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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Lycan on August 02, 2008, 02:50:45 pm

Title: Here's a question for you to try out.
Post by: Lycan on August 02, 2008, 02:50:45 pm: I just finished a question that I thought was pretty cool so I thought I might give you guys a go.

Prove mathematically that if the speed of a moving particle moving along a curve is constant, the acceleration is perpendicular to the velocity.

The topic is vector calculus if you want a clue.
Title: Here's a question for you to try out.
Post by: phagist_ on August 02, 2008, 07:01:28 pm: Constant velocity means acceleration is zero, not constant speed.
Title: Here's a question for you to try out.
Post by: Collin Li on August 02, 2008, 07:07:45 pm: Constant velocity means that the acceleration is zero and that there is constant speed. (Ohh, okay, I get what you mean now, phagist_)

On the other hand, constant speed does not mean constant velocity.
Title: Here's a question for you to try out.
Post by: Collin Li on August 02, 2008, 07:09:17 pm: Quote from: dcc on August 02, 2008, 04:07:42 pm
If a moving particle is moving with a constant speed, then the acceleration of the particle is zero.

Hence $\vec{v}.\vec{a} = \alpha \cdot \vec{0} = \vec{0}$ (where $\alpha$ is some constant), hence $\vec{v} \perp \vec{a}$

This result should hold true for any natural number of dimensions of movement.

This doesn't make sense. A dot product should give you a scalar quantity. How come you are getting the zero vector?
Title: Here's a question for you to try out.
Post by: Collin Li on August 02, 2008, 07:17:13 pm: It still remains that a scalar times a vector is a vector, so that does not seem like a correct argument to me.

I would do it by proposing the velocity vector of the form: $(x,\, y)$ ,

and then using the fact that: $C = \sqrt{x^2 + y^2}$ , where $C$ is some constant,

hence we can find $y$ in terms of $x$ , and then you can find the dot product between the velocity vector and the acceleration vector.
Title: Here's a question for you to try out.
Post by: Collin Li on August 02, 2008, 07:29:03 pm: Take: $\textbf{v} = x\,\textbf{i} + y\,\textbf{j}$

Since: $C = \sqrt{x^2 + y^2}$ , where $C$ is a constant,

$\implies y = \pm\sqrt{C^2 - x^2}$

$\implies \textbf{v} = x\,\textbf{i} \pm\sqrt{C^2 - x^2}\,\textbf{j}$

$\implies \textbf{a} = \frac{dx}{dt}\,\textbf{i} \pm \frac{d}{dt}\left(\sqrt{C^2 - x^2}\right)\,\textbf{j}$

$\implies \textbf{a} = \frac{dx}{dt}\,\textbf{i} \mp \frac{x}{\sqrt{C^2 - x^2}}\frac{dx}{dt}\,\textbf{j}$

Therefore:

$\textbf{v}\cdot\textbf{a} = x\frac{dx}{dt} - \sqrt{C^2 - x^2}\left[\frac{x}{\sqrt{C^2 - x^2}}\right]\frac{dx}{dt}$

(Note the plus/minus and minus/plus signs will always produce a negative sign)

$\tf\;\;\textbf{v}\cdot\textbf{a} = x\frac{dx}{dt} - x\frac{dx}{dt} = 0$
Title: Here's a question for you to try out.
Post by: dcc on August 02, 2008, 07:34:46 pm: Im sorry, I misread the question as velocity, rather then speed.
Title: Here's a question for you to try out.
Post by: Mao on August 02, 2008, 07:36:54 pm: Quote from: Lycan on August 02, 2008, 02:50:45 pm
I just finished a question that I thought was pretty cool so I thought I might give you guys a go.

Prove mathematically that if the speed of a moving particle moving along a curve is constant, the acceleration is perpendicular to the velocity.

The topic is vector calculus if you want a clue.

