Curvilinear Motion of a particle

[Hootenanny]

Oh man...

Is it $$v(t)=\frac{d\vec{r}}{dt}=[\frac{dx}{dt}\hat i+2x\frac{dx}{dt}\hat j]\frac{m}{s}$$?

Correct! And can you now find the acceleration...?

Aside:

It is common practice in physics and applied mathematics to use Newton's notation to make things a little easier to read;

$$\frac{dx}{dt} = \dot{x}\hspace{1cm};\hspace{1cm}\frac{d^2x}{dt^2} = \ddot{x}(t)$$

So your velocity would become;

$$\underline{v}(t) = \underline{\dot{r}}(t)= \dot{x}(t)\hat{i}+ 2x\cdot\dot{x}(t)\hat{j}$$

But either notation is equally correct, so just use whichever you feel most comfortable with.
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[Saladsamurai]

I am thinking product rule is needed for the j component?$$\vec{a}(t)=\frac{d\vec{v}}{dt}=\frac{d^2x}{dt}\hat{i}+[2x*\frac{d^2x}{dt}+2*\frac{dx}{dt}]\hat{j}$$

Or am I way off the mark now?

 

[Hootenanny]

I am thinking product rule is needed for the j component?$$\vec{a}(t)=\frac{d\vec{v}}{dt}=\frac{d^2x}{dt}\hat{i}+[2x*\frac{d^2x}{dt}+2*\frac{dx}{dt}]\hat{j}$$

Or am I way off the mark now?

Nope, your spot on again ! So, now you need to find the maximum of this function...

 

[Saladsamurai]

Nope, your spot on again ! So, now you need to find the maximum of this function...

Please tell me that requires further differentiation. . . Wouldn't I now have to find the 1st derivative of this function, set it to zero and solve; then use the 2nd derivative to find concave up/down?

I am going fruit loops here! Thanks Hootenanny from trudging through this with me

 

[Hootenanny]

Please tell me that requires further differentiation. . . Wouldn't I now have to find the 1st derivative of this function, set it to zero and solve; then use the 2nd derivative to find concave up/down?

Indeed you would, but let's just start with the first derivative for now (and don't forget you have a condition on v)

I am going fruit loops here! Thanks Hootenanny from trudging through this with me

No problem, it's a pleasure

 

[Saladsamurai]

Before I go any further here, I have a question. I still do not have x(t). How am I going to cope with that? I have only differentiated the general function of time. I still don't have an explicit function x(t)=something. So where is this all going?

 

[Hootenanny]

As promised I'm back again and after some addition thought (on the train this morning), I offer a few more hints. You have correctly found the acceleration vector;

$$\underline{a}(t) = \ddot{x}(t)\underline{i} + \left[2x\cdot\ddot{x}(t) + 2\dot{x}(t)\right]\underline{j}$$

And you are asked to maximise the magnitude of the acceleration vector;

$$a(t) = \sqrt{\left[\ddot{x}(t)\right]^2 + \left[2x\cdot\ddot{x}(t) + 2\dot{x}(t)\right]^2}$$

However, you are asked to find the point on the curve at which maximum acceleration occurs. Therefore, you must maximise a(t) with respect to x(t). Hence, you should solve;

$$\frac{d}{dx}a(t) = \frac{d}{dx}\sqrt{\left[\ddot{x}(t)\right]^2 + \left[2x\cdot\ddot{x}(t) + 2\dot{x}(t)\right]^2} := 0$$

I would also be useful to note that the point at which a(t) is maximal coincides with the point at which the function under the root is maximal (although the value of the maximum may be different). Do you see why? Hence, you can find the point on the curve at which the magnitude of the acceleration is maximal by solving;

$$\frac{d}{dx}a^2(t) = \frac{d}{dx}\left\{\left[\ddot{x}(t)\right]^2 + \left[2x\cdot\ddot{x}(t) + 2\dot{x}(t)\right]^2\right\} := 0$$