Find Horizontal/Vertical Tangents of Polar Curve r=cos(theta)+sin(theta)

[Pull and Twist]

How do I find the polar coordinates of the points on the polar curve r=cos(theta)+sin(theta), 0(greater than or equal to)(theta)(less than or equal to)(pi), where the tangent line is horizontal or vertical?

I know that I need to convert the coordinates to x & y and then take the derivative of the two (dy/dx) but can't seem to figure it out.

Using the chain rule I was able to figure out dx/d(theta)=2(-sin(theta))+cos^2(theta)-sin^2(theta) and dy/d(theta)=2cos(theta)+cos^2(theta)-sin^2(theta) but have no idea if I am on the correct path or what I need to do next. I thought about using the double angle trig identity but am not sure if it applies in this situation. It has been awhile since I had to utilize trig.

Any help would be awesome. Thank you.

 

[MarkFL]

Consider the following theorem:

[box=green]

Slope of Tangent Line​

---

If $f$ is a differentiable function of $\theta$, then the slope of the tangent line to the graph of $r=f(\theta)$ at a point $(r,\theta)$ on the graph is:

$$\d{y}{x}=\frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}=\frac{f(\theta)\cos(\theta)+f'(\theta)\sin(\theta)}{f'(\theta)\cos(\theta)-f(\theta)\sin(\theta)}\tag{1}$$

Provided that \(\displaystyle \d{x}{\theta}\ne0\).[/box]

For this problem, we are given:

\(\displaystyle f(\theta)=r=\cos(\theta)+\sin(\theta)\) where (I assume you mean) \(\displaystyle 0\le\theta\le\pi\)

So, first, let's compute $f'(\theta)$...what do you get?

 

[Pull and Twist]

If I am doing it correctly, f′(θ)=cos(θ)-sin(θ)

 

[MarkFL]

If I am doing it correctly, f′(θ)=cos(θ)-sin(θ)

Yes, good...that's correct. So, using (1) that I posted above, what do we require for a horizontal tangent line?

 

[Pull and Twist]

That the slope is 0... in other words, f′(θ)=0

 

[MarkFL]

That the slope is 0... in other words, f′(θ)=0

We actually need \(\displaystyle \d{y}{x}=0\)...which means the numerator on the far right side of (1) needs to be zero:

\(\displaystyle f(\theta)\cos(\theta)+f'(\theta)\sin(\theta)=0\)

So, plug in $f(\theta)$ and $f'(\theta)$ that you computed into this expression...what do you get?

 

[Pull and Twist]

I get; 2sin(θ)cos(θ)+cos²(θ)-sin²(θ)

Again, assuming I did it correctly.

 

[MarkFL]

I get; 2sin(θ)cos(θ)+cos²(θ)-sin²(θ)

Again, assuming I did it correctly.

Yes, you did it correctly! (Sun)

We do want this to be equal to zero:

\(\displaystyle 2\sin(\theta)\cos(\theta)+\cos^2(\theta)-\sin^2(\theta)=0\)

Now, can you think of any familiar trigonometric identities that will allow you to simplify the expression on the left?

 

[Pull and Twist]

This is where I got stuck earlier because I'm rusty on my trig identities. What I want to do is apply the double angle formula to the cos²(θ)−sin²(θ) portion so I can get;

2sin(θ)cos(θ)+cos2(θ)=0

Then distribute that first 2...

sin2(θ)cos2(θ)+cos2(θ)=0

Factor out the cos2(θ)

2cos(θ)(2sin(θ)+1)=0

Then set both equal to 0.

2cos(θ)=0 ... cos(θ)=0/2 ... θ=pi/2

2sin(θ)+1=0 ... sin(θ)=-1/2 ... θ=-pi/6 ... θ=11pi/6

Not sure though...

