Deriving decomposition of transverse acceleration

[philipfisher]

Homework Statement

The question asks one to derive the acceleration vector,

$$\vec{a} = [\frac{d^2r}{dt^2} - r(\frac{d\Theta}{dt})^2]\vec{u}_r + [\frac{1}{r}\frac{d}{dt}(r^2\frac{d\Theta }{dt})]\vec{u}_\Theta$$

from the velocity vector.

$$\vec{v} = \vec{u}_r\frac{dr}{dt} + \vec{u}_\Theta \frac{d\Theta }{dt}$$

Homework Equations

See above.

The Attempt at a Solution

I have had no problem deriving this version of the acceleration vector,

$$\vec{a} = [\frac{d^2r}{dt^2} - r(\frac{d\Theta}{dt})^2]\vec{u}_r + [2\frac{dr}{dt}\frac{d\Theta }{dt}+r\frac{d^2\Theta }{dt^2}]\vec{u}_\Theta$$

from the velocity vector. But I'm stumped on how one simplifies my version of the transverse component of acceleration,

$$2\frac{dr}{dt}\frac{d\Theta }{dt}+r\frac{d^2\Theta }{dt^2}$$

into the textbook's version.

$$\frac{1}{r}\frac{d}{dt}(r^2\frac{d\Theta }{dt})$$

Any chance someone could give me a hint? I've googled for an explanation without any success. I've also looked in two different textbooks, both of which just state that,

$$2\frac{dr}{dt}\frac{d\Theta }{dt}+r\frac{d^2\Theta }{dt^2} = \frac{1}{r}\frac{d}{dt}(r^2\frac{d\Theta }{dt})$$

without explaining why. I'm guessing the process of simplification is so basic that the texts just assume it needs no explaining, but the simplification eludes me and I'd really appreciate whatever help anyone could provide. Thanks.

 

[ap123]

$$\frac{1}{r}\frac{d}{dt}(r^2\frac{d\Theta }{dt})$$

You can just use the standard product rule here (combined with the chain rule), remembering that r is a function of t.

 

[cepheid]

Welcome to PF,

It's a case where, if you work backwards, you'll see that this relation is true:$$\frac{1}{r}\frac{d}{dt}(r^2\frac{d\Theta }{dt})$$

Differentiate using the product rule:$$= \frac{1}{r}\frac{d\Theta }{dt}\frac{d}{dt}(r^2) + \frac{1}{r}(r^2)\frac{d}{dt}\left(\frac{d\Theta }{dt}\right)$$ $$=\frac{1}{r}2r\frac{dr}{dt}\frac{d\Theta}{dt} + r\frac{d^2\Theta}{dt^2}$$

In the last step, in the leftmost term, r² was differentiated using the chain rule

Edit: eventually you'll get adept at "reverse" differentiating things that are exact derivatives in your head, and something like this will be clear by inspection.

 

[philipfisher]

Thank you. I'd tried the process of working backward by differentiating the textbook answer, but completely forgot the need to use the chain rule when doing so. I really appreciate the help.

 

[Nickie]

Hello,

Thank you for your question. The simplification that you are looking for involves using the product rule for differentiation. Let's start with your version of the transverse component of acceleration:

2\frac{dr}{dt}\frac{d\Theta }{dt}+r\frac{d^2\Theta }{dt^2}

We can rewrite this as follows:

2\frac{d}{dt}(r\frac{d\Theta }{dt})+r\frac{d^2\Theta }{dt^2}

Now, using the product rule for differentiation, we can write:

2(\frac{d}{dt}r)\frac{d\Theta }{dt}+2r\frac{d^2\Theta }{dt^2}+r\frac{d^2\Theta }{dt^2}

Simplifying this, we get:

2(\frac{dr}{dt}\frac{d\Theta }{dt}+r\frac{d^2\Theta }{dt^2})+r\frac{d^2\Theta }{dt^2}

Now, notice that the term in parentheses is the same as the one we started with. So we can replace it with the simplified version:

2(\frac{1}{r}\frac{d}{dt}(r^2\frac{d\Theta }{dt}))+r\frac{d^2\Theta }{dt^2}

And finally, using the definition of the transverse component of acceleration, we can write:

\frac{1}{r}\frac{d}{dt}(r^2\frac{d\Theta }{dt})+r\frac{d^2\Theta }{dt^2}

I hope this helps! Please let me know if you have any further questions or need clarification.