# Applications of Calculus

• May 30th 2008, 09:39 PM
Gusbob
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 $\displaystyle 15ms^{-1}$, find its velocity after 3s.

What I've done:

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

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

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 $\displaystyle \frac{dx}{dt} = \sqrt{3x^2 + 2x + 225} \Rightarrow \frac{dt}{dx} = \frac{1}{\sqrt{3x^2 + 2x + 225}}$ but it doesn't work well.
• May 30th 2008, 09:43 PM
Mathstud28
Quote:

Originally Posted by Gusbob
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 $\displaystyle 15ms^{-1}$, find its velocity after 3s.

What I've done:

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

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

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 $\displaystyle \frac{dx}{dt} = \sqrt{3x^2 + 2x + 225} \Rightarrow \frac{dt}{dx} = \frac{1}{\sqrt{3x^2 + 2x + 225}}$ but it doesn't work well.

I think you are overcomplicating things.

Consider

$\displaystyle v(t)=\int{a(t)dt}+v_0$
• May 30th 2008, 09:52 PM
Gusbob
Thank you. I've tried that before, but didn't derive the vē equation first. But now that I have it, it works (Clapping)
• May 30th 2008, 09:55 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
I think you are overcomplicating things.

Consider

$\displaystyle v(t)=\int{a(t)dt}+v_0$

Quote:

Originally Posted by Gusbob
Thank you. I've tried that before, but didn't derive the vē equation first. But now that I have it, it works (Clapping)

Just to make sure I wasnt ambiguous $\displaystyle v_0$ is the constant of integration with the integration of a(t), so dont add an extra c, the c is $\displaystyle v_0$
• May 30th 2008, 10:57 PM
CaptainBlack
Quote:

Originally Posted by Gusbob
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 $\displaystyle 15ms^{-1}$, find its velocity after 3s.

What I've done:

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

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

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 $\displaystyle \frac{dx}{dt} = \sqrt{3x^2 + 2x + 225} \Rightarrow \frac{dt}{dx} = \frac{1}{\sqrt{3x^2 + 2x + 225}}$ but it doesn't work well.

$\displaystyle a=\frac{dv}{dt}=3x-1$

so differentiate this again:

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

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

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

and so:

$\displaystyle x(t)=\frac{A}{\sqrt{3}}e^{\sqrt{3}t}-\frac{B}{\sqrt{3}}e^{-\sqrt{3}t}+C$

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

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

RonL
• May 30th 2008, 11:05 PM
CaptainBlack
Quote:

Originally Posted by Mathstud28
I think you are overcomplicating things.

Consider

$\displaystyle v(t)=\int{a(t)dt}+v_0$

What you mean is:

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

and how do you proceed from this point?

RonL
• May 30th 2008, 11:39 PM
Gusbob
Quote:

Originally Posted by CaptainBlack
$\displaystyle a=\frac{dv}{dt}=3x-1$

so differentiate this again:

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

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

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

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)
• May 30th 2008, 11:44 PM
Chris L T521
Quote:

Originally Posted by Gusbob
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. :D
• May 31st 2008, 01:05 AM
mr fantastic
Quote:

Originally Posted by Gusbob
[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 $\displaystyle a = \frac{d}{dx} \left( \frac{v^2}{2} \right)$ or $\displaystyle v \frac{dv}{dx}$ 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.
• May 31st 2008, 02:32 AM
CaptainBlack
Quote:

Originally Posted by Gusbob
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