 

[MarkFL]

This is where I got stuck earlier because I'm rusty on my trig identities. What I want to do is apply the double angle formula to the cos²(θ)−sin²(θ) portion so I can get;

2sin(θ)cos(θ)+cos2(θ)=0

Then distribute that first 2...

sin2(θ)cos2(θ)+cos2(θ)=0

Factor out the cos2(θ)

2cos(θ)(2sin(θ)+1)=0

Then set both equal to 0.

2cos(θ)=0 ... cos(θ)=0/2 ... θ=pi/2

2sin(θ)+1=0 ... sin(θ)=-1/2 ... θ=-pi/6 ... θ=11pi/6

Not sure though...

You cannot distribute that 2, for in general, we have:

\(\displaystyle a\sin(x)\ne\sin(ax)\) and \(\displaystyle a\cos(x)\ne\cos(ax)\)

What you want are the following identities:

Step 1:

[box=red]

Double-Angle Identities for Sine and Cosine​

---

$$\sin(2x)=2\sin(x)\cos(x)\tag{2}$$
$$\cos(2x)=\cos^2(x)-\sin^2(x)\tag{3}$$[/box]

Step 2:

[box=blue]

Linear Combination Identity​

---

$$a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin\left(x+\tan^{-1}\left(\frac{b}{a}\right)\right)\tag{4}$$

where $a,b>0$[/box]

 

[Pull and Twist]

So I would go with;

2sin(θ)cos(θ)+cos2(θ)−sin2(θ)=0

becomes;

sin(2θ)+cos(2θ)=0 ... 2sin(θ)+2cos(θ)=0

?

I've don't ever remember seeing the Linear Combination Identity in Pre-Calc.

 

[Pull and Twist]

Using Linear Combination Identity would give me??

2√2*sin(θ+pi/4)=0

sin(θ+pi/4)=0/2√2

θ+pi/4=0/sin

θ=0-pi/4

θ=-pi/4

θ=7pi/4

?

 

[MarkFL]

So I would go with;

2sin(θ)cos(θ)+cos2(θ)−sin2(θ)=0

becomes;

sin(2θ)+cos(2θ)=0 ... 2sin(θ)+2cos(θ)=0

?

I've don't ever remember seeing the Linear Combination Identity in Pre-Calc.

You are correct here:

\(\displaystyle \sin(2\theta)+\cos(2\theta)=0\)

But, you cannot pull the 2's out in front of the trig. functions, for the reason I cited earlier.

So, using the linear combination identity, we may write (after dividing through by $\sqrt{2}$ and using $\tan^{-1}(1)=\dfrac{\pi}{4}$):

\(\displaystyle \sin\left(2\theta+\frac{\pi}{4}\right)=0\)

Now, using the fact that if given:

\(\displaystyle \sin(x)=0\implies x=k\pi\) where \(\displaystyle k\in\mathbb{Z}\) (this means $k$ is an arbitrary integer)

What do we get as the general solution for $\theta$?

 

[Pull and Twist]

I have no idea... I always want to get θ by itself normally...

2θ+pi/4=sin^-10

2θ=0-pi/4

θ=-pi/8

Which I don't believe is right.

 

[MarkFL]

The general solution would come from:

\(\displaystyle 2\theta+\frac{\pi}{4}=k\pi\)

\(\displaystyle 2\theta=k\pi-\frac{\pi}{4}=\frac{\pi}{4}\left(4k-1\right)\)

Thus:

\(\displaystyle \theta=\frac{\pi}{8}\left(4k-1\right)\)

Now, using the restriction given on $\theta$, we may state:

\(\displaystyle 0\le\frac{\pi}{8}\left(4k-1\right)\le\pi\)

Multiply through by \(\displaystyle \frac{8}{\pi}\):

\(\displaystyle 0\le4k-1\le8\)

Add through by 1:

\(\displaystyle 1\le4k\le9\)

Divide through by 4:

\(\displaystyle \frac{1}{4}\le k\le\frac{9}{4}\)

Since $k$ is an integer, we then find:

\(\displaystyle k\in\{1,2\}\)

Hence, our two solutions are:

\(\displaystyle \theta=\frac{\pi}{8}\left(4(1)-1\right)=\frac{3\pi}{8}\)

\(\displaystyle \theta=\frac{\pi}{8}\left(4(2)-1\right)=\frac{7\pi}{8}\)

Now, using $f(\theta)$, plug these angles into get the two points $(r,\theta)$.