$|\vec{v}|=k$

$\vec{v}\cdot \vec{v}=|\vec{v}|^2=k^2$

$\frac{d}{dt}(\vec{v}\cdot \vec{v})=\frac{d}{dt}(k^2)$

    Assuming $\vec{v}$ is defined in $\mathbb{R}^3$

    $\vec{v}\cdot \vec{v}=x(t)\times x(t)+y(t)\times y(t)+z(t)\times z(t)$

    $\implies \frac{d}{dt}\left[x(t)\times x(t)+y(t)\times y(t)+z(t)\times z(t)\right]=x(t)\times \frac{dx}{dt}+\frac{dx}{dt}\times x(t)+y(t)\times \frac{dy}{dt}+\frac{dy}{dt}\times y(t) +z(t)\times \frac{dz}{dt}+\frac{dz}{dt}\times z(t)$ (product rule)

    $\implies \frac{d}{dt}(\vec{v}\cdot \vec{v})=\left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)\cdot \left(x(t),y(t),z(t)\right) + \left(x(t),y(t),z(t)\right) \cdot\left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)$

   (this can be shown for n-dimensions)

   $\implies \boxed{\frac{d}{dt}(\vec{v}\cdot \vec{v})=\frac{d\vec{v}}{dt}\cdot\vec{v}+\vec{v}\cdot \frac{d\vec{v}}{dt}}$ (basically, the product rule)

$\frac{d\vec{v}}{dt}\cdot\vec{v}+\vec{v}\cdot \frac{d\vec{v}}{dt}=0$

$2\vec{a}\cdot\vec{v}=0$

$\therefore \vec{a} \perp \vec{v}$ , given that neither $\vec{a}$ nor $\vec{v}$ are 0

hence, in general, where the magnitude of velocity is a constant, acceleration is perpendicular to the direction of motion

EDIT: and for dcc's sake, this holds true where velocity is defined and continuous (and in this case, its magnitude is also constant)
Title: Re: Here's a question for you to try out.
Post by: Collin Li on August 02, 2008, 07:50:44 pm: Moderator action: Unmerged by popular request.
Title: Re: Here's a question for you to try out.
Post by: enwiabe on August 02, 2008, 07:51:29 pm: coblin's request*
Title: Re: Here's a question for you to try out.
Post by: Collin Li on August 02, 2008, 07:52:25 pm: Ahmad, dcc and I versus Mao.
Title: Re: Here's a question for you to try out.
Post by: dcc on August 02, 2008, 07:55:31 pm: how about:

$\vec{v}(t) = \sum_{n = 1}^\infty \dfrac{\left(6 - 6 \cos n \pi \right) \sin \left(n \pi t \right)}{n \pi} \ \vec{i}$
Title: Re: Here's a question for you to try out.
Post by: Mao on August 02, 2008, 07:56:51 pm: Quote from: dcc on August 02, 2008, 07:55:31 pm
how about:

$\vec{v}(t) = \sum_{n = 1}^\infty \dfrac{\left(6 - 6 \cos n \pi \right) \sin \left(n \pi t \right)}{n \pi} \ \vec{i}$

is that constant magnitude?
Title: Re: Here's a question for you to try out.
Post by: dcc on August 02, 2008, 08:00:32 pm: (http://omploader.org/vbjlz/saw.jpg)

The speed is constant (and sometimes undefined).
Title: Re: Here's a question for you to try out.
Post by: Mao on August 02, 2008, 08:02:26 pm: then for intervals where the velocity is continuous (and constant), acceleration will be perpendicular to the direction of motion.
:)
Title: Re: Here's a question for you to try out.
Post by: dcc on August 02, 2008, 08:05:52 pm: Im still not convinced that saying that $\textbf{a}.\textbf{v} = 0 \implies \textbf{a} \perp \textbf{v}$

What about $\textbf{v} = \textbf{a} = \textbf{0}$ , does that mean $\textbf{0} \perp \textbf{0}$ still holds true?
Title: Re: Here's a question for you to try out.
Post by: Mao on August 02, 2008, 08:16:09 pm: Quote from: dcc on August 02, 2008, 08:05:52 pm
Im still not convinced that saying that $\textbf{a}.\textbf{v} = 0 \implies \textbf{a} \perp \textbf{v}$

What about $\textbf{v} = \textbf{a} = \textbf{0}$ , does that mean $\textbf{0} \perp \textbf{0}$ still holds true?

that is a valid point (as for all vectors)

but when neither are zero vectors, that would be true.
Title: Re: Here's a question for you to try out.
Post by: dcc on August 02, 2008, 08:21:59 pm: Quote from: Mao on August 02, 2008, 08:16:09 pm
Quote from: dcc on August 02, 2008, 08:05:52 pm
Im still not convinced that saying that $\textbf{a}.\textbf{v} = 0 \implies \textbf{a} \perp \textbf{v}$

What about $\textbf{v} = \textbf{a} = \textbf{0}$ , does that mean $\textbf{0} \perp \textbf{0}$ still holds true?

that is a valid point (as for all vectors)

but when neither are zero vectors, that would be true.

