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**christelleO****Guest**

Hi,

Please help, I don`t see how the following is derived:

We know dV/dt = a from the definition of acceleration

Multiplying by sides by v (velocity) yields v dv/dt = av

Left side can be reduced to d/dt(V^2/2)

My question is: How does v dv/dt become d/dt( V^2/2)?

Can someone show me all intermediary steps and what rules are at play and why they are used.

Any help will be appreciated

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,408

hi christelleO,

Welcome to the forum.

I'll show how the right hand side becomes the left. It amounts to the same thing.

If you has t^2 /2 and you wanted to differentiate with respect to t, that would be straight forward. You would get 2t / 2 = t

But you want to differentiate v^2 /2. It hasn't got a 't' in it, although we suspect that v is a function of t.

So you have to use the chain rule:

dy/dx = dy/du . du/dx

In this problem y = v^2 /2. x = t. and u = v so the chain rule becomes

d(v^2/2)/dt = d(v^2/2)/dv . dv/dt

What this amounts to is this. Differentiate with respect to v and then times by dv/dt.

d(v^2/2)/dt = v.dv/dt

hope that helps,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**christelle****Member**- Registered: 2015-07-23
- Posts: 1

Hi Bob,

Thank you for putting me out of my misery, so quickly.

It's simple when you know how, isn't it, and now thanks to you, I do.

Much appreciated

Christelle

PS: love the Galileo quote.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,408

hi Christelle

You're welcome!

I see you have joined.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**kpstatic****Member**- Registered: 2018-10-31
- Posts: 1

I'm still having trouble seeing where (v^2/2) came from without integrating v. Also what is the purpose of multiplying v to both side from a physics standpoint. Its my first year in calculus based physics and i'm trying to understand both the calc and physics of it all. Thanks.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,408

hi kpstatic

Welcome to the forum.

In mechanics, to switch between acceleration, a, velocity, v and distance, s it is necessary to differentiate and integrate as

I seem to remember this bit of algebraic manipulation arises when dealing with simple harmonic motion, because a is defined as a function of s rather than t. So a way is needed to express a in terms that can be integrated.

If you post back more precisely where you've got to in your course and why you came searching here, I'll try to provide more help.

Bob

ps. s is used for distance as d is likely to be used for differentiating. x may also be used.

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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