Then, for the vertical tangent lines, what condition from (1) do we require?

 

[Pull and Twist]

f′(θ)cos(θ)−f(θ)sin(θ)=0

 

[MarkFL]

f′(θ)cos(θ)−f(θ)sin(θ)=0

Correct...now can you follow a similar method to solve this?

By the way, you don't nee the linear combination identity...for example, in the first part of the problem, we have:

\(\displaystyle \sin(\theta)+\cos(2\theta)=0\)

This implies:

\(\displaystyle \tan(2\theta)=-1\)

Which in turn implies:

\(\displaystyle 2\theta=-\frac{\pi}{4}+k\pi\)

Which is the same equation we obtain using the linear combination identity...:D

So, can you proceed?

 

[Pull and Twist]

And like before;

-2sin(θ)cos(θ)-cos2(θ)+sin2(θ)=0

-(2sin(θ)cos(θ)+cos2(θ)-sin2(θ))=0

-(sin(2θ)+cos(2θ))=0

- - - Updated - - -

Correct...now can you follow a similar method to solve this?

By the way, you don't nee the linear combination identity...for example, in the first part of the problem, we have:

\(\displaystyle \sin(\theta)+\cos(2\theta)=0\)

This implies:

\(\displaystyle \tan(2\theta)=-1\)

Which in turn implies:

\(\displaystyle 2\theta=-\frac{\pi}{4}+k\pi\)

Which is the same equation we obtain using the linear combination identity...:D

So, can you proceed?

So how would you proceed without getting the same answer again?

 

[MarkFL]

And like before;

-2sin(θ)cos(θ)-cos2(θ)+sin2(θ)=0

-(2sin(θ)cos(θ)+cos2(θ)-sin2(θ))=0

-(sin(2θ)+cos(2θ))=0

- - - Updated - - -
So how would you proceed without getting the same answer again?

Well, that's not correct...what we want, as you stated is

\(\displaystyle f'(\theta)\cos(\theta)-f(\theta)\sin(\theta)=0\)

Thus, we find:

\(\displaystyle \left(\cos(\theta)-\sin(\theta)\right)\cos(\theta)-\left(\sin(\theta)+\cos(\theta)\right)\sin(\theta)=0\)

\(\displaystyle \cos^2(\theta)-\sin(\theta)\cos(\theta)-\sin^2(\theta)-\sin(\theta)\cos(\theta)=0\)

\(\displaystyle \cos^2(\theta)-\sin^2(\theta)-2\sin(\theta)\cos(\theta)=0\)

\(\displaystyle \cos(2\theta)-\sin(2\theta)=0\)

Can you rewrite this in terms of $\tan(2\theta)$?

 

[Pull and Twist]

I have no idea... I didn't quite grasp how you're going from the sin cos function to the tangent one... are you using an identity i am not familiar with or are you dividing by the inverse to create the tangent?

 

[MarkFL]

I have no idea... I didn't quite grasp how you're going from the sin cos function to the tangent one... are you using an identity i am not familiar with or are you dividing by the inverse to create the tangent?

Divide through by $\cos(2\theta)$...what do you get?

Note that this means we require $\cos(2\theta)\ne0$...:D

 

[Pull and Twist]

How did I overlook that... I feel silly.

1−sin(2θ)/cos(2θ)=0/cos(2θ)

1−tan(2θ)=0

tan(2θ)=1

- - - Updated - - -

Then

tan(2θ)=1

implies

2θ=pi/4+k(pi)

 

[Pull and Twist]

Then you can factor out pi/4 to get...

