MHB What is the rate of change of s with respect to time when y=π/2 radians?

[karush]

https://www.physicsforums.com/attachments/2883
It was found earlier that
$ \frac{dx}{dt}=\frac{\pi}{30} \frac{rad}{sec} $ and $\frac{dy}{dx}=\frac{\pi}{60} \frac{rad}{sec}$

(C) since $s$ and $x$ and are related by the Law of Cosines;
What is the rate of change of with respect to time when $y=\frac{\pi}{2}$ radians?
Indicate units of measure.

ok I did this
$\displaystyle s^2=200-200\cdot\cos⁡ {\frac{\pi}{30}}\cdot\text {t}$
Let
$u = 200 - 200\cdot\cos\left({\frac{\pi}{30}}\right)\cdot\text{t}$
then
$\frac{du}{dt}=200\cos{\left(\frac{\pi}{30}\right)}$

provided I went the right direction not sure what to do next

 

[Jameson]

Hmm, so I almost agree with your set up using the law of cosines. I get this:

$$s^2=10^2+10^2-2(10)(10) \cos(x)$$

From here we can probably differentiate both sides with respect to time, treating $s$ and $x$ as functions of $t$. If that is the correct approach then it would lead to:

$$2s \frac{ds}{dt}=200 \sin(x) \cdot \frac{dx}{dt}$$

This doesn't contain anything about $y$ though so I would use an alternate form of the law of cosines where $y$ is the angle you take the cosine of. However, I don't know what "What is the rate of change of with respect to time" means because it's not clear to me what rate of change we are supposed to find.

I hope this is some food for thought until someone else can step in. I've done many problems like this before but am missing something in the setup or maybe from an earlier part.

 

[karush]

op correction

What is the rate of change of with respect to time when [FONT=MathJax_Math]y[FONT=MathJax_Main]=[FONT=MathJax_Math]π[FONT=MathJax_Main]2 radians per second?
small correction on OP

also, [FONT=MathJax_Math]$$\frac{dy}{dt}=\frac{\pi}{60}\frac{rad}{sec}$$

 

[Jameson]

Rate of change of what though? $\frac{ds}{dt}$? $y$ isn't expressed in radians per second either, so did you mean $\frac{dy}{dt}=\frac{\pi}{2}$?

 

[karush]

yes, since s is a length

sorry I should close this ... to much typo error in it.

thanks for help... pretty sure I know how to do it. once I find the original problem.

 

[DavidCampen]

\[let \,\theta = x,\,r = 10\] then
\[s =(r^2 sin^2\theta + (r-rcos\theta)^2)^\frac{1}{2}\]
\[s = r (sin^2\theta + 1 -2cos\theta + cos^2\theta)^\frac{1}{2}\]
\[\frac{s}{r} = (2 - 2cos\theta)^\frac{1}{2}= (4sin^2\frac{\theta}{2})^\frac{1}{2}=2sin\frac{\theta}{2}\]
\[s = 2rsin\frac{\theta}{2}\]
\[\frac{ds}{d\theta}=rcos\frac{\theta}{2}\]
\[\frac{d\theta}{dt} = \frac{\pi}{30}\frac{rad}{sec}\]
\[\frac{ds}{dt} = \frac{ds}{d\theta}\frac{d\theta}{dt}\]
When \[y = \frac{\pi}{2}\] then \[\theta = 0\]
so \[\frac{ds}{d\theta} = rcos\frac{\theta}{2} = r\]
and \[\frac{ds}{dt} = r\frac{d\theta}{dt} = \frac{r\pi}{30}\frac{rad}{sec}\]

 

[karush]

wow, that was a great help..thanks for all the steps.

So let me try this if $\displaystyle x=\frac{\pi}{2}$ then
$\displaystyle
s^2=200-200\cdot\cos{\left(x⁡\right)}
$
$\displaystyle
2s \frac{ds}{dt}= -200 \sin⁡{\right(x\left)} \frac{dx}{dt}
$
$\displaystyle
2s \frac{ds}{dt}= -200 \cdot \sin⁡{|left(\frac{\pi}{2}\right)}\cdot\frac{dx}{dt}
$
$\displaystyle
\frac{2}{\sqrt{200}}\cdot\frac{ds}{dt}=
-200\sin\left({\frac{\pi}{2}}\right)\cdot\frac{\pi}{30}
$

$\displaystyle
\frac{ds}{dt}\approx 148 \frac{cm}{sec}
$

I think anyway this ans seem too large