And Another Question About Partial Derivatives

[Maor Hadad]

Homework Statement

<br /> \frac{d}{dt}\left(\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}\right)=0\Rightarrow\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=const\Rightarrow\dot{q}=A\Rightarrow q=At+B<br />

Homework Equations

Why it ok to say that:
<br /> \frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=const\Rightarrow\dot{q}=A<br />

The Attempt at a Solution

<br /> \left(\frac{\dot{q}}{const}\right)^{2}-\dot{q}-1=0\Rightarrow?<br />

 

[Samy_A]

Homework Statement

<br /> \frac{d}{dt}\left(\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}\right)=0\Rightarrow\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=const\Rightarrow\dot{q}=A\Rightarrow q=At+B<br />

Homework Equations

Why it ok to say that:
<br /> \frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=const\Rightarrow\dot{q}=A<br />

The Attempt at a Solution

<br /> \left(\frac{\dot{q}}{const}\right)^{2}-\dot{q}-1=0\Rightarrow?<br />

How did you get the last equation?

Anyway, if you solve $\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=C$ for $\dot{q}$, what do you get as result?

 

[Maor Hadad]

If you mean
<br /> \left(\frac{\dot{q}}{const}\right)^{2}-\dot{q}-1=0\Rightarrow?<br />
It's by dividing two sides by const and multiplying by
<br /> \sqrt{1+\left(\dot{q}\right)^{2}}<br />
and than squaring both sides.
If I'll continue I'll need to solve quadratic equation which it's results will depend on the value on C.
It can be either one, two or non complex or real numbers..

 

[Samy_A]

If you mean
<br /> \left(\frac{\dot{q}}{const}\right)^{2}-\dot{q}-1=0\Rightarrow?<br />
It's by dividing two sides by const and multiplying by
<br /> \sqrt{1+\left(\dot{q}\right)^{2}}<br />
and than squaring both sides.

I don't see how this operation gives you the term $\dot q$.

If I'll continue I'll need to solve quadratic equation which it's results will depend on the value on C.
It can be either one, two or non complex or real numbers..

Maybe. Why would that be a problem? You need to prove that $\dot q$ is a constant, ie doesn't depend on $t$.

 

[epenguin]

solve $\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=C$ for $\dot{q}$, what do you get as result?

Do that again and do it right!

 

[Maor Hadad]

Yes, you're right. It dosen't! my professor accidently wrote it as:
<br /> \frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)}}<br />

and that's how I tried to solve it...

thank you!

 

[Ray Vickson]

Homework Statement

<br /> \frac{d}{dt}\left(\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}\right)=0\Rightarrow\frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=const\Rightarrow\dot{q}=A\Rightarrow q=At+B<br />

Homework Equations

Why it ok to say that:
<br /> \frac{\dot{q}}{\sqrt{1+\left(\dot{q}\right)^{2}}}=const\Rightarrow\dot{q}=A<br />

The Attempt at a Solution

<br /> <br />

This just says that if $\dot{q}/\sqrt{\dot{q}^2+1}$ is constant, the quantity $\dot{q}$ must also be a constant. That is obvious, but if you still cannot see it, imagine plotting a graph of $y = \dot{q}/\sqrt{\dot{q}^2+1}$ in a $(\dot{q},y)$ plot. You can see that if $\dot{q}/\sqrt{\dot{q}^2+1}$ is changed, $\dot{q}$ must be changed as well.

 

[Maor Hadad]

Hi,
I understand now what my problem was.
I can also 'see' why it's true through the algebra, but I think the fact the independent variable is a derivatie of time, makes it a bit un-intuitiable for me.
I tried plotting it but it came as something $\delta$ like, so it didn't helped so much .
but thank you for the help :)

and if don't mind I had another question about partial derivatives I'll really appreciate if you'll take a short glimpse at it :)

and even if you don't - thank you again!

 

[Ray Vickson]

Hi,
I understand now what my problem was.
I can also 'see' why it's true through the algebra, but I think the fact the independent variable is a derivatie of time, makes it a bit un-intuitiable for me.
I tried plotting it but it came as something $\delta$ like, so it didn't helped so much .
but thank you for the help :)

and if don't mind I had another question about partial derivatives I'll really appreciate if you'll take a short glimpse at it :)

and even if you don't - thank you again!

Which message are you responding to? You need to use the "Quote" button, in order to keep straight what messages are being addressed.

 

[Maor Hadad]

Hi, It's: A Question About Partial Derivatives,
thanks a lot!

Which message are you responding to? You need to use the "Quote" button, in order to keep straight what messages are being addressed.