MHB Angle change rate is a function of the instantaneous angle

[zshaban]

Dear All,

I have an integration probelm and I hope to find a solution for.

I assume that an objects moves vertically in the +ive direction of the $y$ axis, from $x=a$, with constant speed $v$. I monitor the rate of change of the angle $\theta\left({t}\right)$. I have derived this rate to be:

$\frac{d\theta\left({t}\right)}{dt}=\frac{v}{a} \cos^2\left({\theta\left({t}\right)}\right)$.

I would like to solve this equation to get $\theta\left({t}\right)$. Does this make sense? Do you have any idea of how to do it?

Many thanks
Zohair

 

[chisigma]

Dear All,

I have an integration probelm and I hope to find a solution for.

I assume that an objects moves vertically in the +ive direction of the $y$ axis, from $x=a$, with constant speed $v$. I monitor the rate of change of the angle $\theta\left({t}\right)$. I have derived this rate to be:

$\frac{d\theta\left({t}\right)}{dt}=\frac{v}{a} \cos^2\left({\theta\left({t}\right)}\right)$.

I would like to solve this equation to get $\theta\left({t}\right)$. Does this make sense? Do you have any idea of how to do it?

Many thanks
Zohair

Wellcome on MHB zshaban!...

You can solve the ODE...

$\displaystyle \frac{d \theta}{d t} = \frac{v}{a}\ \cos^{2} \theta\ (1)$

... separating the variables as follows...

$\displaystyle \frac{d \theta}{\cos^{2} \theta} = \frac{v}{a}\ d t \implies \tan \theta = \frac{v}{a} (t + c) \implies \theta = \tan^{-1} (\frac{v\ t}{a} + c)\ (2)$

Kind regards

$\chi$ $\sigma$

 

[zshaban]

Wellcome on MHB zshaban!...

You can solve the ODE...

$\displaystyle \frac{d \theta}{d t} = \frac{v}{a}\ \cos^{2} \theta\ (1)$

... separating the variables as follows...

$\displaystyle \frac{d \theta}{\cos^{2} \theta} = \frac{v}{a}\ d t \implies \tan \theta = \frac{v}{a} (t + c) \implies \theta = \tan^{-1} (\frac{v\ t}{a} + c)\ (2)$

Kind regards

$\chi$ $\sigma$

Thank you, chisigma! That was really helpful.

 

[DavidCampen]

I solve it this way:
\[vt = r\sin\theta,\,\,\,\,a=r\cos\theta\]
\[\tan\theta=\frac{vt}{a}\]
\[\theta=\tan^{-1}\frac{vt}{a}=\theta(t)\]
The domain of $\theta(t)=\pm\infty,$ the range is $\pm\frac{\pi}{2}$.

I don't understand the reason for treating this as an integration/differential equation problem.

 

[zshaban]

I solve it this way:
\[vt = r\sin\theta,\,\,\,\,a=r\cos\theta\]
\[\tan\theta=\frac{vt}{a}\]
\[\theta=\tan^{-1}\frac{vt}{a}=\theta(t)\]
The domain of $\theta(t)=\pm\infty,$ the range is $\pm\frac{\pi}{2}$.

I don't understand the reason for treating this as an integration/differential equation problem.

Thanks David. The intention was to consider a motion that makes an angle $\alpha$ with the $x-axis$, which is more complicated than the case shown above. I just wanted to find the tool to solve similar but more general cases, so I posted a simplified case.

Thanks again for your post.

Cheers,
Zohair