Vertical and horizontal tangent

[Petrus]

I am currently working with parametric equation and trying to solve finding points on the curve where the tangent is horizontal or vertical.
When I do with trigometry I get problem...
And I need help to understand this. I know what vertical and horizontal means.
exemple this one i am working with
\(\displaystyle x=2\cos\theta\), \(\displaystyle y=sin2\theta\)
Vertical tangent:
We know that vertical tangent is when \(\displaystyle \frac{dx}{d\theta}=0\) and \(\displaystyle \frac{dy}{d\theta} \neq 0\)
and we got \(\displaystyle \frac{dx}{d\theta}= -2\sin\theta\) so we got \(\displaystyle 0=-2\sin\theta\) and when I solve it I get \(\displaystyle \theta=0\)

Horizontal tangent
We know that horizontal tangent is when \(\displaystyle \frac{dy}{d\theta}=0\) and \(\displaystyle \frac{dx}{d\theta} \neq 0\)
so we got \(\displaystyle \frac{dy}{d\theta}=2\cos2\theta\) and then we will get \(\displaystyle 0=2\cos2\theta\) and i get \(\displaystyle \theta=\frac{1}{2}\)
I understand after we solved for \(\displaystyle \theta\) we put it in the x and y to get the point. I got problem solving those equation with trigometry.

Regards,

 

[Sudharaka]

I am currently working with parametric equation and trying to solve finding points on the curve where the tangent is horizontal or vertical.
When I do with trigometry I get problem...
And I need help to understand this. I know what vertical and horizontal means.
exemple this one i am working with
\(\displaystyle x=2\cos\theta\), \(\displaystyle y=sin2\theta\)
Vertical tangent:
We know that vertical tangent is when \(\displaystyle \frac{dx}{d\theta}=0\) and \(\displaystyle \frac{dy}{d\theta} \neq 0\)
and we got \(\displaystyle \frac{dx}{d\theta}= -2\sin\theta\) so we got \(\displaystyle 0=-2\sin\theta\) and when I solve it I get \(\displaystyle \theta=0\)

Horizontal tangent
We know that horizontal tangent is when \(\displaystyle \frac{dy}{d\theta}=0\) and \(\displaystyle \frac{dx}{d\theta} \neq 0\)
so we got \(\displaystyle \frac{dy}{d\theta}=2\cos2\theta\) and then we will get \(\displaystyle 0=2\cos2\theta\) and i get \(\displaystyle \theta=\frac{1}{2}\)
I understand after we solved for \(\displaystyle \theta\) we put it in the x and y to get the point. I got problem solving those equation with trigometry.

Regards,

Hi Petrus, :)

Here is another approach along the lines of the same method used http://www.mathhelpboards.com/f10/find-tangent-4348/.

When the tangent is horizontal,

\[\frac{dy}{dx}=0\]

When the tangent is vertical, \(\dfrac{dy}{dx}\) is undefined.

Method:

1) Find \(\dfrac{dy}{dx}\). You should get, \(\dfrac{dy}{dx}=\dfrac{2\sin^{2}{\theta}-1}{\sin{\theta}}\).

2) Find the value of \(\theta\) (using (1)) when \(\dfrac{dy}{dx}=0\Rightarrow 2\sin^{2}{\theta}-1=0\). Then you'll get all the points on which \(\dfrac{dy}{dx}=0\).

For the second part where the tangent is vertical \(\dfrac{dy}{dx}\) is undefined. Therefore, \(\sin\theta=0\). Hope you can continue. :)

 

[Petrus]

Hello Sudharaka,
This is what I learned from todays lecturne and this is how I would solve, unfortently I could not understand and do with your method but here is how I did.
\(\displaystyle x=2\cos\theta\), \(\displaystyle y=\sin2\theta\)

Horizontal tangent line:
\(\displaystyle \frac{dy}{d\theta}=0 <=> 2\cos2\theta=0\)
if we look at unit circle we can see that

\(\displaystyle 2\theta=\frac{\pi}{2} <=> \theta=\frac{\pi}{4}\)
\(\displaystyle 2\theta=\frac{3\pi}{2} <=> \theta=\frac{3\pi}{4}\)
remember we can always add \(\displaystyle 2\pi\) but we want to stay in the range \(\displaystyle 0\leq \theta \leq 2\pi\)
so we got also
\(\displaystyle 2\theta=\frac{5\pi}{2} <=> \theta= \frac{5\pi}{4}\)
\(\displaystyle 2\theta=\frac{7\pi}{2} <=> \theta = \frac{7\pi}{4}\)
and if we put this point in x and y we get our horizontal tangent line
\(\displaystyle x=\pm\sqrt{2}, y=\pm1\) (4 points)

Vertical tangent line:

\(\displaystyle \frac{dx}{d\theta}=0 <=> -2\sin\theta=0\)
if we solve it we get
\(\displaystyle \theta=0\)
\(\displaystyle \theta=\pi\)
and to get our x and y point we put that in the function and get the point
\(\displaystyle x=\pm2, y=0\)

Regards,

 

[Sudharaka]

