MHB What is the rate of change, in radians p/s

[karush]

View attachment 2456
Point $$P$$ moves at $a$ constant rate along the semicircle centered at $$O$$ from $$M$$ to $$N$$.
The radius of the semicircle is $10cm$, and it takes $30s$ for $P$ to move from $M$ to $N$. $∠POM$ has measure $x$ radians, $∠OPM$ has measure $y$ radians, and $MP=S$ cm as indicated in the figure.

$a.$ What is the rate of change, in radians $p/s$, of $x$ with respect to time?

$b.$ What is the rate of change, in radians $p/s$ of $y$ with respect to time?

there is $c$ and $d$ but will do $a$ and $b$ first

Not real sure how to set this up, but by observation $P$ moves at a constant rate of

$\frac{10cm}{30s}$ and x is increasing and y is decreasing as $P$ moves from $M$ to $N$.
I don't think s is relevant for questions a and b.

 

[MarkFL]

a) Since $P$ moves at a constant rate, then so too must $x$ and we may state:

$$\frac{dx}{dt}=\frac{\Delta x}{\Delta t}$$

What are the initial and final measures of $x$ and $t$?

 

[DavidCampen]

Work in radians/second not cm/second.

 

[karush]

What are the initial and final measures of $x$ and $t$?

well for $x$ the initial measure would be $0\pi$ or just $0$ and the final value would be $\pi$ as $t$ goes from $0$ to $30s$

so presume that $P=10x$ in radians.
So if x is $\frac{\pi}{4}$ then $P$ is $\frac{5\pi}{2}$

I guess we are relating the arc length of $MN$ to $x$ and $y$ is $\pi - x$

frankly I have hard time with related rates ...(Wasntme)

 

[DavidCampen]

First write down the rate of change of x specified in radians/second. Note that x and y are not Cartesian coordinates but angles. The problem statement has told you the rate of change of x; it is a constant value, you do not have to calculate it.

It is unfortunate that the problem statement uses x and y for angles when those variables are usually used to represent Cartesian coordinates. The problem is not very difficult but many things were done to obscure that.

If you drop a perpendicular line from point P to the line MN and call the point of intersection Q then you can use trig functions to calculate the length of PQ.

When the problem statement says "radians p/s", what does "p/s" mean?

 

[karush]

sorry I took so long to reply to this but need to get back to it.

my understanding of question a is

$$\displaystyle$$
$$\frac{dx}{dt}=\frac{\pi}{30}\text{radians per second}$$

 

[Prove It]

sorry I took so long to reply to this but need to get back to it.

my understanding of question a is

$$\displaystyle$$
$$\frac{dx}{dt}=\frac{\pi}{30}\text{radians per second}$$

Yes that is correct.

 

[Prove It]

Now as for part b), since the triangle is isosceles, that means the remaining angle is also equal to y, and since angles in a triangle add to $\displaystyle \begin{align*} 180^{\circ} \end{align*}$ or $\displaystyle \begin{align*} \pi ^C \end{align*}$, that means

$\displaystyle \begin{align*} x + 2y &= \pi \\ 2y &= \pi - x \\ y &= \frac{1}{2} \left( \pi - x \right) \\ \frac{\mathrm{d}}{\mathrm{d}t} \left( y \right) &= \frac{\mathrm{d}}{\mathrm{d}t} \left[ \frac{1}{2} \left( \pi - x \right) \right] \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ \frac{1}{2} \left( \pi - x \right) \right] \, \frac{\mathrm{d}x}{\mathrm{d}t} \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= -\frac{1}{2} \cdot \frac{\pi}{30} \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= -\frac{\pi}{60} \end{align*}$

 

[karush]

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

$\displaystyle
s^2 = 10^2 + 10^2 - 2\cdot 10 \cdot10\cdot \cos{x}
$

$\displaystyle
s^2 = 200 - 200 \cdot \cos{\frac{\pi}{30}\cdot t}
$

$\displaystyle
s = \sqrt{200 - 200 \cdot \cos{\frac{\pi}{30}\cdot t}}
$

just seeing if I am going in the right direction...

 

[Prove It]

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

$\displaystyle
s^2 = 10^2 + 10^2 - 2\cdot 10 \cdot10\cdot \cos{x}
$

$\displaystyle
s^2 = 200 - 200 \cdot \cos{\frac{\pi}{30}\cdot t}
$

$\displaystyle
s = \sqrt{200 - 200 \cdot \cos{\frac{\pi}{30}\cdot t}}
$

just seeing if I am going in the right direction...

Yes that is correct, but the s^2 equation will be easier to work with when you differentiate to find ds/dt.

 

[karush]

I am going to come back to this with a new thread, also in $c$ it should read
"when $y=\frac{\pi}{2}$