How to partially diffenentiate a derivative

[JohnFromAus]

I am stuck with the following which comes from an attempt to find acceleration in polar coordinates. The velocity in the x direction is

[math] V = cos\phi dr/dt - sin\phi (r d\phi/dt)

[/math]

and I need to find dV/dt

 

so I think I need [math]\delta V/\delta r \ and\ \delta V/\delta \phi

[/math]

Then I have terms like [math] \delta (dr / dt) \delta r[/math] which I dont know how to handle.

 

I have a book which gives the same formula for V and then the answer on the next line - I cannot "join the dots"!

 

Have spent a bit of time looking on the internet but only found a different answer to the one in my book - could not get that either.

 

Any help appreciated - either hints, references - or the missing steps!

John

PS its some 40+ years since I was really on top of my calculus so I could be completely screwed up!

OK it should be PARTIALLY

[D H]

No partial derivatives are needed here.

 

It might help to understand where that velocity expression comes about in the first place. The x coordinate of the position vector is [math]x=r\cos\phi[/math]. Differentiating with respect to time using the chain rule and product rule, [math]\frac{dx}{dt} = \frac {dr}{dt}\,\cos \phi - \frac{d\phi}{dt}\,r\sin\phi[/math]. Differentiate again to yield the acceleration in the x direction:

 

[math]\frac{d^2x}{dt^2} =

\frac {d^2r}{dt^2} \, \cos \phi

- \frac{d^2\phi}{dt^2}\,r\sin\phi

- \frac {d\phi}{dt}\left(2 \frac {dr}{dt} \sin \phi

+ \frac{d\phi}{dt}\,r\cos\phi\right)[/math]

[JohnFromAus]

DH

Im still confused.

Surley the [math] cos\phi \ and \ the \ -rsin\phi [/math] in the expression for dx/dt are just the partial derivatives of [math]x=cos\phi[/math]

 

However can now see where answer comes from - must go back to my book and check up on all this - particularly the use of "[math]\delta[/math]" and "d" which seem to be used pretty much as the same thing in some web sites.

Thanks

John

[D H]

Im still confused.

Surley the [math] cos\phi \ and \ the \ -rsin\phi [/math] in the expression for dx/dt are just the partial derivatives of [math]x=cos\phi[/math]

 

The objective is to calculate the total derivative of the velocity wrt time. One doesn't need partial derivatives to do this. It can be done with simple freshman calculus. Here is a bit more verbose computation of the time derivative of [math]x=r\cos\phi[/math]:

 

[math]

\begin{aligned}

\frac{dx}{dt} &= \frac d{dt}(r\cos\phi) & \text{from}\ x=r\cos\phi \\

&= \frac {dr}{dt}\,\cos\phi + r\frac d{dt}(\cos\phi) & \text{product rule} \\

&= \frac {dr}{dt}\,\cos\phi - r\,\sin\phi \frac {d\phi}{dt} & \text{chain rule}

\end{aligned}

[/math]

 

The second derivative calculation is just a bit more messy, but can be done directly. Simply differentiate the left hand side of [math]\dot x = \dot r \cos \phi - r\sin\phi \dot \phi[/math]. Note that I changed to dot notation for time derivatives in anticipation of the development that follows.

 

You can use partial derivatives to attack this problem. In general, if [math]u=f(x_1,x_2,\cdots,x_n,t)[/math],

[math]\frac {du}{dt} =

\frac{\partial f}{\partial x_1}\,\frac{dx_1}{dt} +

\frac{\partial f}{\partial x_2}\,\frac{dx_2}{dt} + \cdots +

\frac{\partial f}{\partial x_n}\,\frac{dx_n}{dt} +

\frac{\partial f}{\partial t}[/math]

We want to find the total time derivative of [math]v=\dot r \cos \phi - r\sin\phi \dot \phi[/math]. Thus our function [math]f[/math] involves four variables: [math]r[/math], [math]\phi[/math], [math]\dot r[/math], and [math]\dot \phi[/math]. The partial derivatives are

[math]

\begin{aligned}

\frac{\partial v}{\partial r} &= -\sin\phi\dot\phi \\[-1pt]

\frac{\partial v}{\partial \phi} &= -(\dot r \sin \phi + r\cos\phi \dot \phi) \\[-1pt]

\frac{\partial v}{\partial \dot r} &= \cos \phi \\[-1pt]

\frac{\partial v}{\partial \dot \phi} &= -r\sin\phi \\[-1pt]

\frac{\partial v}{\partial t} &= 0

\end{aligned}

[/math]

 

Applying the above rule for computing the total derivative,

[math]\dot v= -\sin\phi \dot \phi \dot r - (\dot r \sin \phi + r\cos\phi \dot \phi)\dot \phi + \cos \phi \ddot r - r\sin\phi \ddot \phi[/math]

[JohnFromAus]

Thanks D H that clears up my problems - for now anyway!

John

 

Just for your info it was when you pointed out that r and dr/dt could be treated as separate variables that the penny dropped!

John