\pi^2}{L^2} must also satisfy the conditions

    T_n\!''+\frac{n^2\pi^2a^2}{L^2}T_n=0,~~~T_n\!'(0)=  0

    After some calculations [which I have omitted], we get the general solution:

    T_n(t)=A_n\cos\left(\frac{n\pi at}{L}\right)+B_n\sin\left(\frac{n\pi at}{L}\right)

    To apply the condition, we need to find T_n\!'(t):

    \begin{aligned}T_n\!'(t)&=-A_n\frac{n\pi a}{L}\sin\left(\frac{n\pi at}{L}\right)+B_n\frac{n\pi a}{L}\cos\left(\frac{n\pi at}{L}\right) \\ &=\frac{n\pi a}{L}\bigg[-A_n\sin\left(\frac{n\pi at}{L}\right)+B_n\cos\left(\frac{n\pi at}{L}\right)\bigg]\end{aligned}

    Applying the condition T'(0)=0, we get:

    T_n\!'(0)=\frac{n\pi a}{L}\bigg[-A_n\sin(0)+B_n\cos(0)\bigg]=B_n\frac{n\pi a}{L}=0

    Therefore, B_n=0. Thus, T_n(t) can be defined as the non-trivial solution

    T_n(t)=\cos\left(\frac{n\pi a}{L}\right)

    Since y(x,t)=X(x)T(x), we can say that

    y_n(x,t)=X_n(x)T_n(t)=\sin\bigg(\frac{n\pi x}{L}\bigg)\cos\bigg(\frac{n\pi at}{L}\bigg),~~~~n\in\mathbb{Z}^+

    Each of these terms satisfy the wave equation \frac{\partial^2y}{\partial t^2}=a^2\frac{\partial^2y}{\partial x^2} and the homogeneous boundary condition given at the start of this part of the problem. Thus, by the principle of superposition, we define y_n(x,t) as the inifinte series:

    y_n(x,t)=\sum_{n=1}^{\infty}A_n\sin\bigg(\frac{n\p  i x}{L}\bigg)\cos\bigg(\frac{n\pi at}{L}\bigg)

    All we need to do now is find \left\{A_n\right\} such that it satisfies the nonhomogeneous condition

    y(x,0)=\sum_{n=1}^{\infty}A_n\sin\bigg(\frac{n\pi x}{L}\bigg)=f(x)

    for 0<x<L. However, this will be the Fourier Sine Series of f(x) if we choose

    A_n=\frac{2}{L}\int_0^L f(x)\sin\bigg(\frac{n\pi x}{L}\bigg)\,dx

    Therefore, the Formal Series Solution to Problem A is

    y(x,t)=\sum_{n=1}^{\infty}A_n\sin\bigg(\frac{n\pi x}{L}\bigg)\cos\bigg(\frac{n\pi at}{L}\bigg)

    where

    A_n=\frac{2}{L}\int_0^L f(x)\sin\bigg(\frac{n\pi x}{L}\bigg)\,dx

    ----------------------------------------------

    Problem B

    Let us start from ***:

    The homogeneous intial condition

    y(x,0)=X(x)T(0)=0

    implies that T(0)=0. Thus, the solution T_n(t) associated with the eigenvalue \lambda_n=\frac{n^2\pi^2}{L^2} must also satisfy the conditions

    T_n\!''+\frac{n^2\pi^2a^2t^2}{L^2}T_n=0,~~~T_n(0)=  0

    After some calculations [which I have omitted again...], the general solution takes the form

    T_n(t)=A_n\cos\bigg(\frac{n\pi at}{L}\bigg)+B_n\sin\bigg(\frac{n\pi at}{L}\bigg)

    Applying the initial condition, we get

    T_n=A_n=0

    Thus, T_n(t) can be defined as the non-trivial solution

    T_n(t)=\sin\bigg(\frac{n\pi at}{L}\bigg)

    Since y(x,t)=X(x)T(t), we can say that

    y_n(x,t)=X(x)T(t)=\sin\bigg(\frac{n\pi at}{L}\bigg)\sin\bigg(\frac{n\pi x}{L}\bigg)

    Each of these terms satisfy the wave equation \frac{\partial^2y}{\partial t^2}=a^2\frac{\partial^2y}{\partial x^2} and the homogeneous boundary condition given at the start of this part of the problem. Thus, by the principle of superposition, we define y_n(x,t) as the inifinte series:

    y_n(x,t)=\sum_{n=1}^{\infty}B_n\sin\bigg(\frac{n\p  i x}{L}\bigg)\sin\bigg(\frac{n\pi at}{L}\bigg)

    Find y_t(x,y):

    y_t(x,t)=\sum_{n=1}^{\infty}B_n\frac{n\pi a}{L}\cos\bigg(\frac{n\pi x}{L}\bigg)\sin\bigg(\frac{n\pi at}{L}\bigg)