Hello Sudharaka,
This is what I learned from todays lecturne and this is how I would solve, unfortently I could not understand and do with your method but here is how I did.
\(\displaystyle x=2\cos\theta\), \(\displaystyle y=\sin2\theta\)

Horizontal tangent line:
\(\displaystyle \frac{dy}{d\theta}=0 <=> 2\cos2\theta=0\)
if we look at unit circle we can see that

\(\displaystyle 2\theta=\frac{\pi}{2} <=> \theta=\frac{\pi}{4}\)
\(\displaystyle 2\theta=\frac{3\pi}{2} <=> \theta=\frac{3\pi}{4}\)
remember we can always add \(\displaystyle 2\pi\) but we want to stay in the range \(\displaystyle 0\leq \theta \leq 2\pi\)
so we got also
\(\displaystyle 2\theta=\frac{5\pi}{2} <=> \theta= \frac{5\pi}{4}\)
\(\displaystyle 2\theta=\frac{7\pi}{2} <=> \theta = \frac{7\pi}{4}\)
and if we put this point in x and y we get our horizontal tangent line
\(\displaystyle x=\pm\sqrt{2}, y=\pm1\) (4 points)

Vertical tangent line:

\(\displaystyle \frac{dx}{d\theta}=0 <=> -2\sin\theta=0\)
if we solve it we get
\(\displaystyle \theta=0\)
\(\displaystyle \theta=\pi\)
and to get our x and y point we put that in the function and get the point
\(\displaystyle x=\pm2, y=0\)

Regards,

Note that I have updated my method as it contained some errors. Your method is correct and there's no problem with it. Essentially both methods are using the same fact which is when the tangent is horizontal,

\[\frac{dy}{dx}=0~~~~~~~(1)\]

But since,

\[\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\]

We get,

\[\frac{dy}{d\theta}=0~~~~~~~~~(2)\]

I have used (1) whereas you have used (2). The same goes for the vertical tangent. Hope this shreds light on the matter.

 

[Petrus]

Note that I have updated my method as it contained some errors. Your method is correct and there's no problem with it. Essentially both methods are using the same fact which is when the tangent is horizontal,

\[\frac{dy}{dx}=0~~~~~~~(1)\]

But since,

\[\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\]

We get,

\[\frac{dy}{d\theta}=0~~~~~~~~~(2)\]

I have used (1) whereas you have used (2). The same goes for the vertical tangent. Hope this shreds light on the matter.

I did not see you did update it and now when I see it, it's make sense:) I understand now :) one question. I get \(\displaystyle \frac{dy}{d\theta}=2\cos2\theta\) while you get something diffrent

Regards,

 

[Sudharaka]

I did not see you did update it and now when I see it, it's make sense:) I understand now :) one question. I get \(\displaystyle \frac{dy}{d\theta}=2\cos2\theta\) while you get something diffrent

Regards,

In my method (post #2) I haven't found \(\frac{dy}{d\theta}\) anywhere. I have found \(\frac{dy}{dx}\) instead. Of course to calculate \(\frac{dy}{dx}\) I have used the chain rule.

\[\frac{dy}{dx}=\frac{dy}{d\theta}\frac{d\theta}{dx}=\frac{2\cos 2\theta}{-2\sin\theta}=\frac{2\sin^{2}\theta-1}{\sin\theta}\]

Get sense? :)

 

[Petrus]

In my method (post #2) I haven't found \(\frac{dy}{d\theta}\) anywhere. I have found \(\frac{dy}{dx}\) instead.

Hello Sudharaka,
Don't get me wrong I know you solved \(\displaystyle \frac{dy}{dx}\)
I maybe said it wrong, my apologize.
What I mean was
\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)
then you used double angle identity to simplify \(\displaystyle \cos2\theta= 1-2sin^2\theta\)
and that was my answer to my question!

Thanks Sudharaka for taking your time and helping me!

Regards,

- - - Updated - - -

In my method (post #2) I haven't found \(\frac{dy}{d\theta}\) anywhere. I have found \(\frac{dy}{dx}\) instead. Of course to calculate \(\frac{dy}{dx}\) I have used the chain rule.

\[\frac{dy}{dx}=\frac{dy}{d\theta}\frac{d\theta}{dx}=\frac{2\cos 2\theta}{-2\sin\theta}=\frac{2\sin^{2}\theta-1}{\sin\theta}\]

Get sense? :)

Yeah I just figoure that out!:D Thanks Sudaharaka!

 

[Sudharaka]

Hello Sudharaka,
Don't get me wrong I know you solved \(\displaystyle \frac{dy}{dx}\)
I maybe said it wrong, my apologize.
What I mean was
\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)
then you used double angle identity to simplify \(\displaystyle \cos2\theta= 1-2sin^2\theta\)
and that was my answer to my question!

Thanks Sudharaka for taking your time and helping me!

Regards,

- - - Updated - - -Yeah I just figoure that out!:D Thanks Sudaharaka!

You are welcome. I hope you understood and clarified all your doubts, if not please don't hesitate to ask. :)

 

[Petrus]

You are welcome. I hope you understood and clarified all your doubts, if not please don't hesitate to ask. :)

Hello Sudharaka.
Thanks for your kindness! I understand this now and I am really gratefull!

Regards,