2θ=(pi)/4(1+4k)

Divide by the 2...

θ=(pi)/8(1+4k)

Then use what we know about the restrictions on θ again to get...

-1/4≤k≤7/4

?

- - - Updated - - -

Then...

θ=(pi)/8(4(1)+1)=5(pi)/8

θ=(pi)/8(4(2)+1)=9(pi)/8

 

[MarkFL]

Correct, so what are the only values $k$ can be, and then what are your solutions?

edit: I see you posted again while I was posting...you should find from the compound inequality you correctly derived, that:

\(\displaystyle k\in\{0,1\}\)

Notice that one of the solutions you obtained is greater than $\pi$.

 

[Pull and Twist]

Correct, so what are the only values $k$ can be, and then what are your solutions?

edit: I see you posted again while I was posting...you should find from the compound inequality you correctly derived, that:

\(\displaystyle k\in\{0,1\}\)

Notice that one of the solutions you obtained is greater than $\pi$.

I just noticed that too...

θ=(pi)/8(4(0)+1)=(pi)/8

θ=(pi)/8(4(1)+1)=5(pi)/8


Those would be the two points... then if I plug a point in the original r(θ) function I should be able to get the corresponding point for that given θ.

It does come out super funky though... (View attachment 3423

That's for the first horizontal tangent point we got.

 

[MarkFL]

Another way to express that point is:

\(\displaystyle (r,\theta)=\left(\sqrt{1+\frac{1}{\sqrt{2}}},\frac{3\pi}{8}\right)\)

and the other horizontal tangent point is:

\(\displaystyle (r,\theta)=\left(-\sqrt{1-\frac{1}{\sqrt{2}}},\frac{7\pi}{8}\right)\)

 

[Pull and Twist]

Another way to express that point is:

\(\displaystyle (r,\theta)=\left(\sqrt{1+\frac{1}{\sqrt{2}}},\frac{3\pi}{8}\right)\)

and the other horizontal tangent point is:

\(\displaystyle (r,\theta)=\left(-\sqrt{1-\frac{1}{\sqrt{2}}},\frac{7\pi}{8}\right)\)

Much better looking... for the vertical tangents I got the same r value for both which matches the r value of the 3(pi)/8 horizontal line. Is that correct?

 

[MarkFL]

Much better looking... for the vertical tangents I got the same r value for both which matches the r value of the 3(pi)/8 horizontal line. Is that correct?

I get for the vertical tangent, the points:

\(\displaystyle (r,\theta)=\left(\sqrt{1+\frac{1}{\sqrt{2}}},\frac{\pi}{8}\right),\,\left(\sqrt{1-\frac{1}{\sqrt{2}}},\frac{5\pi}{8}\right)\)

 

[Pull and Twist]

I get for the vertical tangent, the points:

\(\displaystyle (r,\theta)=\left(\sqrt{1+\frac{1}{\sqrt{2}}},\frac{\pi}{8}\right),\,\left(\sqrt{1-\frac{1}{\sqrt{2}}},\frac{5\pi}{8}\right)\)

Is there any way you can work out the r for 5pi/8 so I can see why it came out wrong for me?? Thank you, you've been an amazing help. I feel like I've got a much better understanding of this.

 

[MarkFL]

Is there any way you can work out the r for 5pi/8 so I can see why it came out wrong for me?? Thank you, you've been an amazing help. I feel like I've got a much better understanding of this.

I just plugged those angles into a calculator, for the given function. This is the result for \(\displaystyle \theta=\frac{5\pi}{8}\):

sin'('5pi'/'8')''+'cos'('5pi'/'8')' - Wolfram|Alpha

 

[Pull and Twist]

Awesome, thank you so much! You've been a major help.

 

[MarkFL]

Awesome, thank you so much! You've been a major help.

It is a pleasure to help someone who shows their work and makes a genuine effort. (Yes)

I look forward to seeing you around. :D