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Math Help - Applications of Calculus

  1. #1
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    Applications of Calculus

    Question:
    The acceleration of a particle is given by a = 3x+1 (where a is acceleration, x is displacement from origin). If the particle is initially at the origin and moving at a velocity of  15ms^{-1}, find its velocity after 3s.

    What I've done:

     a = v\frac{dv}{dx} = 3x -1
    \frac{1}{2} v^2 = \frac{3}{2}x^2 + x + c

    when t=0, x=0, v = 15
     v^2 = 3x^2 + 2x + 225

    I'm not sure where to go on from here. I know if I find t in terms of x, I can substitute it into that equation. I've tried  \frac{dx}{dt} = \sqrt{3x^2 + 2x + 225} \Rightarrow \frac{dt}{dx} = \frac{1}{\sqrt{3x^2 + 2x + 225}} but it doesn't work well.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Gusbob View Post
    Question:
    The acceleration of a particle is given by a = 3x+1 (where a is acceleration, x is displacement from origin). If the particle is initially at the origin and moving at a velocity of  15ms^{-1}, find its velocity after 3s.

    What I've done:

     a = v\frac{dv}{dx} = 3x -1
    \frac{1}{2} v^2 = \frac{3}{2}x^2 + x + c

    when t=0, x=0, v = 15
     v^2 = 3x^2 + 2x + 225

    I'm not sure where to go on from here. I know if I find t in terms of x, I can substitute it into that equation. I've tried  \frac{dx}{dt} = \sqrt{3x^2 + 2x + 225} \Rightarrow \frac{dt}{dx} = \frac{1}{\sqrt{3x^2 + 2x + 225}} but it doesn't work well.
    I think you are overcomplicating things.

    Consider

    v(t)=\int{a(t)dt}+v_0
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    Thank you. I've tried that before, but didn't derive the vē equation first. But now that I have it, it works
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    I think you are overcomplicating things.

    Consider

    v(t)=\int{a(t)dt}+v_0
    Quote Originally Posted by Gusbob View Post
    Thank you. I've tried that before, but didn't derive the vē equation first. But now that I have it, it works
    Just to make sure I wasnt ambiguous v_0 is the constant of integration with the integration of a(t), so dont add an extra c, the c is v_0
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    Quote Originally Posted by Gusbob View Post
    Question:
    The acceleration of a particle is given by a = 3x+1 (where a is acceleration, x is displacement from origin). If the particle is initially at the origin and moving at a velocity of  15ms^{-1}, find its velocity after 3s.

    What I've done:

     a = v\frac{dv}{dx} = 3x -1
    \frac{1}{2} v^2 = \frac{3}{2}x^2 + x + c

    when t=0, x=0, v = 15
     v^2 = 3x^2 + 2x + 225

    I'm not sure where to go on from here. I know if I find t in terms of x, I can substitute it into that equation. I've tried  \frac{dx}{dt} = \sqrt{3x^2 + 2x + 225} \Rightarrow \frac{dt}{dx} = \frac{1}{\sqrt{3x^2 + 2x + 225}} but it doesn't work well.
    a=\frac{dv}{dt}=3x-1

    so differentiate this again:

    \frac{d^2v}{dt^2}=3v

    which is a constant coefficient homogeneous ODE, which you should be able to solve, to get:

    v(t)=Ae^{\sqrt{3}t}+Be^{-\sqrt{3}t}

    and so:

     <br />
x(t)=\frac{A}{\sqrt{3}}e^{\sqrt{3}t}-\frac{B}{\sqrt{3}}e^{-\sqrt{3}t}+C<br />

    Now putting both of these back into the equation \frac{dv}{dt}=3x-1 shows that C=0.

    Now apply the initial condition to find A and B and then its simple.

    RonL
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    Quote Originally Posted by Mathstud28 View Post
    I think you are overcomplicating things.

    Consider

    v(t)=\int{a(t)dt}+v_0
    What you mean is:

    v(t)=\int_0^t {a(\tau)~d\tau}+v_0

    and how do you proceed from this point?

    RonL
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    a=\frac{dv}{dt}=3x-1

    so differentiate this again:

    \frac{d^2v}{dt^2}=3v

    which is a constant coefficient homogeneous ODE, which you should be able to solve, to get:

    v(t)=Ae^{\sqrt{3}t}+Be^{-\sqrt{3}t}

    Thank you for that, but I have no idea what "constant coefficient homogeneous ODE" is.

    And for Mathstud28's method, I integrated a = 3x + 1 in respect to t, treating 3x as a constant. I then substituted t = 3 to get the a relation between v and x at that time. Then solved simultaneously with the equation I derived. Is that a valid method? or is it incredible coincidence that I got the right answer to 1 dp. (16.1)
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Gusbob View Post
    Thank you for that, but I have no idea what "constant coefficient homogeneous ODE" means.
    I explain this in my Differential Equations Tutorial. Read the section on Undetermined coefficients (Post #6), and see if this helps out.
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  9. #9
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    Quote Originally Posted by Gusbob View Post
    [snip]

    And for Mathstud28's method, I integrated a = 3x + 1 in respect to t, treating 3x as a constant. [snip]
    Sorry, but that's completely wrong. x is the displacement of the object. I assume that the object is actually moving and so x is changing. Therefore .......

    x is NOT a constant and CANNOT be treated as one.

    This is not the only error. Correct me if I'm wrong but doesn't the original question say a = 3x + 1? Why then do you say a = 3x - 1 in your very first line of working.

    No offence, but this whole thread is a comedy of errors as a consequence of post #2 (and to a lesser extent the confusion of what a is actually meant to be).

    CaptainB's approach is correct and efficient. But you don't have background to use this approach (not your fault). So:

    The bottom line is that you must first use one of the expressions a = \frac{d}{dx} \left( \frac{v^2}{2} \right) or v \frac{dv}{dx} to get v = v(x). You did this but then lost your way.

    You then have to do the following:

    dx/dt = v(x) => dt/dx = 1/v(x).

    Integrate to get x = x(t).

    dx/dt gives you v as a function of time.

    Substitute t = 3 and you're done.
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by Gusbob View Post
    Thank you for that, but I have no idea what "constant coefficient homogeneous ODE" is.

    And for Mathstud28's method, I integrated a = 3x + 1 in respect to t, treating 3x as a constant. I then substituted t = 3 to get the a relation between v and x at that time. Then solved simultaneously with the equation I derived. Is that a valid method? or is it incredible coincidence that I got the right answer to 1 dp. (16.1)
    x is not a constant so no its not valid.

    RonL
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