MHB Tangents at the pole (polar)

[Amad27]

In a polar function,

$r = 1 - 2\cos(t)$ what are the tangents at the pole, considering $t$ an angle?

I am not sure what the pole is BUT!

$x = \cos(t) - 2\cos^2(t)$

$y = \sin(t) - \sin(2t)$ $dx/dt = -\sin(t) + 4\cos(t)\sin(t)$

$dy/dt = \cos(t) - 2\cos(2t)$

$dy/dx = \frac{\cos(t) - 2\cos(2t)}{-\sin(t) + 4\cos(t)\sin(t)}$

 

[Opalg]

In a polar function,

$r = 1 - 2\cos(t)$ what are the tangents at the pole, considering $t$ an angle?

I am not sure what the pole is

The pole in polar coordinates is the origin. So you need to find the values of $t$ for which $r=0$.

 

[Amad27]

The pole in polar coordinates is the origin. So you need to find the values of $t$ for which $r=0$.

That can't make sense. How can you find an angle for which the length of a line is equal to 0. That line doesn't exist. It is a single coordinate.

 

[Opalg]

That can't make sense. How can you find an angle for which the length of a line is equal to 0. That line doesn't exist. It is a single coordinate.

Think about it more carefully and you will see how it makes sense. There are two values of the parameter $t$ at which the curve passes through the origin. Your job is to find the equations of the tangents at those points.

 

[Amad27]

Think about it more carefully and you will see how it makes sense. There are two values of the parameter $t$ at which the curve passes through the origin. Your job is to find the equations of the tangents at those points.

Wait, so the pole is $(r, \theta) = (0, 0)$

But then the denominator becomes 0 for $dy/dx$

Also, I have a major question.

QUESTION:
When you have a polar equation, $r = 1-2\cos(\theta)$

Why is the derivative $dy/dx$ and not $dr/d\theta$

 

[Opalg]

Wait, so the pole is $(r, \theta) = (0, 0)$

But then the denominator becomes 0 for $dy/dx$

The equation of the curve is $r = 1 - 2\cos\theta.$

If $r=0$ then $1 - 2\cos\theta = 0$, which means that $\cos\theta = \frac12$, and therefore $\sin\theta= \pm\frac{\sqrt3}2.$

You have correctly calculated that $\frac{dy}{dx} = \frac{\cos\theta - 2\cos(2\theta}{-\sin\theta + 4\cos\theta\sin\theta}$. All you now have to do is to insert the above values for $\cos\theta$ and $\sin\theta$ into that formula. That will give you the gradients of the curve when it passes through the origin (and the denominator does not become 0 at those points).

Also, I have a major question.

QUESTION:
When you have a polar equation, $r = 1-2\cos(\theta)$

Why is the derivative $dy/dx$ and not $dr/d\theta$

The question asks for the tangents to the curve. You probably find it easier to describe these using cartesian coordinates rather than polar coordinates.

 

[Amad27]

The equation of the curve is $r = 1 - 2\cos\theta.$

If $r=0$ then $1 - 2\cos\theta = 0$, which means that $\cos\theta = \frac12$, and therefore $\sin\theta= \pm\frac{\sqrt3}2.$

You have correctly calculated that $\frac{dy}{dx} = \frac{\cos\theta - 2\cos(2\theta}{-\sin\theta + 4\cos\theta\sin\theta}$. All you now have to do is to insert the above values for $\cos\theta$ and $\sin\theta$ into that formula. That will give you the gradients of the curve when it passes through the origin (and the denominator does not become 0 at those points).The question asks for the tangents to the curve. You probably find it easier to describe these using cartesian coordinates rather than polar coordinates.

Geometrically, it still does not make sense. To create an angle, you need two rays. If $r= 0$ then there can't be an angle?

Since $\cos(\theta) = 1/2 \implies \pi/3$

$-\sin(\theta) + 2\sin(2\theta)$

At $\pi/3$ this gives $= -\sqrt{3}/2 + \sqrt{3}$

?

 

[Opalg]

Geometrically, it still does not make sense. To create an angle, you need two rays. If $r= 0$ then there can't be an angle?

Since $\cos(\theta) = 1/2 \implies \pi/3$

$-\sin(\theta) + 2\sin(2\theta)$

At $\pi/3$ this gives $= -\sqrt{3}/2 + \sqrt{3}$

?

Perhaps a picture will help. The polar equation $r = 1-2\cos\theta$ gives the black curve. You need to find the tangents at the origin, which are the two green lines.

[graph]vmgasyxoyt[/graph]

 

[Amad27]

Perhaps a picture will help. The polar equation $r = 1-2\cos\theta$ gives the black curve. You need to find the tangents at the origin, which are the two green lines.

Hey there,

So there is an intuition.

You have

$y = f(\theta)$
$ x = g(\theta)$

$\implies \theta = g^{-1}(x)$
$y = f(g^{-1}(x))$

y' = f'(g^{-1}(x)}\cdot \frac{1}{g'(g^{-1}(x))}$$

So, really you are finding the derivative of a xy curve, which is just parametric. So in really the derivative at theta something, means you are finding the derivative of a xy function, not with respect to theta. Really. It is focused on $x$ isn't it?

Can you please read the last paragraph carefully, and respond to it, share thoughts. Possibly confirm. Please. I would really appreciate this favor.