Udemy Quiz Question on Radians and Quadrants

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Why is the answer "on the positive vertical axis" and not "in the third quadrant?" the question on the quiz is: Where is an angle of 10.5π radians in standard position located?

Is there a difference between 10.5 radians and 10.5π radians?

I am so stupid, this question makes me want to cry.
Question: Where is an angle of 10.5π radians in standard position located?

A. On the positive vertical axis
B. On the negative vertical axis
C. In the second quadrant
D. In the fourth quadrant

I thought I had to divide 10.50 by 3.14, which I thought yielded 3.34π. Then I subtracted 3.34π by 2π to get 1.34π. That is 3rd quadrant.

I also tried 3.34π times 180° which yielded 601.2°. That is also in the third quadrant. Clearly not an answer choice.

I hate how stupid I am. I don't know why I can't understand such a simple question. Please explain in simple terms, step-by-step.
 
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Simone said:
Question: Where is an angle of 10.5π radians in standard position located?

A. On the positive vertical axis
B. On the negative vertical axis
C. In the second quadrant
D. In the fourth quadrant

I thought I had to divide 10.50 by 3.14, which I thought yielded 3.34π. Then I subtracted 3.34π by 2π to get 1.34π. That is 3rd quadrant.
Correct IF THERE WAS NOT A MULTIPLIER OF ##\pi = 3.14159##. But there is a multiplier of ##\pi##, so it is easier. ##10.5\pi / \pi = 10.5## That is 5 full circles plus one-fourth of a circle in the counterclockwise direction. That is on the positive vertical axis.
 
FactChecker said:
Correct IF THERE WAS NOT A MULTIPLIER OF ##\pi = 3.14159##. But there is a multiplier of ##\pi##, so it is easier. ##10.5\pi / \pi = 10.5## That is 5 full circles plus one-fourth of a circle in the counterclockwise direction. That is on the positive vertical axis.
How are 10.5 radians and 5.25 rotations? By dividing 10.5π by π it becomes radians without pi... so would I then still divide by 3.14 since the multiplier is gone? I feel I am missing something fundamental here, im sorry.

Why did you divide by π instead of 2π? How did you know to divide by any multiple of π in the first place? How did you get 5.25 rotations from 10.5? What do you mean by a multiplier of π? Is there a difference between 10.5 radians and 10.5π radians? If so, what is it?
 
Simone said:
Is there a difference between 10.5 radians and 10.5π radians? If so, what is it?
The difference is that ##10.5\pi## has a multiplier of ##\pi=3.14159265358979....##. That is quite different from 10.5
##10.5\pi = 32.9867228626928##.
Now you can proceed as you did before and divide 32.9867228626928 by 3.14159265358979, but it is easier to just leave ##\pi## there symbolically so that it cancels cleanly with no calculations: ##10.5\pi / \pi = 10.5##.
 
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Simone said:
Why did you divide by π instead of 2π? How did you know to divide by any multiple of π in the first place? How did you get 5.25 rotations from 10.5? What do you mean by a multiplier of π? Is there a difference between 10.5 radians and 10.5π radians? If so, what is it?
For a problem like this, it's better to just visualize the Unit Circle instead of trying to plug in numbers for ##\pi## -- Just remember that there are ##2 \pi## radians in a a circle, and the rest is easy:

1671741182205.png

https://precalctrig.weebly.com/4-unit-circle.html
 
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… And just in case you ever wonder why such impractical way to measure angles has been used:

”In calculus and most other branches of mathematics beyond practical geometry, angles are measured in radians. This is because radians have a mathematical naturalness that leads to a more elegant formulation of some important results.
Results in analysis involving trigonometric functions can be elegantly stated when the functions' arguments are expressed in radians.”

Copied from
https://en.wikipedia.org/wiki/Radian

:cool:🎄
 
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Simone said:
How are 10.5 radians and 5.25 rotations?
10.5 π radians represents 5.25 rotations. 10.5 radians represents and angle of about 601.1 degrees.
Simone said:
By dividing 10.5π by π it becomes radians without pi... so would I then still divide by 3.14 since the multiplier is gone? I feel I am missing something fundamental here, im sorry.
Why would you want to divide 10.5π by π?
 
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Simone said:
By dividing 10.5π by π it becomes radians without pi... so would I then still divide by 3.14 since the multiplier is gone?
No. ##\pi## is the 3.14 (actually it is 3.14159265358979.... ) So rather than multiplying out the ##10.5\pi## and then dividing by ##\pi##, it is simpler to just symbolically cancel the ##\pi##s.
 
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FactChecker said:
No. ##\pi## is the 3.14 (actually it is 3.14159265358979.... ) So rather than multiplying out the ##10.5\pi## and then dividing by ##\pi##, it is simpler to just symbolically cancel the ##\pi##s.
I did not like the concept at 16, why do it like that? Why not just degrees?
Perhaps it was not explained that well? Other guys got it so it was probably just me!
 
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pinball1970 said:
why do it like that? Why not just degrees?
So that the fourth derivative of the ##\sin## function is the ##\sin## function instead of the fourth power of ##\frac{\pi}{180}## times the ##\sin## function. Well, that and a few thousand other things where radians keep factors of ##\pi## and ##180## out of our formulas.
 
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jbriggs444 said:
So that the fourth derivative of the ##\sin## function is the ##\sin## function instead of the fourth power of ##\frac{\pi}{180}## times the ##\sin## function.
Fourth derivative? Is that snap crackle or pop?
 
pinball1970 said:
Fourth derivative? Is that snap crackle or pop?
1671899022531.png

It will be one of those. Let us see which....

If we normalize a particle in simple harmonic motion to an angular frequency of one positive radian per time unit, a phase of zero at time zero and an amplitude of one distance unit then the position function will be ##x = \sin t##

The velocity will be ##v = \cos t## (first derivative of ##\sin t##).

The acceleration will be ##a = -\sin t## (second derivative of ##\sin t##)

The jerk will be ##j = -\cos t## (third derivative of ##\sin t##)

The snap will be ##s = \sin t## (fourth derivative of ##\sin t##).

Conveniently, there are no factors of ##\pi## or ##180## anywhere except in the step where we normalized things.

Edit: Changed position from "##s##" to "##x##" to avoid the conflict with snap.
 

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