    All we need to do now is find \left\{B_n\right\} such that it satisfies the nonhomogeneous condition

    y_t(x,0)=\sum_{n=1}^{\infty}B_n\frac{n\pi a}{L}\cos\bigg(\frac{n\pi x}{L}\bigg)\sin\bigg(\frac{n\pi at}{L}\bigg)=g(x)

    for 0<x<L. However, this will be the Fourier Sine Series of g(x) if we choose

    B_n\frac{n\pi a}{L}=\frac{2}{L}\int_0^L g(x)\sin\bigg(\frac{n\pi x}{L}\bigg)\,dx\implies B_n=\frac{2}{n\pi a}\int_0^L g(x)\sin\bigg(\frac{n\pi x}{L}\bigg)\,dx

    Therefore, the Formal Series Solution to Problem B is

    y(x,t)=\sum_{n=1}^{\infty}B_n\sin\bigg(\frac{n\pi x}{L}\bigg)\sin\bigg(\frac{n\pi at}{L}\bigg)

    where

    B_n=\frac{2}{n\pi a}\int_0^L g(x)\sin\bigg(\frac{n\pi x}{L}\bigg)\,dx

    -------------

    Thus, the Formal Series Solution to the One Dimensional Wave Equation is

    \color{red}\boxed{\begin{aligned}y(x,t)&=\sum_{n=1  }^{\infty}\bigg[A_n\cos\bigg(\frac{n\pi a}{L}t\bigg)+B_n\sin\bigg(\frac{n\pi a}{L}t\bigg)\bigg]\sin\bigg(\frac{n\pi}{L}x\bigg)\\ \text{where} \\<br />
A_n&=\frac{2}{L}\int_0^L f(x)\sin\bigg(\frac{n\pi x}{L}\bigg)\,dx \\<br />
B_n&=\frac{2}{n\pi a}\int_0^L g(x)\sin\bigg(\frac{n\pi x}{L}\bigg)\,dx\end{aligned}}



    That was a lot of work...

    --Chris
    Last edited by Chris L T521; January 3rd 2009 at 06:25 AM.
    Follow Math Help Forum on Facebook and Google+

  10. #25
    Member
    Joined
    Jul 2008
    From
    Sofia, Bulgaria
    Posts
    75
    Quote Originally Posted by TheEmptySet View Post
    Okay this is similiar to the one above. I think this is a cool problem

    \frac{d^2x}{dt^2}+0.4\frac{dx}{dt}+9.04x=6e^{-\frac{t}{5}}\cos(3t), \\\ x(0)=x'(0)=0

    here's my suggestion - easy and simple

    x''+0,4x'+9,4x = 3e^(-t/5)cos(3t)

    ok, x_h is easy so I'll omit it

    for x_p, we have to go from R to C and I'll use the differential operator
    D = d/dt

    x''+0,4x'+9,4x = 3e^(-t/5)*е^(3it)

    (D^2 + 0,4D + 9,4)x = 3e^([3i-1/5]t) and let p(D) be the char. polynomial

    p(D)x = 3e^([3i-1/5]t)

    so, going back to R, a particular solution is:

    x_p = Re{ 3([3i-1/5]t)/p(3i-1/5) } , where p(3i-1/5) is not 0, atherwise it's: x_p = Re{ 3t([3i-1/5]t)/p'(3i-1/5) }

    then the well-known procedure - finding the general solution and the 2 constants
    Follow Math Help Forum on Facebook and Google+

  11. #26
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Chris L T521 View Post
    That was a lot of work...
    Another equation is here.

    Have you ever learned Sturm-Liouville Theory? (It is about eigenvalues and eigenfunctions of a boundary value problem).
    Follow Math Help Forum on Facebook and Google+

  12. #27
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by ThePerfectHacker View Post
    Another equation is here.

    Have you ever learned Sturm-Liouville Theory? (It is about eigenvalues and eigenfunctions of a boundary value problem).
    No I haven't. I plan on learning it on my own when I find the time. If I don't learn it, I'm taking a class in the spring on the last part of Diff Equ [Laplace Trans, Fourier Series and Transforms, Power Series, Sturm-Liouville and other topics]...

    If I did know how to apply Sturm-Liouville Theory, by how much would the amount of work needed to solve this equation reduce?

    --Chris
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Secondary school's competition - marathon.
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: August 30th 2010, 04:45 AM
  2. Small limit marathon
    Posted in the Calculus Forum
    Replies: 40
    Last Post: July 20th 2008, 09:40 PM
  3. Limit Marathon
    Posted in the Calculus Forum
    Replies: 104
    Last Post: June 15th 2008, 11:47 AM
  4. Geometry Marathon
    Posted in the Geometry Forum
    Replies: 49
    Last Post: July 16th 2007, 01:44 PM

Search Tags


/mathhelpforum @mathhelpforum