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

In summary, the rate of change in radians per second is a measure of how quickly an angle is changing over time. It is calculated by dividing the change in angle by the change in time, and a positive rate of change indicates an increase in angle, while a negative rate of change indicates a decrease. This concept is used in various fields, including physics, engineering, and astronomy, to analyze and predict the motion of rotating objects and natural phenomena.
  • #1
karush
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Point \(\displaystyle P\) moves at $a$ constant rate along the semicircle centered at \(\displaystyle O\) from \(\displaystyle M\) to \(\displaystyle N\).
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.
 
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  • #2
a) Since $P$ moves at a constant rate, then so too must $x$ and we may state:

\(\displaystyle \frac{dx}{dt}=\frac{\Delta x}{\Delta t}\)

What are the initial and final measures of $x$ and $t$?
 
  • #3
Work in radians/second not cm/second.
 
  • #4
MarkFL said:
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)
 
  • #5
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?
 
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  • #6
sorry I took so long to reply to this but need to get back to it.

my understanding of question a is

\(\displaystyle \displaystyle\)
\(\displaystyle \frac{dx}{dt}=\frac{\pi}{30}\text{radians per second}\)
 
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  • #7
karush said:
sorry I took so long to reply to this but need to get back to it.

my understanding of question a is

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

Yes that is correct.
 
  • #8
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

$\displaystyle \begin{align*} x + 2y &= \pi \\ 2y &= \pi - x \\ y &= \frac{1}{2} \left( \pi - x \right) \\ \frac{\mathrm{d}}{\mathrm{d}t} \left( y \right) &= \frac{\mathrm{d}}{\mathrm{d}t} \left[ \frac{1}{2} \left( \pi - x \right) \right] \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ \frac{1}{2} \left( \pi - x \right) \right] \, \frac{\mathrm{d}x}{\mathrm{d}t} \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= -\frac{1}{2} \cdot \frac{\pi}{30} \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= -\frac{\pi}{60} \end{align*}$
 
  • #9
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...
 
  • #10
karush said:
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.
 
  • #11
I am going to come back to this with a new thread, also in $c$ it should read
"when $y=\frac{\pi}{2}$
 

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

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

The rate of change in radians per second is a measure of how quickly a given angle is changing over time, specifically in terms of radians (a unit of measurement for angles) and seconds (a unit of measurement for time).

How is the rate of change in radians p/s calculated?

The rate of change in radians per second is calculated by dividing the change in the angle, measured in radians, by the change in time, measured in seconds. This gives the average rate of change over the given time interval.

What does a positive rate of change in radians p/s indicate?

A positive rate of change in radians per second indicates that the angle is increasing over time. This means that the angle is getting larger, or turning in a counterclockwise direction.

What does a negative rate of change in radians p/s indicate?

A negative rate of change in radians per second indicates that the angle is decreasing over time. This means that the angle is getting smaller, or turning in a clockwise direction.

How is the rate of change in radians p/s used in real-world applications?

The rate of change in radians per second is used in many fields, including physics, engineering, and astronomy. It can be used to calculate the speed and acceleration of rotating objects, such as gears or planets, and to model and predict various natural phenomena, such as the motion of waves or the rotation of celestial bodies.

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