Stationary point of y = 5 - 6sinx - 3x

[Monkeyseat]

The curve y = 5 - 6sinx - 3x is defined for 0 < x < Pi.

Find the x-coordinate of the stationary point of the curve.

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Solution:

dy/dx = -6cosx - 3

Stationary point when dy/dx = 0.

0 = -6cosx - 3
6cosx = -3
cos x = -3/6
cos x = -1/2

Principal value = cos^-1 (-1/2)
Principal value = 2.094

Okay so I have the correct answer, but the book says to leave in the form of 2pi/3. I know my answer is equal to that but how can I show it in this way?

From this graph:

I cut the axis off but the thick red line I drew on crosses y at -0.5 I know it meets the x-axis inbetween pi/2 and 3pi/4 but how do I prove it is 2pi/3? Do I have to draw a CAST diagram?

Thanks for the help. I know it's a 'basic' question but I don't know how to show 2.094 as 2pi/3.

 

[galactus]

Just write it in that form instead of in decimal form. They're the same.

Your solutions are actually \(\displaystyle \frac{2(3C+1){\pi}}{3}\) or \(\displaystyle \frac{2(3C-1){\pi}}{3}\)

If we enter in C=0, we get \(\displaystyle \frac{2{\pi}}{3} \;\ or \;\ \frac{-2{\pi}}{3}\)

If we enter in C=1, we get \(\displaystyle \frac{8\pi}{3} \;\ or \;\ \frac{4\pi}{3}\)

And so on. See, it repeats every \(\displaystyle 2\pi\) interval. That is its period.

It would appear you want the first one where C=0.

What we have is \(\displaystyle cos^{-1}(\frac{-1}{2})\). Which is what you have. This has a solution of \(\displaystyle \frac{2\pi}{3}\) and then repeats every 2pi interval.

for that matter, you can see this on the unit circle. Then, because cos has a period of 2Pi just add and subtract 2Pi from \(\displaystyle \frac{2\pi}{3}\).

See what I mean?.

 

[Monkeyseat]

Not quite.

I have cos^-1 (-1/2) which is 2.094 radians. I know it is the same value as 2pi/3 but is there not a way to prove this or find this by showing it on the graph or something else? :?

I guess I'm just looking into too much. I just don't know how someone got cos^-1(-1/2) and then went "Ah! That's the same as 2pi/3!" How did they know 2.094 was the same as that?

 

[galactus]

Because \(\displaystyle \frac{2\pi}{3}=2.09439510239.........\)

That's about it.

 

[Monkeyseat]

I know that. I just thought there would be a logical way to get to the answer in the form 2pi/3. I guess not.

If you did cos^-1 (-1/2) you would say the answer was 2.094 wouldn't you? The only way I knew the answer could also be wrote as 2pi/3 was because I checked the answers. I mean, does anyone know off the top of their head that 2.094 equals 2pi/3? :?

Say in an exam you had this question and you did cos^-1 (-1/2) and got 2.094. How would you know that was 2pi/3?

That is why I thought there would be some way to find the answer in the form 2pi/3 without having to check the answers!

 

[Deleted member 4993]

Not quite.

I have cos^-1 (-1/2) which is 2.094 radians. I know it is the same value as 2pi/3 but is there not a way to prove this or find this by showing it on the graph or something else? :?

I guess I'm just looking into too much. I just don't know how someone got cos^-1(-1/2) and then went "Ah! That's the same as 2pi/3!"

From unit circle - or basic cosine values:

\(\displaystyle \cos({\pm\frac{2\cdot\pi}{3} \, + 2\cdt n\cdot\pi}) = -\frac{1}{2}\)

How did they know 2.094 was the same as that?

Using calculator - or long hand knowing value of ? = 3.14159 and multiplying/dividing

 

[Monkeyseat]

Not quite.

I have cos^-1 (-1/2) which is 2.094 radians. I know it is the same value as 2pi/3 but is there not a way to prove this or find this by showing it on the graph or something else? :?

I guess I'm just looking into too much. I just don't know how someone got cos^-1(-1/2) and then went "Ah! That's the same as 2pi/3!"

From unit circle - or basic cosine values:

\(\displaystyle \cos({\pm\frac{2\cdot\pi}{3} \, + 2\cdt n\cdot\pi}) = -\frac{1}{2}\)

How did they know 2.094 was the same as that?

Using calculator - or long hand knowing value of ? = 3.14159 and multiplying/dividing

Thanks for the reply, I don't know what the unit circle is at the moment but I get the gist of what you are saying. So there was no real logical approach to figuring it out and they just got 2pi/3 by checking values on a calculator by using trial and improvement?

Say in an exam you had this question and you did cos^-1 (-1/2) and got 2.094. How would you know that this was equal to 2pi/3? I know 2.094 would probably suffice as an answer but to know that this number was equal to 2pi/3, you would have to sit there for 5 minutes and try and figure it out via trial and improvement.

I doubt you could do that for every question like this - some of them you might not even be able to write in terms of pi! Also, it would be hard to tell which ones you could write in terms of pi wouldn't it?

 

[galactus]

Just divide 2.094 by pi and we get .66654...

Which, we can see is close to 2/3. So, \(\displaystyle \frac{2\pi}{3}\)

 

[Deleted member 4993]

Thanks for the reply, I don't know what the unit circle is at the moment but I get the gist of what you are saying.

So there was no real logical approach to figuring it out and they just got 2pi/3 by checking values on a calculator by using trial and improvement?

This is very logical approach - derived from fundamental geometry/trigonometry.

Say in an exam you had this question and you did cos^-1 (-1/2) and got 2.094. How would you know that this was equal to 2pi/3? I know 2.094 would probably suffice as an answer but to know that this number was equal to 2pi/3, you would have to sit there for 5 minutes (anybody who remembers high school trigonometry should take 5 seconds - and trigonometry is one of the prerequisite for these courses - isn't it?) and try and figure it out via trial and improvement.

I doubt you could do that for every question like this - some of them you might not even be able to write in terms of pi! Also, it would be hard to tell which ones you could write in terms of pi wouldn't it?

 

[Monkeyseat]

Thanks for the reply, I don't know what the unit circle is at the moment but I get the gist of what you are saying.

So there was no real logical approach to figuring it out and they just got 2pi/3 by checking values on a calculator by using trial and improvement?

This is very logical approach - derived from fundamental geometry/trigonometry.

Say in an exam you had this question and you did cos^-1 (-1/2) and got 2.094. How would you know that this was equal to 2pi/3? I know 2.094 would probably suffice as an answer but to know that this number was equal to 2pi/3, you would have to sit there for 5 minutes (anybody who remembers high school trigonometry should take 5 seconds - and trigonometry is one of the prerequisite for these courses - isn't it?) and try and figure it out via trial and improvement.

I doubt you could do that for every question like this - some of them you might not even be able to write in terms of pi! Also, it would be hard to tell which ones you could write in terms of pi wouldn't it?

Okay. Thanks both of you for the help.