So the question perhaps should of been framed:

"Prove that if the speed of a particle is constant and the particle is accelerating, then the acceleration of the particle is perpendicular to the velocity."
Title: Re: Here's a question for you to try out.
Post by: Mao on August 02, 2008, 08:33:46 pm: "Prove that if the speed of a moving particle is constant and the particle has a non-zero acceleration, then the direction of acceleration of the particle is perpendicular to the velocity."
Title: Re: Here's a question for you to try out.
Post by: Lycan on August 02, 2008, 09:28:31 pm: Eh, the particle was stated to be moving. I also in a 'curve' which was meant to imply non-zero acceleration but then lines are also considered curves aren't they. Anywho, the dot product rule is correct. Although I had:

x^2 + y^2 = c^2 and differentiated both sides in terms of T for a more elegant solution.
Title: Re: Here's a question for you to try out.
Post by: Mao on August 02, 2008, 09:31:26 pm: Quote from: Lycan on August 02, 2008, 09:28:31 pm
Eh, the particle was stated to be moving. I also in a 'curve' which was meant to imply non-zero acceleration but then lines are also considered curves aren't they. Anywho, the dot product rule is correct. Although I had:

x^2 + y^2 = c^2 and differentiated both sides in terms of T for a more elegant solution.

but that only works for the particular case of $\mathbb{R}^2$ (you can similarly show for higher dimensions, etc)
Title: Re: Here's a question for you to try out.
Post by: Lycan on August 03, 2008, 08:40:05 pm: It was just a general idea, of course you can simply extend it to n dimensions.

x^2 + y^2 + z^2 ... = c^2
Title: Re: Here's a question for you to try out.
Post by: Mao on August 03, 2008, 09:10:48 pm: mind posting up your proof?
Title: Re: Here's a question for you to try out.
Post by: Lycan on August 04, 2008, 08:07:10 am: Sure.

Let's just pretend these are all vectors and look nice and stuff.

Let v(t) = x(t)i + y(t)j + z(t)k + ... n dimensions

Speed = ([x(t)]^2 + [y(t)]^2 + [z(t)]^2)^(1/2) + ...= k, where k is a constant

Therefore [x(t)]^2 + [y(t)]^2 + [z(t)]^2 + ... = k^2

d/dt([x(t)]^2 + [y(t)]^2 + [z(t)]^2 + ... ) = d/dt(k^2) = 0

Using chain rule:

2x(t)x'(t) + 2y(t)y'(t) + 2z(t)z'(t) + ... = 0

x(t)x'(t) + y(t)y'(t) + z(t)z'(t) + ... = 0

a(t)= d/dt(v(t)) = x'(t)i + y'(t)j + z'(t)k + ....

a.v = x(t)x'(t) + y(t)y'(t) + z(t)z'(t) + ...= 0

Hence they are perpendicular.
Title: Re: Here's a question for you to try out.
Post by: Mao on August 04, 2008, 06:31:05 pm: lolol, your solution is subjected to the same critique dcc gave to mine xD
Title: Re: Here's a question for you to try out.
Post by: dcc on August 04, 2008, 06:31:59 pm: Quote from: Mao on August 04, 2008, 06:31:05 pm
lolol, your solution is subjected to the same critique dcc gave to mine xD

Not if he is answering the revised question, which implies that neither $\textbf{a}$ or $\textbf{v}$ are zero :P

p.s. I might add, that if you are trying to prove that two vectors are perpendicular using the dot product, then it is simply not enough to state:

$\textbf{a}.\textbf{b} = 0$ as the last line of your proof

rather, you MUST state:

$\textbf{a}.\textbf{b} = 0$ hence $\textbf{a} \perp \textbf{b}$

Otherwise you have not proved as to whether the two vectors are perpendicular :)
Title: Re: Here's a question for you to try out.
Post by: Lycan on August 04, 2008, 06:46:36 pm: I realise that. I was just rushing as I was typing it up before heading off to school.